• Home
  • Textbooks
  • Practice Problem in Physics for the JEE Main and Advanced
  • Laws of Motion and Friction

Practice Problem in Physics for the JEE Main and Advanced

Abhay Kumar

Chapter 4

Laws of Motion and Friction - all with Video Answers

Educators

Section 1

Section A

00:55

Problem 1

In the figure, ends $P$ and $Q$ of an inextensible string move downwards with uniform speed $u$. Pullcys are fixed and massless. What is the speed $v$ of the mass $m$ in upward direction?
Solution
$x_{2}\left|c_{1}\right| 2 \sqrt{y^{2} \mid d^{2}}\left|c_{2}\right| x_{1}-$ constant
$\Rightarrow \frac{d x_{2}}{d t}+0+2 \frac{1}{2 \sqrt{y^{2}+d^{2}}} 2 y \frac{d y}{d t}+0+\frac{d x_{1}}{d t}-0$
$\Rightarrow \frac{d x_{2}}{d t} \mid \frac{d x_{1}}{d t} ? 2\left(\frac{y}{\sqrt{y^{2} \mid d^{2}}}\right) \frac{d y}{d t}-0 \Rightarrow u \backslash u \backslash 2(\cos \theta) v-0$
$\Rightarrow 2 u+(2 \cos \theta) v-0 \quad \Rightarrow u+v \cos \theta-0$
$v-\left|\frac{u}{\cos \theta}\right|$
$\left\lfloor\Lambda s \frac{d x_{1}}{d t}-v_{\mathrm{r}}, \frac{d x_{2}}{d t}-v_{Q}, \frac{d y}{d t}-\quad v_{m}\right\rfloor \quad \therefore v-\frac{u}{\cos \theta}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:41

Problem 2

In the figure $m_{2}>m_{1}$, pulley and string are light. Using constraint method find the relation between accelerations of $m_{1}$ and $m_{2}$.
Solution $x_{1}+c+x_{2}=$ constant
$\Rightarrow \frac{d x_{1}}{d t}+0+\frac{d x_{2}}{d t}=0$
$\Rightarrow \frac{d x_{1}}{d t}+\frac{d x_{2}}{d t}=0 \Rightarrow \frac{d^{2} x_{1}}{d t^{2}}+\frac{d^{2} x_{2}}{d t^{2}}=0$
$m_{z}$
$\Rightarrow a_{1}+a_{2}=0 \Rightarrow a_{1}=\left|-a_{2}\right|=a_{3}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:47

Problem 3

The ring $m_{2}$ and block $m_{1}$, arc held in the position shown in figure and the system is relcased. II $m_{2}>m_{1}$ the ring $m_{2}$ slidcs down along the smooth fixed vertical rod, find $\frac{v}{u}$. Solution
'lhe length of the string is constant,
$$
x+c+\sqrt{y^{2}+d^{2}}-\text { constant }
$$
$\Rightarrow \frac{d x}{d t}+0+\frac{1}{2 \sqrt{y^{2}+d^{2}}} \times 2 y \frac{d y}{d t}+0-0$
As $\frac{d x}{d t}=u$ and $\frac{d y}{d t}=v$
$u+\cos \theta \quad v=0$
$\Rightarrow \quad u+v \cos \theta=0 \quad \Rightarrow v=\left|\frac{-u}{\cos \theta}\right|$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:14

Problem 4

Find the constraint equation between the accelerations of the blocks $m_{1}, m_{2}$ and $m_{3}$.
Solution
For string l: $x_{1}+c+x_{\mathbb{B}}=$ constant
$\Rightarrow \frac{d^{2} x_{1}}{d t^{2}}|0| \frac{d^{2} x_{B}}{d t^{2}}-0 \quad \Rightarrow \quad a_{1} \mid a_{s}-0$
For string 2: $x_{2}-x_{s}+c_{1}+x_{3}-x_{B}=$ constant
$\Rightarrow \quad x_{2}+x_{3}-2 x_{B}+c_{1}=$ constant
$\Rightarrow \frac{d^{2} x_{2}}{d t^{2}}+\frac{d^{2} x_{3}}{d t^{2}}-\frac{2 d^{2} x_{B}}{d t^{2}}-0 \quad \Rightarrow \quad a_{2}+a_{3}-2 a_{n}-0 \quad \ldots$
From \Gammaeqs. (1) and (2), we get:
$\Rightarrow a_{2}+a_{3}-2\left(-a_{1}\right)-0 \quad \Rightarrow 2 a_{1}+a_{2}+a_{3}-0$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:39

Problem 5

Assuming pulleys and strings are light, find constraint relation between the accelerations of the bar $m_{1}$ and the block $m_{2}$ as in the given figure.
Solution
For 1, $x_{1}+x_{E}=$ constant
$\frac{d^{2} x_{1}}{d t^{2}}+\frac{d^{2} x_{n}}{d t^{2}}-0 \quad \Rightarrow a_{1}+a_{B}-0$
For 2 ,
$\left(x_{1}-x_{n}\right)+\left(x_{c}-x_{n}\right)-$ constant
$\Rightarrow x_{1}-2 x_{A}+x_{C}-$ constant
$\Rightarrow \frac{d^{2} x_{1}}{d t^{2}}-2 \frac{d^{2} x_{B}}{d t^{2}}+\frac{d^{2} x_{C}}{d t^{2}}-0 \quad \Rightarrow \quad a_{1}-2 a_{B}+a_{C}-0$
For 3 , $\left(x_{1}-x_{c}\right)+\left(x_{2}-x_{C}\right)=$ constant
(3) $\Rightarrow \quad x_{1}+x_{2}-2 x_{C}=$ constant
$\Rightarrow a_{1}+a_{2}-2 a_{c}=0$
From Eqs. (1) and (2) $a_{1}-2\left(-a_{1}\right)+a_{c}=0$
$\Rightarrow a_{1}+2 a_{1}+a_{C}=0 \quad \Rightarrow \quad 3 a_{1}+a_{c}=0$
$\therefore \quad a_{\ell}=-3 a_{1}$
From ?q. (3) $\Rightarrow a_{1}+a_{2}-2\left(-3 a_{1}\right)=0$
$\Rightarrow a_{1}+a_{2}+6 a_{1}=0 \quad \Rightarrow \quad 7 a_{1}+a_{2}=0$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:53

Problem 6

A block is pulled on a smooth surface with the help of a rope and ropc is pulled with speed $u$ as in figure. Find the horizontal velocity of the block. (Assume the contact docsn't losc during motion.)
Solution
$x+c+\sqrt{y^{2}+d^{2}}-$ constant
$\Rightarrow \frac{d x}{d t}+\frac{d \sqrt{y^{2} \mid d^{2}}}{d t}-0$
$\Rightarrow \quad u+\frac{1}{2 \sqrt{y^{2} \mid d^{2}}} \times 2 y \frac{d y}{d t}-0$
$\Rightarrow u+\frac{y}{\sqrt{y^{2} \mid d^{2}}}\left(\frac{d y}{d t}\right)-0 \Rightarrow u+\sin \theta(-v)-0$ as $\left(\frac{d y}{d t}\right)--v$
$\Rightarrow \sin \theta(v)-u \quad \therefore \frac{v}{u}-\operatorname{cosec} \theta$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:39

Problem 7

Using the constraint equation find the acceleration of the block $m_{2}$ if the acceleration of the block $m_{1}$ is $a$ as in figure. Solution For $1, a_{1}=a_{A} \quad$... (1)
For $2, x_{A}+\left(x_{A}-x_{2}\right)=$ constant $\Rightarrow 2 x_{A}-x_{2}=$ constant
$\Rightarrow \quad 2 \frac{d^{2} x_{A}}{d t^{2}} \frac{d^{2} x_{2}}{d t^{2}}-0 \quad \Rightarrow \quad 2 a_{A} \quad a_{2}-0$
$\Rightarrow 2 a_{1}-a_{3}=0 \Rightarrow 2 a_{1}=a_{2} \quad \therefore \quad$ accelcration of block $m_{2}=2 a$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:56

