Question

Matching Numerical 2: In the figure, block 1 of mass $m_1$ slides from rest along a frictionless ramp from height $h = 3.3 m$ and then collides elastically with stationary block 2, which has mass $m_2 = 5m_1$. After the collision, block 2 slides into a region where the coefficient of kinetic friction $\mu_k = 0.10$ and comes to a stop in distance $d$ within that region. Match the quantity on the left with the correct value in SI units on the right. Each value on the right is used only once. 21) speed of block 1 just before the collision A) 3.7 22) speed of block 2 just after the collision B) 1.5 23) speed of block 1 just after the collision C) 8.0 24) $d$ 25) maximum vertical height up the ramp block 1 reaches after the collision D) 5.4 E) 2.7

          Matching Numerical 2: In the figure, block 1 of mass $m_1$ slides from rest along a frictionless ramp from height $h = 3.3 m$ and then collides elastically with stationary block 2, which has mass $m_2 = 5m_1$. After the collision, block 2 slides into a region where the coefficient of kinetic friction $\mu_k = 0.10$ and comes to a stop in distance $d$ within that region. Match the quantity on the left with the correct value in SI units on the right. Each value on the right is used only once.
21) speed of block 1 just before the collision
A) 3.7
22) speed of block 2 just after the collision
B) 1.5
23) speed of block 1 just after the collision
C) 8.0
24) $d$
25) maximum vertical height up the ramp block 1 reaches after the collision
D) 5.4
E) 2.7

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matching numerical 2 in the figure block 1 of mass m1 slides from rest along a frictionless ramp from height h 33 m and then collides elastically with stationary block 2 which has mass m2 68445

Added by Roberto J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vishal Gupta

10/17/2024

Step 1

The potential energy at the top of the ramp is converted into kinetic energy at the bottom. $m_1gh = \frac{1}{2}m_1v_1^2$ $v_1 = \sqrt{2gh} = \sqrt{2(9.8)(3.3)} = \sqrt{64.68} \approx 8.04 m/s$ This is approximately 8.0 m/s. Show more…

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Matching Numerical 2: In the figure, block 1 of mass $m_1$ slides from rest along a frictionless ramp from height $h = 3.3 m$ and then collides elastically with stationary block 2, which has mass $m_2 = 5m_1$. After the collision, block 2 slides into a region where the coefficient of kinetic friction $\mu_k = 0.10$ and comes to a stop in distance $d$ within that region. Match the quantity on the left with the correct value in SI units on the right. Each value on the right is used only once. 21) speed of block 1 just before the collision A) 3.7 22) speed of block 2 just after the collision B) 1.5 23) speed of block 1 just after the collision C) 8.0 24) $d$ 25) maximum vertical height up the ramp block 1 reaches after the collision D) 5.4 E) 2.7

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Transcript

00:01 Hi, here in this given problem, this is, first of all, this is the smooth inclined plane.

00:09 The block one slides down it.

00:15 It meets with the nether block, block two at the bottom.

00:21 Then this is the rough surface at which, after traveling through the distance deep, both of them combined with each other.

00:31 They come to rest.

00:34 Height of this block 1 initially that was edge and it was having some initial velocity also this height h this is given as 1 .5 meter and for the first block initially speed v1 i this is given as 4 .2 meter per second so first of all we will find speed gained by the first block at the bottom we will say it to be equal to v1f for which we will apply conservation of mechanical energy it says initial gravitational potential energy of the block 1 m1 gh plus its kinetic energy half m1 i square should be equal to its kinetic energy half m1 v1 f square at the bottom canceling mass of the block from both the sides we get 9 .8 multiplied by h which is 1 .6 plus half times of v if which is 4 .2 square is equal to 1 by 2 v1 f square which is missing we have to find it in the first part of the given problem so this v1f, that will be given by square root of 2 times of 9 .8 into 1 .6 plus simply square of 4 .2, because this 2 will be cancelled.

02:37 So this v1f, that is calculated to be equal to 7 .0 meter per second.

02:46 Speed gained by the first block when it reaches to the ground.

02:50 Answer for the first part.

02:53 Then in the second part of the problem, we have to find speed of the combined blocks just after block 1 hits with the second block...

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