Question

A block of mass $m=1\text{kg}$ and a pan of equal mass are connected by a string going over a smooth light pulley as shown in the figure. Initially the system is at rest when a particle of mass $1\text{kg}$ falls on the pan and sticks to it. If the particle strikes the pan with a speed $v=3\text{m/s}$. Find the speed with which the system moves just after collision

• A. $4\text{m/s}$
• B. $3\text{m/s}$
• C. $2\text{m/s}$
• D. $1\text{m/s}$

Answer 1

The correct option is D $1\text{m/s}$

Let, the speed of system just after collision is $v_f$.

As there is a sudden change in the speed of the block, the tension must change by a large amount during the collision.

Let $N$ be the normal force between the particle and the pan and $T$ be the tension in the string.


Now, impulse imparted by normal force $N$ to the particle is in upward direction, and it is equal to change in momentum.

Thus, $\int Ndt=mv-m{v}_f...\left(i\right)$

Impulse imparted to pan,

$\int (N-T)dt=m{v}_f..\left(ii\right)$

And impulse imparted to block,

$\int Tdt=m{v}_f...\left(iii\right)$

From equations $\left(ii\right)$ and $\left(iii\right)$,

$\int Ndt=2m{v}_f...\left(iv\right)$

Now, from equations $\left(i\right)$ and $\left(iv\right)$,

$2m{v}_f=mv-m{v}_f$

$\therefore v_f=\frac{v}{3}=\frac{3}{3}=1\text{m/s}$

Hence, $\left(\text{D}\right)$ is the correct answer.

Alternate solution: Let $v_f$ be the downward velocity of the system just after collision. Apply conservation of momentum, $m\times v_i=(m+m+m)v_f$ $\therefore v_f=v_i/3=3/3=1\text{m/s}$