Problem 8

In the figurc, assuming pullcys and string are light, if the dircetions of aceclerations arc given then find theconstraint oquation. Solution $x_{1}+x_{2}+2 x_{3}=$ constant
$\frac{d^{2} x_{1}}{d t^{2}}, \frac{d^{2} x_{2}}{d t^{2}} ? 2 \frac{d^{2} x_{3}}{d t^{2}}-0$
Since $x_{1}$ and $x_{2}$ are assumed to be decreasing with time,
therefore, $\frac{d^{2} x_{1}}{d t^{2}}-a_{1}$ and $\frac{d^{2} x_{2}}{d t^{2}}-a_{2}$
and $x_{3}$ is assumed to be increasing with time, therefore,

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:34

Problem 9

Assuming pulleys and string are light, find the constraint equation for accelerations of $m_{1}$ and $m_{2}$ Solution $6 x_{1}+c+5 x_{2}+c^{\prime}=$ constant
$\Rightarrow \quad 6 \frac{d^{2} x_{1}}{d t^{2}}+5 \frac{d^{2} x_{2}}{d t^{2}}-0$
$\Rightarrow 6 a_{1}+5 a_{2}=0$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:37

Problem 10

Assuming pullcys and string are light, find the constraint cquation for accelcrations of $m_{1}, m_{2}$ and $m_{3}$
Solution
$x_{1}+x_{2}+x_{3}=$ constant
$\frac{d^{2} x_{1}}{d t^{2}}+\frac{d^{2} x_{2}}{d t^{2}}+\frac{d^{2} x_{3}}{d t^{2}}-0 \Rightarrow a_{1 A}+a_{2}+a_{3}-0$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:31

Problem 11

An inextensible string $A B$ is tied to a block $B$ of negligible dimensions and passes over a small pullcy $C$ :
so that the frec end $A$ hangs $h_{1}$ unil above the ground on which the block $B$ rests. In this initial position shown in figure, the frec cnd $A$ is $h$ unit below $C$. If now the end $A$ moves horizontally with a velocity
$u$, obtain an cxpression for the velocity of the block at any time $t$.
Solution In time $t$, the end $A$ moves to the position $A_{1}$. So that $A A_{1}=u l .$ The block $B$ moves upwards to the position $B_{1}$. Let $B B_{1}=y$.
'Ihen length $A_{1} C=h+y \quad$ ln the $\Delta A C A_{1},(h+y)^{2}=h^{2}+(u i)^{2}$
$\begin{array}{llll}\text { or } h^{2}+y^{2}+2 h y=h^{2}+u^{2} t^{2} & \therefore & y^{2}+2 h y-u^{2} t^{2}=0 & \ldots(1)\end{array}$
\Lambdafter solving equation (1), we get
$y--h+\sqrt{h^{2}+u^{2} t^{2}}$
Ihis is the equation for the displacement $y$ of the block. Velocity of the block:
$v-\frac{d y}{d t}-\frac{d}{d t}\left[h \mid\left(h^{2} \mid u^{2} t^{2}\right)^{1 / 2}\right]-\frac{1}{2}\left(h^{2} \mid u^{2} t^{2}\right)^{-1 / 2} \times 2 u^{2} t$
or $\quad v-\frac{u^{2} t}{\left(h^{2}+u^{2} t^{2}\right)^{1 / 2}}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:51

Problem 12

A rod of length $l$ is inclined at an angle with the floor against a smooth vertical wall. If the end $A$ moves with velocity $v_{1}$ and the rod makes an angle $\alpha$ with horizontal, what is the velocity ol $B$ at that instant?
Solution $\sqrt{x^{2}+y^{2}}-l-$ constant
$\Rightarrow x^{2}+y^{2}=l^{2}=$ constant
$\Rightarrow \quad 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}-0 \Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}-0$
$\Rightarrow \quad x v_{1}-y v_{2}-0 \quad$ As $\left(\frac{d y}{d t}\right)--v_{2}$
$\Rightarrow x v_{1}=y v_{2}$
$\Rightarrow \frac{v_{1}}{v_{2}}-\frac{y}{x} \Rightarrow \frac{v_{1}}{v_{3}}-\tan \alpha \Rightarrow v_{2}-v_{1} \cot \alpha$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:35

Problem 13

As shown in figure, the velocity of rod at any instant in downward direction is $u$. Then what will be the velocity of triangular wedge in horizontal direction at that instant?
Solution $\tan \theta-\frac{y}{x} \Rightarrow y-x \tan \theta \Rightarrow \frac{d y}{d t}-\tan \theta \frac{d x}{d t}$
$\Rightarrow u^{-} v \tan \theta \Rightarrow v-\frac{u}{\tan \theta} \Rightarrow v-u \cot \theta$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:33

Problem 14

A racing car travelling along a track at a constant speed of $40 \mathrm{~m} / \mathrm{scc}$. A camcraman is recording the cvent from a distance $40 \mathrm{~m}$ directly away from $A$ the track as in figure. In order to kecp the car under vicw, with what angular velocity the camcra should be rotaled after 1 sce from the start?
Solution
Angular velocity $-\frac{d \theta}{d t}-?$
At $t-1$ sec $\quad \therefore \operatorname{lan} \theta-\frac{x}{40} \quad \Rightarrow \quad x-40 \operatorname{an} \theta$
$d x$
$\Rightarrow \frac{d x}{d t}-\frac{d(40 \tan \theta)}{d t} \Rightarrow \frac{d x}{d t}-40 \sec ^{2} \theta \frac{d \theta}{d t} \Rightarrow \frac{d \theta}{d t}-\frac{d t}{40 \sec ^{2} \theta}$
$\Rightarrow \frac{d \theta}{d t}-\omega-\frac{40}{40 \sec ^{2} \theta} \Rightarrow \frac{d \theta}{d t}-\frac{1}{\sec ^{2} \theta}$ as $\frac{d x}{d t}=v-40 \mathrm{~m} / \mathrm{s}$
$\Lambda t \quad t=1 \mathrm{sec}, x=40 \mathrm{~m}$ and $d=40 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:21

Problem 15

'Iwo unequal masses moving along straight lines are brought to rest by equal resistance forces. If one mass moves twice as long (time) as the other but goes only $1 / 3$ of the distance covered by the other before coming to rest. Find: (i) the ratio of their velocities; and (ii) the ratio of their masses. Solution
(A) $F=m_{1} a_{1}$ and $\frac{v_{1}}{a_{1}}-t_{1}$
(B) $\mathrm{F}=m_{2} a_{2}$ and $\frac{v_{2}}{a_{2}}-t_{2}$
$\Lambda \mathrm{s} \frac{t_{2}}{t_{1}}-2$
$\therefore \frac{v_{2}}{v_{1}} \cdot \frac{a_{1}}{a_{2}}-2$
Again $\quad S_{1}-\frac{v_{1}^{2}}{2 a_{1}}$ and $S_{2}-\frac{v_{2}^{2}}{2 a_{2}}$
$\therefore \frac{S_{2}}{S_{1}^{*}}-\frac{v_{2}^{2}}{v_{1}^{2}} \cdot \frac{a_{1}}{a_{2}}-\frac{1}{3}$
$\ldots(2)$
From \Gammaass. (1) and (2),
$\frac{v_{1}^{2}}{v_{2}^{2}} \cdot \frac{1}{3} \cdot \frac{v_{2}}{v_{1}}-\frac{1}{3} \frac{v_{1}}{v_{2}}$ or $\frac{v_{1}}{v_{2}}-6$
$\therefore \frac{a_{1}}{a_{2}}-12 .$ So $m_{1}: m_{2}-\frac{F}{a_{1}}: \frac{F}{a_{2}}-a_{2}: a_{1}-1: 12$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:43

Problem 16

Calculate the forces of couplings in the figure.
Solution Acecleration of the system $-\frac{F}{m}-\frac{110}{\text { Total mass }}-\frac{110}{55}-2 \mathrm{~m} / \mathrm{sec}^{2}$ $=30 \times 2=60 \mathrm{~N} ; \quad T_{2}=15 \times 2=30 \mathrm{~N} ; \quad T_{3}=5 \times 2=10 \mathrm{~N}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:55

Problem 17

The pulley arrangements of Figure (a) and (b) are idcntical. The mass of the rope is negligible. In figure (a), the mass $m$ is lilled by attaching a mass $2 m$ to the other ond of the ropc. In Figure $(b), m$ is lifled up by pulling the other end of the ropc with a constant downward foree $F=2 m g$. The acccleration of $m$ is the same in both cases. Is it correct?

