Question

A block of mass m and a pan of equal mass m, are tied by a string going over a smooth light pulley as shown in figure. Initially the system is at rest when a particle of mass m falls on the pan and sticks to it. If particle strikes the pan with speed 6 m/s, then find speed with which the system moves just after collision

Answer 1

2 m/s

Answer 2

The system consists of a block of mass $m$ and a pan of mass $m$, connected by a string over a smooth light pulley. Initially, the system is at rest. A particle of mass $m$ falls onto the pan with a speed of 6 m/s and sticks to it. We need to find the speed of the system just after the collision.

Let $v_p$ be the initial speed of the particle just before striking the pan, so $v_p = 6$ m/s. Let $V$ be the speed of the system (block, pan, and particle) just after the collision. Since the string is inextensible and the pulley is smooth, the block will move upwards with speed $V$, and the pan with the stuck particle will move downwards with speed $V$.

We can use the impulse-momentum theorem for each part of the system during the collision. Let's assume the upward direction is positive for the block and the downward direction is positive for the pan and the particle.

Initial velocities: Block: $v_{B,i} = 0$ Pan: $v_{Pan,i} = 0$ Particle: $v_{P,i} = 6$ m/s (downward)

Final velocities (just after collision): Block: $v_{B,f} = -V$ (upward is positive, so downward is negative. Block moves up, so velocity is +V. Let's be consistent with the similar question's approach of considering the direction of motion after collision as positive for that object.) Let's assume upward is positive for the block and downward is positive for the pan and particle. Initial velocities: Block: $v_{B,i} = 0$ Pan: $v_{Pan,i} = 0$ Particle: $v_{P,i} = 6$ (downward)

Final velocities (just after collision): Block: $v_{B,f} = V$ (upward) Pan + Particle: $v_{Pan+P,f} = V$ (downward)

During the very short time interval of the collision, the gravitational force is non-impulsive compared to the impulsive forces (normal force between particle and pan, and tension in the string).

Let $N$ be the magnitude of the impulsive normal force exerted by the pan on the particle (upward). Let $T'$ be the magnitude of the impulsive tension in the string (upward on the pan, upward on the block).

Impulse-momentum theorem for the particle (downward direction is positive): Initial momentum = $m v_{P,i} = m(6)$. Final momentum = $m v_{Pan+P,f} = m V$. (The particle's final velocity is the same as the pan's final velocity). Impulse on particle = Change in momentum. The impulsive force on the particle is the normal force $N$ acting upwards. So, the impulse is $-N \Delta t$. $\int (-N) dt = m V - m(6) = m(V-6)$. $\int N dt = m(6-V)$.

Impulse-momentum theorem for the pan (downward direction is positive): Initial momentum = $m v_{Pan,i} = m(0) = 0$. Final momentum = $m v_{Pan+P,f} = m V$. Forces on the pan are the normal force from the particle $N$ (downward) and the tension $T'$ from the string (upward). The net impulsive force on the pan is $N - T'$ (downward). $\int (N - T') dt = m V - 0 = mV$. $\int N dt - \int T' dt = mV$.

Impulse-momentum theorem for the block (upward direction is positive): Initial momentum = $m v_{B,i} = m(0) = 0$. Final momentum = $m v_{B,f} = m V$. The force on the block is the tension $T'$ (upward). $\int T' dt = m V - 0 = mV$.

Now we have a system of equations for the impulses:

1. $\int N dt = m(6-V)$
2. $\int N dt - \int T' dt = mV$
3. $\int T' dt = mV$

Substitute (3) into (2): $\int N dt - mV = mV$ $\int N dt = 2mV$.

Substitute this expression for $\int N dt$ into (1): $2mV = m(6-V)$. Since $m \neq 0$, we can divide by $m$: $2V = 6-V$. $3V = 6$. $V = \frac{6}{3} = 2$ m/s.

The speed with which the system moves just after collision is 2 m/s. The block moves upwards with 2 m/s, and the pan with the stuck particle moves downwards with 2 m/s.

The final answer is $\boxed{2 m/s}$.

Explanation of the solution:

1. Identify the system just before and just after the collision.
2. Apply the impulse-momentum theorem to each component (particle, pan, block) during the collision. Neglect non-impulsive forces like gravity during the short collision time.
3. Set up equations based on the change in momentum and the impulses of the interactive forces (normal force between particle and pan, tension in the string).
4. Use the constraint of the string (equal speed for block and pan-particle combination) and the relationship between the impulses.
5. Solve the resulting equations for the unknown speed $V$.

Answer: 2 m/s