Question

A block of mass $m$ and a pan of equal mass are connected by a string going over a smooth light pulley as shown. Initially the system is at rest when a particle of mass $m$ falls on the pan and sticks to it. If the particle strikes the pan with a speed $v$, the speed with which the system moves just after collision is

• A. $\frac{v}{3}$
• B. $\frac{v}{4}$
• C. $\frac{v}{5}$
• D. $\frac{v}{2}$

Answer 1

The correct option is A $\frac{v}{3}$

Given, mass of block $=m$
mass of pan $=m$
mass of particle $=m$
Initial speed of particle before striking the pan $=v$
Let speed of system just after collision $=v_0$
Let us suppose, tension in string is $T$ and normal force between particle and pan is $N$.
For impulsive force $F$
$\int Fdt=J=$ impulse = change in momentum


Impulse on particle:
On particle, only normal force $N$ acts
$\therefore \int -Ndt=$ change in momentum
$\Rightarrow \int -Ndt=m{v}_0-mv$
[taking downward direction +ve]
$\Rightarrow \int Ndt=mv-m{v}_0$ ... (1)
Impulse on pan:
On pan, normal force and tension acts
$\therefore \int (N-T)dt=$ change in momentum
$\Rightarrow \int (N-T)dt=m{v}_0-0$
$\Rightarrow \int Ndt-\int Tdt=m{v}_0$ ... (2)


Impulse on block:
On block, tension force acts
$\therefore \int Tdt=$ change in momentum
$\Rightarrow \int Tdt=m{v}_0$ .... (3)
On adding equation (2) and (3)
$\int Ndt=2m{v}_0$ ... (4)
On comparing equation (1) and (4)
$mv-m{v}_0=2m{v}_0$
$\Rightarrow v_0=\frac{v}{3}$
System moves with $\frac{v}{3}$ just after collision.