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:40

Problem 18

Two masses $m$ and $2 m$ are connected by a massless string which passes over a pulley as shown in the figure. The masses are initially held with equal lengths on either side of the pulley. Find the velocities of the masses at the instant the lighter mass moves up a distancc of $6.54 \mathrm{~m}$. The string is suddenly cut at that instant. Calculate the time taken by cach mass to reach the ground. $\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right)$ Solution Acecleration of the system $-\frac{(2 m \quad m) g}{2 m 1 m}-\frac{g}{3}-3.27 \mathrm{~m} / \mathrm{scc}^{2}$
$2 m$ moves down and $m$ moves up with acceleration $3.27 \mathrm{~m} / \mathrm{s}^{2}$ from rest. If $v$ is velocity after moving $6.54 \mathrm{~m}$, $v^{2}=2 a S=2 \times 3.27 \times 6.54 \quad \therefore v=6.54 \mathrm{~m} / \mathrm{s}$
Initially, both are at height of $13.08 \mathrm{~m}$ from the ground. When they moved $6.54 \mathrm{~m}$, height of $m$ from ground $=13.08+6.54=19.62 \mathrm{~m}$ and height of $2 \mathrm{~m}$ from ground $=13.08-6.54=6.54 \mathrm{~m}$. When the string is cut, $m$ has upward velocity $6.54 \mathrm{~m} / \mathrm{s}$ and $2 m$, the same downward velocity.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:58

Problem 19

Find the tension of strings connccting the masses shown in the figure. The pullcys are massless and the surface friction is zero. Find the relation betwecn the masses $m, m^{\prime}$ and $M$, if $M$ is in cqilibrium.
1
Solution
'lension is the same throughout. If $M$ moves down, $m$ moves to the
$M$
right and $m$ ' up. If $m$ moves to the right by $x, m^{\prime}$ moves up by $y$, $M$ moves down by $\frac{x+y}{2}$. So, the acceleration of $M, m$, and $m^{\prime}$ are, respectively, $\frac{a+a^{\prime}}{2}, a$ and $a^{\prime}$.
So, $T-m a, M g \quad 27-\frac{M\left(a+a^{\prime}\right)}{2}, T \quad m^{\prime} g^{-} m^{\prime} a^{\prime}$
$\therefore \quad a \mid a^{\prime}-\frac{2 \operatorname{Mg} 4 T}{M}-2 g \quad \frac{4 T}{M}-\frac{T}{m}, \frac{T-m^{\prime} g}{m^{\prime}}$
$\therefore \quad 2 g \quad \frac{4 T}{M}-\frac{T}{m} \mid \frac{T}{m^{\prime}}-g \quad \therefore 3 g-T\left\{\frac{4}{M} \mid \frac{1}{m}, \frac{1}{m^{\prime}}\right\} \quad \therefore \quad T-\frac{3 g}{\frac{4}{M}+\frac{1}{m}+\frac{1}{m^{\prime}}}$
If $M$ is in cquilibrium, $2 T=M g$
$\therefore \quad \frac{M g}{2}-\frac{3 g}{4}{M+\frac{1}{M}+\frac{1}{m^{\prime}}} \quad \therefore 4+\frac{M}{m}+\frac{M}{m^{\prime}}-6$
or $M f\left[\frac{1}{m}+\frac{1}{m^{\prime}}\right]-2$
$\therefore \quad M-\frac{2 m m^{\prime}}{m \mid m^{\prime}}$ Thus, $M$ is the harmonic mean of $m$ and $m^{\prime}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:10

Problem 20

Find the acceleration of the system shown in the figure. The pulleys are smooth and the threads are of negligible masses.
Solution
'The tension $-\frac{2 m_{1} m_{2}}{m_{1}+m_{2}} \times$ (effective acceleration)
$\therefore \quad T_{1}-\frac{2 \times 6 \times 4}{6+4}(g-a)-4.8(g-a)$
$T_{2}-\frac{2 \times 9 \times 1}{911}(g \mid a)-1.8(g \mid a)$
also, $T=2 T_{1}+2(g-a)=2 \times 4.8(g-a)+2(g-a)=11.6(g-a)$
and $T=2 T_{2}+2(g+a)=2 \times 1.8(g+a)+2(g+a)=5.6(g+a)$
$\therefore \quad 11.6(g-a)=5.6(g+a)$
$\Rightarrow 6 g-17.2 a \quad \therefore \quad a-\frac{6 g}{17.2}-\frac{6 \times 9.8}{17.2}-\frac{588}{172}-3.4 \mathrm{~m} / \mathrm{s}^{2}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:19

Problem 21

In the arrangement shown in the figure, the mass of body 1 is $\eta=4$ times that of body 2 . 'The height $h=20 \mathrm{~cm}$. The masses of the pulleys and the threads as well as the friction are negligible. $\Lambda$ t a certain moment, body 2 is released and the arrangement is set in motion. What is the maximum height that body 2 will go up to? Solution If mass 1 , moves down $1 \mathrm{em}$, mass 2 moves up $2 \mathrm{~cm}$. So, its accelcration will be twice that of 1 .
$\eta^{m g}-2 T=\eta m a \quad T-m g=m 2 a$
$2 T-2 m g=4 m a$
$\therefore \eta m g-2 m g=(\eta+4) m a$ or $\quad a-\frac{(\eta-2) g}{\eta+4}$
$\therefore$ acceleration of body $2-\frac{2(\eta-2) g}{\eta+4}$ With this acceleration, it rises to a height $2 h$. Velocity on reaching $2 h$ is given by $v^{2}-2 a s-\frac{2 \times 2 g(\eta \quad 2)}{\eta+4} 2 h$
On reaching this height, it rises again until its velocity is zero.
$\therefore$ 'The further height reached $-h^{\prime}-\frac{v^{2}}{2 g}-\frac{2 \times 2 g(\eta \quad 2)}{\eta+4} \frac{2 h}{2 g}-\frac{4(\eta \quad 2)}{\eta 14} h$
$\therefore$ Total height reached $-2 h+\frac{4(\eta 2) h}{(\eta+4)}-2 h\left\{1+\frac{2(\eta \quad 2)}{\eta+4}\right\}-2 h\left\{\frac{\eta|4| 2 \eta 4}{\eta+4}\right\}$
$-\frac{6 n h}{\eta 14}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:51

Problem 22

$\Lambda$ trolley of mass $100 \mathrm{~kg}$, starting from rest, describes $100 \mathrm{~m}$ in 10 second. \Lambdat that instant, i.e., at the commencement of the $11^{\text {th }}$ second, two packets, each of mass $12.5 \mathrm{~kg}$, are gently placed in the trolley. I low far does it move in the next 10 seconds assuming that the forces on the trolley remain the same throughout? Solution Since $s-u t+\frac{1}{2} a t^{2}$
$100-0 \div \frac{1}{2} a 100 \quad \therefore \quad a-2 \mathrm{~m} / \mathrm{sec}^{2}$
The velocity at the cnd of 10 seconds $=v=u+a t=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}$ $F=m a=100 \times 2=200 \mathrm{~N}$
During the next 10 seconds, acceleration $-\frac{200}{125}-1.6 \mathrm{~m} / \mathrm{sec}^{2}$ The velocity al the commenecment of the 11 th socond $v^{\prime}$ is given by $m u=m^{\prime} v^{\prime}$ (Taw of conscrvation of momentum) $100 \times 20=125 \times v^{\prime}$
$\therefore \quad v^{\prime}-\frac{100 \times 20}{125}-16 \mathrm{~m} / \mathrm{scc}$
$\therefore$ The distance travelled in the next 10 seconds $=s^{\prime}-u t+\frac{1}{2} a t^{2}$ $-16 \times 10+\frac{1}{2} \times 1.6 \times 100-240 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:07

