Question
Two masses $m_{1}=1 \mathrm{~kg}$ and $m_{2}=2 \mathrm{~kg}$ are connected bya light inextensible string and suspended by means of a weightless pulley as shown in the figure. Assuming that both the $1 \mathrm{k}$masses start from rest. the distance travelled by the centre of mass in two seconds is $\left(\right.$ Take $\left.g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$(a) $\frac{20}{9} \mathrm{~m}$(b) $\frac{40}{9} \mathrm{~m}$(c) $\frac{2}{3} \mathrm{~m}$(d) $\frac{1}{3} \mathrm{~m}$
Step 1
The acceleration of the system is given by $a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}}g$. Substituting the given values, we get $a = \frac{2 - 1}{1 + 2} \cdot 10 = \frac{10}{3} \, m/s^{2}$. Show more…
Show all steps
Your feedback will help us improve your experience
Dheeraj Sharma and 93 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The two bodies of masses $m_{1}$ and $m_{2}\left(m_{1}>m_{2}\right)$ respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and released. The acceleration of the centre of mass is (a) $\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} g$ (b) $\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g$ (c) $g$ (d) zero
Two masses, $m_{1}=1 \mathrm{kg}$ and $m_{2}=3 \mathrm{kg}$ , are connected by a string that is draped over a pulley, as shown above. Mass 1 is positioned on a frictionless horizontal surface, while mass 2 hangs freely. The masses are released from rest. Determine the acceleration of mass 2 . (A) 2.5 $\mathrm{m} / \mathrm{s}^{2}$ (B) 3.3 $\mathrm{m} / \mathrm{s}^{2}$ (C) 5.0 $\mathrm{m} / \mathrm{s}^{2}$ (D) 6.7 $\mathrm{m} / \mathrm{s}^{2}$ (E) 7.5 $\mathrm{m} / \mathrm{s}^{2}$
The two bodies of mass $m_{1}$ and respectively are tied to the ends of a massless string, which passes over a light and frietionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is (a) $\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} g$ (b) $\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2}$ (c) $g$ (d) zero
Centre of Mass
Round 2
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD