Question

There are two men A and B of mass m each and a cart of mass M. cart and both men are moving with v velocity towards the right. A and B are standing on the left and right extreme ends of the cart. A jumps outside the cart with a speed u relative to speed of cart in leftward direction. Find speed of A wrt ground (by conserving momentum)

Answer 1

v - u \frac{m + M}{2m + M}

Answer 2

To find the speed of man A with respect to the ground after he jumps, we will use the principle of conservation of linear momentum.

1. Define the System and Initial State: The initial system consists of Man A, Man B, and the Cart, all moving together.

• Mass of Man A = m
• Mass of Man B = m
• Mass of Cart = M
• Total initial mass of the system = M_initial = m + m + M = 2m + M
• Initial velocity of the system = v (towards the right)

Let's assume the rightward direction is positive. Initial momentum of the system: $P_{initial} = M_{initial} \times v = (2m + M)v$

2. Define the Final State: After Man A jumps, the system splits into two parts:

• Man A (mass m, velocity $v_A$ with respect to the ground)
• Man B + Cart (combined mass m + M, velocity $v_{CB}$ with respect to the ground)

The problem states that Man A jumps with a speed u relative to the speed of the cart in the leftward direction. The relative velocity of A with respect to the cart is given by: $\vec{v}_{A/Cart} = \vec{v}_A - \vec{v}_{CB}$

Since A jumps in the leftward direction relative to the cart, and we've taken right as positive: $v_{A/Cart} = -u$ So, $v_A - v_{CB} = -u$ This gives us a relationship between $v_A$ and $v_{CB}$: $v_A = v_{CB} - u$

Final momentum of the system: $P_{final} = m v_A + (m + M)v_{CB}$

3. Apply Conservation of Momentum: According to the principle of conservation of linear momentum, the total momentum of the system remains constant if no external forces act on it. $P_{initial} = P_{final}$ $(2m + M)v = m v_A + (m + M)v_{CB}$

Now, substitute the expression for $v_A$ from step 2 into the momentum equation: $(2m + M)v = m (v_{CB} - u) + (m + M)v_{CB}$ $(2m + M)v = m v_{CB} - mu + (m + M)v_{CB}$ $(2m + M)v = (m + m + M)v_{CB} - mu$ $(2m + M)v = (2m + M)v_{CB} - mu$

4. Solve for $v_{CB}$: Rearrange the equation to solve for $v_{CB}$: $(2m + M)v_{CB} = (2m + M)v + mu$ $v_{CB} = v + \frac{mu}{2m + M}$

5. Solve for $v_A$ (Speed of A with respect to the ground): Now substitute the expression for $v_{CB}$ back into the equation for $v_A$: $v_A = v_{CB} - u$ $v_A = \left(v + \frac{mu}{2m + M}\right) - u$ $v_A = v + u \left(\frac{m}{2m + M} - 1\right)$ $v_A = v + u \left(\frac{m - (2m + M)}{2m + M}\right)$ $v_A = v + u \left(\frac{m - 2m - M}{2m + M}\right)$ $v_A = v + u \left(\frac{-m - M}{2m + M}\right)$ $v_A = v - u \frac{m + M}{2m + M}$

This is the velocity of man A with respect to the ground. The question asks for the "speed", which is the magnitude of the velocity. Speed of A with respect to the ground = $|v_A| = \left|v - u \frac{m + M}{2m + M}\right|$

The final answer is $v - u \frac{m + M}{2m + M}$.

Explanation of the solution: The problem is solved using the principle of conservation of linear momentum. The initial momentum of the combined system (Man A + Man B + Cart) is calculated. The final momentum is expressed in terms of the velocities of Man A and the combined Man B + Cart system. A crucial step is to relate the velocity of Man A to the velocity of the cart using the given relative speed and direction. By substituting this relative velocity relationship into the momentum conservation equation, the velocities of the cart and Man A with respect to the ground are determined.

Answer: The speed of A with respect to the ground is $v - u \frac{m + M}{2m + M}$. (Note: This is the velocity, and its magnitude is the speed. Assuming positive direction is right, a negative result implies motion to the left).