Problem 23

Two blocks of mass $m=5 \mathrm{~kg}$ and $M=10 \mathrm{~kg}$ are connected by a light string passing over a pulley $B$ as shown. $\Lambda$ nother light string connects the centre of pulley $B$ to the floor and passes over another pulley $A$ as shown. $\Lambda$ n upward force $F$ is applied at the centre of pulley $A$. Both the pulleys are massless. Find the acceleration of block $m$ and $M$, if $F$ is:
(a) $500 \mathrm{~N}$
(b) $300 \mathrm{~N}$
(c) $100 \mathrm{~N}\left(\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution string string
$T_{0} \quad T_{0}$
T $T$
(a) $T^{\prime}=F / 4=125 \mathrm{~N}$
$\Lambda \mathrm{s} T>m y$ and $\mathrm{Mg}$, both the blocks will accelerate upwards. \Lambdacceleration of $m, a_{1}=\frac{T-m g}{m}=\frac{125-50}{5}=15 \mathrm{~m} / \mathrm{s}^{2}$
\Lambdacceleration of $M, a_{2}=\frac{T-M g}{M}=\frac{125-100}{10}=2.5 \mathrm{~m} / \mathrm{s}^{2}$
(b) $7^{\prime}=F / 4=75 \mathrm{~N}$
\Lambdas $T<M g$ and $T>m g, M$ will remain stationary on the floor, where as $m$ will move. Accclcration of $i n, a_{1}-\frac{T-m g}{m}-\frac{75-50}{5}-5 \mathrm{~m} / \mathrm{s}^{2}$
(c) $T=F / 4=25 \mathrm{~N}$
$\begin{array}{ll}\text { weights of blocks are } & m g=50 \mathrm{~N}\end{array} \quad M g=100 \mathrm{~N}$
\Lambdas $7<m g$ and $M g$ both, the blocks will remain stationary on the floor.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:21

Problem 24

In the arrangement shown, the blocks $A, B, C$ and $D$ have masses $m_{1}, m_{2}, m_{3}$ and $m_{4}$ respectively. The springs are weightless and have forec constant $k$ and the string and pullcy are light and smooth. The system is maintained in cquilibrium by the thread $D G$ connccting block $D$ to the ground. I the thread is cut at a certain moment, determine the accelerations of the blocks immediately aflerwards. Solution Considering the cquilibrium of blocks $T_{2}-T_{1}=m_{1} g$
(1)
$T_{2}^{\prime}-T_{3}^{2}=m_{3} g$
$T_{1}^{-1}=m_{2} g \quad \ldots(2)$
$T_{3}^{3}-T_{1}=m_{1} g$
From (1) and (2), $T_{2}=\left(m_{1}+m_{2}\right) g$
From (3) and (4), $T_{2}^{-}-T_{4}=\left(m_{3}+m_{4}\right) g$
$\therefore T_{4}=\left(m_{1}+m_{2}\right) g-\left(m_{3}+m_{1}\right) g=\left(m_{1}+m_{2}-m_{3}-m_{1}\right) g$
$T$ A

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:27

Problem 25

Both the blocks are resting on a horizontal floor and the pullcy is held such that string remains just taut. At moment $t=0$, a force $F=20 t \mathrm{~N}$ starts acting on the pullcy along vertically upward direction as shown in figurc. Calculate:
(a) velocity of $A$ when $B$ loses contact with the floor.
(b) hcight raised by the pulley upto that instant. $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution
(a) Let 'I' be the tension in the string. Then, $27=20$ tor $T=10 t \mathrm{~N}$ Let the block $A$ loses its contact with the floor at time $t=t_{1}$. This happens when the tension in string becomes equal to the weight of $A$. Thus, $T=m g$ or $10 t_{1}=1 \times 10$ or
$t_{1}=1 \mathrm{~s} \quad \ldots(1)$
Similarly, for block $B$, we have $10 t_{2}=2 \times 10 \quad$ or $\quad t_{2}=2 \mathrm{~s} \quad \ldots$ (ii)
\mathrm{\{} i e . , ~ t h e ~ b l o c k ~ $B$ loses contact alter $2 \mathrm{~s}$. For block $A$, at time $t$ such that $t \geq t$ let $a$ be its acceleration in upward direction. Then,
$10 t-1 \times 10=1 \times a=(d v / d t) \quad$ or $\quad d v=10(t-1) d t$
Integrating this cxpression, we get
$\int_{0}^{\mathrm{v}} d v-10 \int_{1}^{t}(t-1) d t$
or $v=5 t^{2}-10 t+5 \quad \ldots(1 \mathrm{v})$
Substituting $t=t_{2}=2 \mathrm{~s} \quad$ or $\quad v=20-20+5=5 \mathrm{~m} / \mathrm{s} \ldots(\mathrm{v})$
(vi) (b) From Eq. (iv), $d y=\left(5 t^{2}-10 t+5\right) d t$
Where $y$ is the vertical displacemcnt of block $A$ at timc $t\left(\geq t_{1}\right)$. Intcgrating, we have
$\int_{v=0}^{<=h} d y-\int_{t=1}^{t=2}\left(5 t^{3}-10 t 15\right) d i \quad \Rightarrow h-5\left[\frac{t^{3}}{3}\right]_{1}^{2} 10\left[\frac{t^{2}}{2}\right]_{1}^{3} 15\lfloor t]_{1}^{3}-\frac{5}{3} \mathrm{~m}$
$\therefore$ lleight raised by pulley upto that instant $-\frac{h}{2}-\frac{5}{6} \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:44

Problem 26

$\Lambda$ cat of mass $m=1 \mathrm{~kg}$ climbs to a rope hung over a light frictionless pulley. The opposite end of the rope is tied to a weight of mass $M=2 \mathrm{~m}$ lying on a smooth horizontal plane. What is the tension of the rope when the cat moves upwards with an acceleration $a=2 \mathrm{~m} / \mathrm{s}^{3}$ relative to the rope'?
Solution Let $a$ be the absolute upward acceleration of the monkey and $a$ ' be the absolute downward acceleration of the rope. $a$ ' is also the tightward acceleration of $M$. Then, $b=a-\left(-a^{\prime}\right)$ (since relative acceleration is the vector difference between the absolute accelerations) or $b-a=a^{\prime}$ Considering upward motion of the cat $\quad T-m g=m a \ldots$ (i) Considering rightward motion of $M$ $T=M a^{\prime}=M(b-a) \quad \ldots($ ii $)$
From (i) and (ii), we get $T=\frac{m M}{m+M}(g+b)=\left(\frac{m \times 2 m}{m+2 m}\right)(10+2)=\frac{2 m}{3} \times 12=8 \mathrm{~N}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:33

Problem 27

Blocks $A$ and $C$ 'start from rest and move to the right
$=$
with acccleration $a_{A}=6 t \mathrm{~m} / \mathrm{s}^{2}$ and $a_{C}=3 \mathrm{~m} / \mathrm{s}^{2}$. After what time does the block $B$ again come to rest?
Solution From constraint relations we can see that acceleration of block is $a_{n}-\left(\frac{36 i)}{2}\right)-1.5-3 t \quad$ or $\frac{d v_{B}}{d t}-1.5-3 t$
or $\int_{0}^{v_{1}} d v_{B}-\int_{0}^{1}(1.5-3 t) d t \quad$ or, $v_{n}-1.5 t-1.5 t^{2}$ or $v_{n}-0$ at $t-1 \mathrm{~s}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:14

Problem 28

$\Lambda$ train moving up an incline of $\sin ^{1}\left(\frac{7}{25}\right)$ goes with an acceleration of $\frac{g}{5} \mathrm{~m} / \mathrm{s}^{2}$. What will be the acceleration of the same train on a level road if it exerts the same traction force' The coefficient of friction between the road and wheels is the same, $0.2$, on both sides.
Solution Traction foree $=m g \sin \alpha+\mu m g \cos \alpha+m a$
$-m g \frac{7}{25}+0.2 m g \frac{24}{25}+\frac{m g}{5}-m g\left\{\frac{7}{25}+\frac{24}{125}+\frac{1}{5}\right\}-m g \times \frac{84}{125}$
Traction foree in the second casc $=\frac{84 m g}{125}-\mu m g=\frac{84 m g}{125}-\frac{1}{5} m g=\frac{59 m g}{125}$
$\therefore$ The acceleration $-\frac{F}{m}-\frac{59 g}{125}$.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:01

Problem 29

A mass $m$ slides down a smooth inclined planc of inclination $\alpha$ and draws unother mass from rest through a distance $d$ metre in $t$ second along a smooth horizontal table which is on level with the top of the plane over which the string passes. Prove that the mass on the table is $m\left[\frac{g t^{2} \sin \alpha-2 d}{2 d}\right]$.
Solution
Let the acceleration of the system be $a$. 'Then for mass $m, m g \sin \alpha-T=m a$
and $m^{\prime} a=T \quad \therefore \quad a=\frac{m g \sin \alpha}{m+m^{\prime}}$The mass $m^{\prime}$ moves a distanee $d$ in $t$ seconds.
$d=\frac{1}{2} a t^{2} \quad \Rightarrow \quad a=\frac{2 d}{t^{2}} \quad \therefore \quad \frac{2 d}{t^{2}}=\frac{m g \sin \alpha}{m+m^{\prime}}$
$\therefore 2 m d+2 m^{\prime} d=m g \sin \alpha t^{2} \Rightarrow 2 m^{\prime} d=m g \sin \alpha t^{2}-2 m d=m\left[g \sin \alpha t^{2}-2 d \mid\right.$
$\therefore \quad m^{\prime}-m\left[\frac{g t^{2} \sin \alpha 2 d}{2 d}\right]-m\left[\frac{g t^{2} \sin \alpha}{2 d}-1\right]$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:55

Problem 30

The system of bodies in the figure starts from rest. Determinc the acecleration of the body $B$ and the tension in the string supporting the body $A$ given the weight of $A=500 N$ and of $B=750 \mathrm{~N}$. The cocfficient of 'friction between the surfaccs is $0.2$ Solution \Lambdapplying Newton's second law of motion to $A$, $500-T-\frac{500}{g} a \quad \ldots$ (i)
Similarly, upplying it to $B$
27 $750 \sin \alpha \quad \mu 750 \cos \alpha-\frac{750}{g} \frac{a}{2}$
c(ii)
since if $A$ moves $x$ down, $B$ moves up the plane $x / 2$. from equation (i) und (ii), we get $\therefore \quad a-\frac{430 \times g}{1375}-3.06 \mathrm{~m} / \mathrm{sec}^{2}$
$T-344 \mathrm{~N}$
$\therefore \quad$ Acecleration of $B-\frac{a}{2}-1.53 \mathrm{~m} / \mathrm{sec}^{2}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:36

Problem 31

In the system of connected bodies in the udjoining figure, the cocfficient of kinctic friction is $0.20$ under bodies $B$ and $C$. Determine the ueceleration of cach body und the tension in the chord supporting $A$.
Solution Suppose, $B$ moves up a distance $x$ and $A$ moves down by a distance $y$. So, $C$ moves down a distance $(2 x-y)$ on its own plane. So, the acceleration of $A, B, C$ are $a_{A}, a_{B},\left(2 a_{B}-a_{A}\right)$, respectively. \Lambdapplying Newton's second law to $A, B, C .$ $27-800 g \sin \alpha-800 \mu g \cos \alpha=800 a_{B}$
where $T$ is the tension of the string.or $27-800 g \times \frac{3}{5}-800 \frac{1}{5} g \times \frac{4}{5}=800 a_{B}$
$27-480 \mathrm{~g}-128 g=800 a_{B} \quad \Rightarrow \quad 27^{\prime}-608 g=800 a_{B}$
(1) Similarly, $400 g-7=400 a_{A} \ldots(2)$
or $800 g-27=800 a_{A}$ \Lambdadding Eqs. (1) and (2), $\quad \therefore 800 g-608 g=800\left(a_{A}+a_{B}\right)$ $\Rightarrow 192 g=800\left(a_{A}+a_{n}\right)$
$\therefore \quad a_{A}+a_{B}=\frac{24 g}{100}$
Now, for $C$ $1000 g \sin \alpha-1000 \mu g \cos \alpha-T=1000 a_{c}=1000\left(2 a_{B}-a_{A}\right)$
or $1000 g \times \frac{3}{5} \quad \frac{1000}{5} g \times \frac{4}{5} \quad T-1000\left(2 a_{B} \quad a_{A}\right)$
or $600 g-160 g-T=1000\left(2 a_{B}-a_{k}\right) \quad \Rightarrow 440 \mathrm{~g}-T=1000\left(2 a_{B}-a_{A}\right)$
and $T-304 g=400 a_{B}(\text { rrom 1 })^{8} \quad \Rightarrow 136 \mathrm{~g}=2400 a_{\mathrm{n}}-1000 a_{A}$
$\frac{136 \mathrm{~g}}{1000}-2.4 a_{B}-a_{A} \ldots(4$
From (3) and (4), $3.4 a_{B}-\frac{136 g}{1000}$ ? $\frac{240 g}{1000}-\frac{376 g}{1000} \quad \therefore a_{B}-\frac{376 g}{1000 \times 3.4}-1.084 \mathrm{~m} / \mathrm{sec}^{2}$
$\therefore a_{A}-2.4 a_{B}-\frac{136 g}{1000}-2.4 \times 1.084-\frac{136 \times 9.8}{1000}-2.6016-1.3328-1.27 \mathrm{~m} / \mathrm{sec}$
$\therefore T^{\prime}=400\left(g-a_{A}\right)=400(9.8-1.27)=400(8.53)=3412 \mathrm{~N}$

Nidhi Singhi
Nidhi Singhi
Numerade Educator
00:33

Problem 32

In the figure, cocflicient of friction between tho two blocks is given as $\mu=1 / 2$. Find the force of friction acting between the $\quad F_{3}=20 \mathrm{~N}$ two blocks. $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution Let acceleration of both the blocks towards left be $a$. I'hen $\quad a=\left(\int-2\right) / 2=(20-f / 4$ or $2 f-4=20-f$ or $f=8 \mathrm{~N}$
Maximum friction between the two blocks can be $F_{1}=$ $f_{\operatorname{lnax}}=\mu n g(m=2 \mathrm{~kg})=(0.5)(2)(10)=10 \mathrm{~N}$
4
Now since $f<f_{\max }$
$F_{2}=20 \mathrm{~N}$
Therefore, friction forec between the two blocks is $8 \mathrm{~N}$.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:25

Problem 33

'Iwo blocks $A$ and $B$ of mass $2 \mathrm{~kg}$ and $4 \mathrm{~kg}$ are placed one over the other as shown in figure. $\Lambda$ time varying horizontal force $F=2 t$ is applied ou the upper block. ( $t$ is in second and $F$ is in newton) Coefficient of friction between $A$ and $B$ is $\mu-\frac{1}{2}$ and the horizontal surlace over which $B$ is placed is smooth $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution
I.imiting friction between $A$ and $B$ is $f_{L}-\mu m_{A} g-\left(\frac{1}{2}\right)(2)(10)-10 \mathrm{~N}$$a_{\mathrm{max}}-\frac{f_{L}}{m_{b}}-\frac{10}{4}-2.5 \mathrm{~m} / \mathrm{s}^{2}$
Thus, both the blocks move together with same acecleration till the common acecleration becomes $2.5 \mathrm{~m} / \mathrm{s}^{2}$, alter that acecleration of $B$ will become constant while that of $A$ will go on increasing. To find the time when the acecleration of both the blocks becomes $2.5 \mathrm{~m} / \mathrm{s}^{2}$ (or when slipping will start between $A$ and $B$ ) we will write $2.5-\frac{F}{\left(m_{A}+m_{B}\right)}-\frac{2 t}{6} \therefore t-7.5 \mathrm{~s}$
Henec, lor $t \leq 7.5 \mathrm{~s} \quad a_{A}-a_{B}-\frac{F}{\left(m_{A} \mid m_{B}\right)}-\frac{2 i}{6}-\frac{t}{3}$
For $t \geq 7.5 \mathrm{~s} a_{B}=2.5 \mathrm{~m} / \mathrm{s}^{2}=$ constant
and $a_{\mathrm{A}}-\frac{F_{L} f_{L}}{m_{A}}$ or
$a_{A}-\frac{2 t-10}{2}$
or $a_{A}-i \quad 5$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:45

Problem 34

Threc blocks $A, B$ and $C$ of equal mass $m$ are paced one over the other on a smooth horizontal ground as in the figure. Cocflicicnt of friction bclween any two blocks of $A, B$ and $C$ is $1 / 2$. Find the maximum valuc of mass of block $D$ so that the blocks $A, B$ and $C$ move withoul slipping over cach other.
Solution Blocks $A$ and $C$ both move due to friction. But less friction is available to $A$ compared with $C$ because normal reaction between $A$ and $B$ is less. Maximum friction between $A$ and $B$ can be $f_{\max }=\mu m_{A} g=(1 / 2) m g$
Maximum accelcration of $A$ can be $a_{\max }-\frac{f_{\text {naax }}}{m}-\frac{g}{2}$ Further $a_{\max }-\frac{m_{D} g}{3 m \mid m_{D}} \Rightarrow \frac{g}{2}-\frac{m_{D} g}{3 m \mid m_{D}} \quad m_{D}-3 m$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:16

Problem 35

Figure shows a small block $\Lambda$ of mass $m$ kept at the left end of a plank $B$ of mass $M=2 m$ and length 1 . 'The system can slide on a horizontal road. 'Ihe system is started towards right with the initial velocity $v$. The friction coefficients between the road and the plank is $1 / 2$ and that between the plank and the block is $1 / 4$. Find:
(a) the time elapsed before the block separates from the plank.
(b) displacement of block and plank relative to ground till that moment.
Solution There will be relative motion betwecn block and plank and plank and road. So at cach surface limiting friction will act. The direction of friction forces al different surfaces are as shown in figure Here, $f_{1}-\frac{m g}{4}$
and $f_{2}-\left(\frac{1}{2}\right)(m+2 m) g-\frac{3 m g}{2}$Relardation of $A$ is $a_{1}-\frac{f_{1}}{m}-\frac{g}{4}$
and retardation of $B$ is $a_{2}-\frac{f_{2}-f_{1}}{2 m}-\frac{5}{8} g$. $\quad f_{2}$.
Since, $\quad a_{2}>a_{1}$
Relative acceleration of $A$ with respect to $B$ is $a_{r}-a_{2} \quad a_{1}-\frac{3}{8} g$
Initial velocity of both $A$ and $B$ is $v$. So, there is no relative initial velocity. I lence,
(a) \Lambdapplying
$s=\frac{1}{2} a t^{2} \quad \Rightarrow \quad l=\frac{1}{2} a_{r} t^{2}=\frac{3}{16} g t^{2}$
$\therefore \quad t=4 \sqrt{\frac{l}{3 g}}$
(b) Displacement o $\Gamma$ block $s_{A}=u_{A} t-\frac{1}{2} a_{A} t^{2} \Rightarrow s_{A}=4 v \sqrt{\frac{1}{3 g}}-\frac{1}{2} \frac{g}{4}\left(\frac{16 l}{3 g}\right) \quad\left(a_{A}=a_{1}=\frac{g}{4}\right)$
or $s_{A}-4 v \sqrt{\frac{l}{3 g}}-\frac{2}{3} l, \quad$ Displacement of plank $s_{n}-u_{g} t-\frac{1}{2} a_{n} t^{2}$
or $s_{R}-4 v \sqrt{\frac{l}{3 g}}-\frac{1}{2}\left(\frac{5}{8} g\right)\left(\frac{16 l}{3 g}\right) \quad\left(a_{n}-a_{2}-\frac{5}{8} g\right)$
or $s_{B}-4 v \sqrt{\frac{l}{3 g}} \frac{5}{3} l$

Nidhi Singhi
Nidhi Singhi
Numerade Educator
00:58

Problem 36

The cocflicient of friction between the block $A$ of mass $m$ and block $B$ of mass $2 m$ is $\mu-\frac{1}{\sqrt{3}}$. The inclined plane is sinooth. If the system of blocks
$A$ and $B$ is releascd from rest and there is no slipping between $A$ and $B$, then $\theta<\frac{\pi}{\alpha}$. Find the value of $\alpha$.
Solution When there is no slipping, then both the blocks move together with acceleration $a=g \sin \theta$, down the plane. I Iorizontal component of this acceleration is $a_{H}=a \cos \theta$ and vertical component is $a_{y}=a \sin \theta$, where $a=g \sin \theta$
$a_{H}=a \cos \theta=g \sin \theta \cdot \cos \theta$
and $a_{V}=a \sin \theta=g \sin ^{2} \theta$
$N=$ normal reaction between $A$ and $B$. Equations of motion in horizontal and vertical directions give:
$m g-N=m a_{p}$ or $N=m g-m a_{v}=m g-m g \sin ^{2} \theta=m g$
$f=$ friction force $\mu N \geq m a_{H}$
or $\mu m g \cos ^{2} \theta \geq m g \sin \theta \cos \theta \quad \Rightarrow \quad \mu \geq \tan \theta \quad$ or
$\theta \leq \tan ^{-1}(\mu)$
$\Rightarrow \theta<\tan { }^{1}\left(\frac{1}{\sqrt{3}}\right) \Rightarrow \quad \theta<\frac{\pi}{6} \quad \therefore \alpha-6$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:06

Problem 37

In the arrangement shown in the figure, $M_{A}=M_{B}=1 \mathrm{~kg}$. String and pulley are massless. Block $B$ is resting on a smooth horizontal surface, while friction coefficient between blocks $A$ and $B$ is $\mu=0.4$. Find the maximum horizontal force that can be applied so that block $A$ does not slip over the block $B$.
Solution
Net horizontal force on block $B$ is zero. I lence, the given figure (a) can be replaced by figure (b).
(a)
(b)
Maximum value of friction is $f_{\max }=\mu m_{1} g=(0.4)(1)(10)=4 \mathrm{~N}$. Block $B$ moves due to friction. 'lherefore, maximum common acceleration of the two blocks can be $a_{\max }-\frac{f_{\max }}{m_{B}}-\frac{4}{1}-4 \mathrm{~m} / \mathrm{s}^{2}$ and $F_{\max }=\left(m_{A} \mid m_{B}\right) a_{\max }-(1+1)(4)-8 \mathrm{~N}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:14

Problem 38

Find the maximum value of mass of block $C$ of that neither $A$ nor $B$ moves. $\left(m_{A}=100 \mathrm{~kg}, m_{R}=200 \mathrm{~kg}\right.$, Pulley and strings are massless. Cocflicient betwecn $A$ and $B$ and betwcen $B$ and horizontal surlace is $\mu=0.4$ )
Solution
Maximum friction that can be obtained between $A$ and $B$ is $f_{1}=\mu m_{A} g=(0.4)(100)(10)=400 \mathrm{~N}$
and maximum friction between $B$ and ground is $f_{3}=\mu\left(m_{A}+m_{S}\right) g=(0.4)(100+200)(10)=1200 \mathrm{~N}$
Drawing diagrams of $A, B$ and $C$ in limiting case,
Fquilibrium of $A$ gives $T_{1}=f_{1}=400 \mathrm{~N}$
Fquilibrium of $B$ gives $2 T_{1}+f_{1}+f_{2}=T_{2} \quad \Rightarrow \quad T_{2}=2(400)+400+1200=2000 \mathrm{~N}$
$\ldots(\mathrm{ii})$
and cquilibrium of $C$ gives $m_{C} g=T_{2} \quad \Rightarrow \quad 10 m_{C}=2000 \quad \therefore \quad m_{C}=200 \mathrm{~kg}=2 \times 10^{2} \mathrm{~kg}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:15

Problem 39

An open car of mass $m_{0}$ is running on smooth horizontal rails under rain falling vertically which it catches and retains in it. If $v_{0}$ is the initial velocity of the car and $\mu$ the mass of water falling into the car per second, find the distanec travelled by it in time $t$.
Solution
$m \frac{d v}{d t} \mid v \frac{d i n}{d t}-0 \quad$ So, $\left(m_{0}+\mu t\right) \frac{d v}{d t} \mid \mu v-0 \quad \therefore \quad \frac{d v}{v}-\frac{-\mu d t}{m_{o} \mid \mu t}$
Integrating, $\ln v=-\ln \left(m_{0}+\mu i\right)+C \quad$ So, $C=\ln v_{0}+\ln m_{0} \quad$ because when $t=0, v=\nu_{0}$
$\therefore \quad \ln \frac{v}{v_{0}}--\ln \left(m_{0}+\mu t\right)+\ln m_{0}-\ln \frac{m_{a}}{m_{o}+\mu t}$
$\therefore \quad v=v_{s} \frac{m_{\theta}}{m_{o}+\mu t}=\left(\frac{d s}{d t}\right) \quad \therefore \frac{d t}{m_{o}+\mu t}=\frac{d s}{m_{o} v_{0}} ;$ Integrating, we get
$\therefore \quad \frac{1}{\mu} \ln \left(m_{o}+\mu \theta\right)=\frac{s}{m_{0} v_{0}}+D$
when $t=0, s=0, \quad \therefore \quad \frac{1}{\mu} \ln m_{o}=D \quad \therefore \quad \frac{1}{\mu} \ln \left(m_{o}+\mu t\right)=\frac{S}{m_{o} v_{o}}+\frac{1}{\mu} \ln m_{o}$
$\Rightarrow \frac{1}{\mu} \ln \frac{\left(m_{0} \mid \mu t\right)}{m_{y}}-\frac{s}{m_{o} v_{o}} \quad \therefore \quad s-\frac{m_{0} v_{\theta}}{\mu} \ln \left(1+\frac{\mu t}{m_{0}}\right)$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:09

Problem 40

$\Lambda$ rain drop of mass $m_{o}$ starts falling from rest and it collects water vapour and grows. If it gains $\lambda \mathrm{kg} / \mathrm{s}$, find its velocity at any instant.
Solution
$\frac{d}{d t}(m v)-m g ; \quad \Lambda s \frac{d m}{d t}-\lambda \quad$ or
$m-\lambda t \mid k^{*}$ where $k$ is a constant.
When $t-0, m-m_{o} \quad \therefore m-m_{0}+\lambda t \quad$ So, $\quad \frac{d}{d t}\left\{\left(m_{,}+\lambda t\right) v\right\}-\left(m_{o}+\lambda t\right) g$
Integrating, $\left(m_{o}+\lambda t\right) v=\int\left(m_{o}+\lambda t\right) g d t-\left(m_{o} t+\frac{\lambda t^{2}}{2}\right) g+C$, where $C$ is a constant
Whon $t=0, v=0$ $\therefore \quad C-0 \quad \therefore\left(m_{0}+\lambda t\right) v-\left(m_{0} t+\frac{\lambda i^{2}}{2}\right) g$
$\therefore v-\frac{g\left(m_{0} i+\frac{\lambda t^{2}}{2}\right)}{m_{0} \mid \lambda t}-\frac{g\left(t+\begin{array}{c}\lambda t^{2} \\ 2 m_{0}\end{array}\right)}{1 \mid \begin{array}{c}\lambda t \\ m_{0}\end{array}}$
'lhis means velocity changes with respect to time.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:12

Problem 41

A clcan body o[ mass $100 \mathrm{~g}$ starts with a velocity of $2 \mathrm{~m} / \mathrm{s}$ on a smooth horizontal planc, accumulating dust at the rate of $5 \mathrm{~g} / \mathrm{s}$. Find the velocity at the end of 20 scconds and the distanec travclled during that period.
Solution Herc $\frac{d}{d t}(m v)-0 . \quad$ So, $\quad m \frac{d v}{d t}+v \frac{d m}{d t}-0 .$
After $t$ seconds, mass of the body $=100+5 t$ So, $(100 \mid 5 \mathrm{t}) \frac{d v}{d t}+5 v-0 \quad$ or $\quad \frac{d v}{v}-\frac{5 d t}{100+5 t}$
Integrating $[\ln v]_{200}^{+}-[\ln (10015 t)]_{0}^{30}$
$\therefore \quad \ln \frac{v}{200}=-\ln 200+\ln 100=\ln \frac{100}{200} \therefore \quad \frac{v}{200}=\frac{1}{2} \quad \Rightarrow \quad v=100 \mathrm{~cm} / \mathrm{sec} .=\operatorname{lm} / \mathrm{sec}$
Nlso, $\ln v=-\ln (100+5 t)+\ln C$
When $t=0, \quad v=200 \mathrm{~cm} / \mathrm{sec}$
$\therefore \ln 200=-\ln 100+\ln C \therefore \quad \ln C=\ln 200+\ln 100=\ln (200 \times 100)$
$\therefore \quad \ln v=-\ln (100+5 t)+\ln (200 \times 100)$
$\therefore \quad v-\frac{200 \times 100}{10015 t}-\frac{d s}{d t} \quad \therefore \quad d s-\frac{200 \times 100 d t}{10015 t} \mathrm{~cm}-\frac{200 d t}{10015 t} \mathrm{~m}$
$\therefore \quad \int d s-s-\int_{0}^{20} \frac{200 d t}{10015 t}-40[\ln (100 \mid 5 t)]_{0}^{20}-40\left[\ln \frac{200}{100}\right]-40 \ln 2$
$=40 \times 0.3010 \times 2.303=27.7 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:08

Problem 42

$\Lambda$ flat car of mass $m_{\circ}$ starts moving to the right due to a constant horizontal force $F_{1}$. Sand spills on the flat car from a stationary hopper. 'Ihe velocity of loading is constant and equal to $\mu \mathrm{kg} / \mathrm{s}$. Find the time dependence of the velocity and the acceleration of the flat car in the process of loading. 'The friction is negligibly small.
Solution $m \frac{d v}{d t}$ ? $v \frac{d n}{d t}-F \quad$ Mass of the car at any instant $=m_{0}+\mu t$
$\therefore \quad\left(m_{0}+\mu t\right) \frac{d v}{d t}+v \mu-F \quad \therefore \quad \frac{d v}{F \mu \nu}-\frac{d t}{m_{0} 1 \mu}$
Integrating, we have $\ln (F-\mu v)=-\ln \left(m_{o}+\mu t\right)+C$
When $t=0, v=0, \quad \therefore \quad \ln F=-\ln m_{o}+C \quad \therefore \quad \frac{F \mu v}{F}-\frac{m_{n}}{m_{o}+\mu t}$
$\frac{\mu v}{F}=\frac{\mu t}{m_{o}+\mu t}\left[\right.$ since if $\frac{a}{b}=\frac{c}{d}$, then $\left.\frac{b-a}{b}=\frac{d-c}{d}\right] \quad \Rightarrow \quad v=\frac{F t}{m_{o}+\mu t}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:52

Problem 43

A parlicle moves in a straight linc under tho actions a retarding forec. If its initial speed is $u$ and the retarding forec is $m K v^{3}$ where $m$ is its mass, $v$ is velocity at any instant and $K$ is a constant, show that $\frac{1}{v}-\frac{1}{u}-K x, x$ being the distance covered before its speed is reduced to $v .$ Show that the time taken to travel $x$ is given by $t-\frac{K x^{2}}{2}+\frac{x}{u}$
Solution $m \frac{d v}{d t}-m k v^{3} \quad$ or
$\frac{d v}{d t}-K v^{3} \quad \Rightarrow \frac{d v}{v^{3}}-K d t$
Intcgrating, $-\frac{1}{2 v^{2}}=-K t+C . \quad$ When $t=0, v=u \quad \therefore \quad-\frac{1}{2 v^{2}}=-K t-\frac{1}{2 u^{2}}$
or $\left(\frac{1}{u^{2}}-\frac{1}{v^{2}}\right)=-2 \mathrm{~K} t \quad \Rightarrow \quad 2 K t=\frac{1}{v^{2}}-\frac{1}{u^{2}}$
Also $m \frac{v d v}{d x}--m K v^{3} \quad \therefore \quad-\frac{d v}{v^{2}}-K d x$
Integrating, $\frac{1}{v}-k x \mid C$
When $x-0, v-u \quad \therefore \quad-\frac{1}{u}+\frac{1}{v}-K x$ or $\frac{1}{v}-\frac{1}{u}-K x$
But $2 \mathrm{~K} \mathrm{l}-\left(\frac{1}{v^{2}} \frac{1}{u^{2}}\right)$
$t=\left(\frac{1}{v^{2}}-\frac{1}{u^{2}}\right) \frac{1}{2 K}=\left(\frac{1}{v}-\frac{1}{u}\right)\left(\frac{1}{v}+\frac{1}{u}\right) \frac{1}{2 K}=K x\left(\frac{2}{u}+K x\right) \frac{1}{2 K} ; \quad \therefore \quad t=\frac{x}{u}+\frac{K x^{2}}{2}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:56

Problem 44

Having gonc through a plank of thickness $h$, a bullet changed its velocity from $v_{\mathrm{v}}$ to $v$. Find the time of motion of the bullet in the plank, assuming the resistive forec to be proportional to the square of the velocity.
Solution Resistanec $-m \frac{d v}{d t}--K v^{2} \quad$ or $\quad \frac{d v}{v^{2}}--\frac{K}{m} d t$
Integrating, $-\frac{1}{v}=-\frac{K}{m} t+c . \quad$ When $\mathrm{t}=0, v=v_{o} \quad$ So, $\quad-\frac{1}{v_{0}}=c$
$\therefore \quad-\frac{1}{v}--\frac{K t}{m}-\frac{1}{v_{0}} \quad$ or $\quad \frac{1}{v_{0}}-\frac{1}{v}--\frac{K t}{m} \quad \ldots(1)$
Also $m \frac{d v}{d t}-m v \frac{d v}{d x}--K v^{2} \quad$ or $\quad \frac{d v}{v}--\frac{\mathrm{K} d x}{m}$
$\therefore$ Integrating, $\ln v=-\frac{K}{m} x+D .$ When $v=v_{0}, x=0 \quad \therefore \quad \ln v_{0}=D$
or $\ln v-\frac{-K}{m} x+\ln v_{c} \quad \therefore \quad \ln \frac{v}{v_{0}}-\left[-\frac{K x}{m}\right]_{\theta}^{h}--\frac{K h}{m} \quad \ldots(2)$
From $(1), t-\frac{m}{K}\left(\frac{1}{v}-\frac{1}{v_{o}}\right)-\frac{m}{K}\left(\frac{v_{0}-v}{v_{o} v}\right) ; \quad$ From $(2), h-\frac{m}{K} \log \frac{v_{0}}{v}$
$\therefore \quad t-\frac{\left(v_{0}, v\right)}{v_{0} v} \frac{h}{\log _{0} v_{0}}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:36

Problem 45

$\Lambda$ motor boat of mass $m$ moves in a lake with a velocity $v_{o} . \Lambda t$ the moment $t=0$, the engine is shut down. Assuming the resistance of water to be proportional to the velocity of the boat, $F=-r v(r$ is a positive constant), find :
(a) how long the boat moved with the shut down engine,
(b) the velocity of the boat as a function of the distance covered till the complete stop, and
(c) the mean velocity of the boat over the time interval (beginning with the moment, $t=0$ ) during which the velocity decreases $\eta$ times.
Solution
Accelcration of the boat $--\frac{r v}{m}-\frac{d v}{d t}$ $\therefore \frac{d v}{v}--\frac{r}{m} d t$
Integrating, $\ln v-\frac{r}{m} t \mid C \quad$ When $t-0, v-v_{0}$ So, $C=\ln v_{0}$
$\therefore \quad \ln v--\frac{r}{m} t+\ln v_{o} \quad$ or $\quad \ln \frac{v}{v_{n}}--\frac{r}{m} t$
or $\frac{v}{v_{0}}=e^{\frac{r}{m} t}$
or $\quad v=v_{o} e^{\frac{r}{u x} t}$ i.e., $v \rightarrow 0$ when $t \rightarrow \infty$
So, it stops when $t$ is infinitc only. Also $\frac{v d v}{d x}--\frac{r v}{m} \quad$ or $\quad \frac{d v}{d x}--\frac{r}{m}$
$\therefore v--\frac{r}{m} x+D$
When $\quad x=0, v=v_{0} \quad \therefore \quad v_{\omega}=D \quad$ Henec $v--\frac{r}{m} x+v_{0} \quad \therefore \quad v-v_{\omega}-\frac{r x}{m}$
Total distanec it moves before it comes to stop is given by
$v-v_{o} \quad \frac{r x}{m}-0 \quad \therefore \quad x-\frac{v_{o} m}{r}$
we have to find the distance $x^{\prime}$ covered before the velocity becomes $\frac{v_{0}}{\eta}$. Bocause $v=v_{o}-\frac{r x}{m} \quad \therefore \quad \frac{v_{0}}{\eta}=v_{o}-\frac{r x^{\prime}}{m} \quad$ or $\quad \frac{r x^{\prime}}{m}=v_{o}-\frac{v_{0}}{\eta}=v_{o}\left(1-\frac{1}{\eta}\right)$
$\therefore \quad x^{\prime}=\frac{m}{r}\left(\frac{\eta 1}{\eta}\right) v_{0}$
But $v-v_{0} e^{n} \quad \Rightarrow \frac{v_{0}}{v}-e^{n}$
$--e^{n t} \quad$ or $\quad \eta-e^{n}$
246
(1)$\therefore \frac{r t}{m}-\ln e^{n} \quad \therefore \quad t-\frac{m \ln e^{\eta}}{r}$
$\Lambda$ verage velocity $-\frac{\text { distance travelled }}{\text { time }}-\frac{x^{\prime}}{t}-\frac{\frac{m}{r}\left(\frac{\eta}{\eta}\right) v_{o}}{m \ln e^{n}}-\frac{(\eta \quad \mathrm{l}) v_{a}}{\eta \ln e^{n}}$
0

Nidhi Singhi
Nidhi Singhi
Numerade Educator
01:07

Problem 46

A water pipe has an intemal diameter of $10 \mathrm{~cm}$. Water flows through it at the rate of $20 \mathrm{~m} / \mathrm{sec}$. The water jet strikes normally on a wall and falls dead Find the forec on the wall.
Solution Mass of water flowing through the tube per second $=A v \rho$ where $A=$ area of cross section and $v=$ velocity of water of density $\rho$. Momentun change/second $=A v \rho v=A v^{2} \rho$
Thus, the forec on the wall $=A v^{2} \rho=\pi \times\left(\frac{5}{100}\right)^{2} \times(20)^{2} \times 1000=3143 \mathrm{~N}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:37

Problem 47

A spherical rain drop, lalling in a constant gravitational ficld, grows by absorption of moisturc from the surroundings at a rate proportional to its surlace area. If it starts with zero radius, find its acceleration.
Solution $\frac{d m}{d t}-K \cdot 4 \pi r^{2}$, where $K$ is constant and $r$ is the radius of the drop at any instant. But $i n=\frac{4}{3} \pi r^{3} \rho=\frac{4}{3} \pi r^{3}$, since $\rho=$ density of water is $\mathrm{l} \mathrm{gm} / \mathrm{cc}$.
$\frac{d m}{d t}=\frac{4}{3} \pi .3 r^{2} \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t} \quad$ and $\frac{d m}{d t}=K \cdot 4 \pi r^{2}$
$\therefore \quad K-\frac{d r}{d t} \quad \therefore r-K t$ (since it starts with rero radius) Since $m \frac{d v}{d t}$ ? $v \frac{d m}{d t}-m g \quad \frac{4}{3} \pi r^{3} \frac{d v}{d t}$ ? $v \cdot K 4 \pi r^{2}-\frac{4}{3} \pi r^{3} g$
$\frac{d v}{d t}+v K \frac{3}{r}-g$ But $\frac{d v}{d t}-a$ and $v-a t$
$\therefore a+a t K \frac{3}{K t}-g$ becomes $4 a-g \quad \therefore a$ -acecleration of drop $-\frac{g}{4}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:34

Problem 48

$\Lambda$ body moves over a plane surface, starting from rest. The frictional force is constant. The tractive force is $(P-K t)$ in gravitational units. The maximum velocity is attained after $t$ seconds. What is the distance covered before it attains the maximum speed if the mass of body is $m$.
Solution Forec $-m a-m \frac{d v}{d t}-(p-K t) g \quad \therefore \quad m v-\int(P-K t) g d t-\left(P t-\frac{K t^{2}}{2}\right) g$The maximum velocity is attained when $\frac{d}{d t}\left(P t-\frac{K t^{2}}{2}\right)$ is 'rero.
or $P-K t=0$ or $t=(T)=\frac{P}{K}$. But $v=\frac{d s}{d t}$
$\therefore \quad m \frac{d s}{d t}-\left(P t-\frac{K t^{2}}{2}\right) g$ or $\quad$ ins $-\left(\frac{P t^{2}}{2}-\frac{K t^{3}}{6}\right) g$
$\therefore \quad s-\frac{1}{i n}\left[\frac{P}{2} \frac{P^{2}}{K^{2}} \quad \frac{K}{6} \frac{P^{3}}{k^{-3}}\right] g-\frac{g}{m}\left[\frac{P^{3}}{2 K^{2}} \quad \frac{P^{3}}{6 K^{2}}\right]-\frac{g}{i n} \frac{P^{3}}{3 K^{2}}-\frac{g}{m} \frac{K^{3} T^{3}}{3 K^{2}}-\frac{g K T^{3}}{3 t h}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator