Question

Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he rolls the ball for the first time. (b) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and B after the ball has made 5 round trips and is held by A. (d) How many times can A roll the ball? (e) Where is the centre of mass of the system "A + B + ball" at the end of the nth trip?

Answer 1

a) 0.5 m/s, b) 10/11 m/s, c) Speed of A = 50/11 m/s, Speed of B = 5 m/s, d) 6, e) 10d/21 from A's initial position

Answer 2

The system consists of A, B, and the ball. The platform is frictionless, so there are no external horizontal forces. Thus, the total momentum of the system is conserved. Initially, A, B, and the ball are at rest, so the initial total momentum is zero. The total momentum remains zero throughout the process.

Let $m_A = m_B = 40$ kg and $m_b = 4$ kg. Let the positive direction be from A towards B. The ball's speed relative to the platform is $v_0 = 5$ m/s.

(a) Find the speed of A after he rolls the ball for the first time.

Initially, $v_A = 0$, $v_B = 0$, $v_b = 0$. A rolls the ball towards B. Let $v_{A1}$ be the velocity of A and $v_{b1}$ be the velocity of the ball after A rolls it. B is still at rest ($v_{B1} = 0$). The ball's velocity relative to the platform is $v_{b1} = 5$ m/s. By conservation of momentum for the system (A + B + ball): $m_A v_{A1} + m_B v_{B1} + m_b v_{b1} = 0$ $40 v_{A1} + 40(0) + 4(5) = 0$ $40 v_{A1} + 20 = 0 \implies v_{A1} = -0.5$ m/s. The speed of A is $|v_{A1}| = 0.5$ m/s.

(b) Find the speed of A after he catches the ball for the first time.

After A rolls the ball, A moves with $v_{A1} = -0.5$ m/s, B is at rest, and the ball moves towards B with $v_{b1} = 5$ m/s. B catches the ball. Let $v_{B1'}$ be the velocity of B and the ball together after the catch. A's velocity $v_{A1'}$ remains $-0.5$ m/s during the catch. By conservation of momentum for the system (A + B + ball): $m_A v_{A1'} + (m_B + m_b) v_{B1'} = 0$ $40(-0.5) + (40+4) v_{B1'} = 0$ $-20 + 44 v_{B1'} = 0 \implies v_{B1'} = 20/44 = 5/11$ m/s. Now B rolls the ball back towards A. Let $v_{B2}$ be the velocity of B and $v_{b2}$ be the velocity of the ball after B rolls it. A's velocity $v_{A2}$ remains $-0.5$ m/s during the roll. The ball's velocity relative to the platform is $v_{b2} = -5$ m/s (towards A). By conservation of momentum for the system (A + B + ball): $m_A v_{A2} + m_B v_{B2} + m_b v_{b2} = 0$ $40(-0.5) + 40 v_{B2} + 4(-5) = 0$ $-20 + 40 v_{B2} - 20 = 0 \implies 40 v_{B2} = 40 \implies v_{B2} = 1$ m/s. The ball moves towards A with $v_{b2} = -5$ m/s. A is moving with $v_{A2} = -0.5$ m/s. A catches the ball. Let $v_{A2'}$ be the velocity of A and the ball together after the catch. B's velocity $v_{B2'}$ remains $1$ m/s during the catch. By conservation of momentum for the system (A + B + ball): $(m_A + m_b) v_{A2'} + m_B v_{B2'} = 0$ $(40+4) v_{A2'} + 40(1) = 0$ $44 v_{A2'} + 40 = 0 \implies v_{A2'} = -40/44 = -10/11$ m/s. The speed of A after he catches the ball for the first time is $|v_{A2'}| = 10/11$ m/s.

(c) Find the speeds of A and B after the ball has made 5 round trips and is held by A.

Let's denote the velocities of A and B after the $n$-th round trip (ball starts with A, goes to B, comes back to A, and is caught by A) as $v_A^{(n)}$ and $v_B^{(n)}$. After 1st round trip (calculated in part b): $v_A^{(1)} = -10/11$ m/s, $v_B^{(1)} = 1$ m/s. Ball is with A.

We observe a pattern for the velocities after $n$ round trips: $v_A^{(n)} = -n \times (10/11)$ m/s $v_B^{(n)} = n \times (1)$ m/s After 5 round trips: $v_A^{(5)} = -5 \times (10/11) = -50/11$ m/s. Speed of A is $|v_A^{(5)}| = 50/11$ m/s. $v_B^{(5)} = 5 \times 1 = 5$ m/s. Speed of B is $|v_B^{(5)}| = 5$ m/s.

(d) How many times can A roll the ball?

The process stops when A cannot catch the ball.

A rolls the ball for the $n$-th time at the start of the $n$-th round trip. The $n$-th round trip ends when A catches the ball for the $n$-th time.

After B rolls the ball towards A (n-th round trip), the relative velocity of the ball with respect to A must be negative for A to catch it. Let $v_A'$ be A's velocity after rolling the ball. The relative velocity of the ball (-5) wrt A is $-5 - v_A'$. For A to catch, $-5 - v_A' < 0 \implies v_A' > -5$. $v_A' = -10(n-1)/11 - 0.5$. So, $-10(n-1)/11 - 0.5 > -5$ $-10(n-1)/11 > -4.5$ $10(n-1)/11 < 4.5$ $20(n-1) < 99$ $n-1 < 99/20 = 4.95$ $n < 5.95$. This means A can catch the ball from B for $n=1, 2, 3, 4, 5$.

So, A can successfully complete the catch for 5 round trips. A rolls the ball for the 1st time, 2nd time, 3rd time, 4th time, and 5th time, and the trip is completed.

A rolls the ball for the 6th time. B catches it (possible). B rolls it back. A's velocity after rolling is $v_A' = -10(5)/11 - 0.5 = -111/22$. Relative velocity of ball (-5) wrt A (-111/22) is $-5 - (-111/22) = -5 + 111/22 = (-110 + 111)/22 = 1/22 > 0$. This means the ball is moving away from A relative to A. A cannot catch the ball. So the process stops when A attempts to catch the ball during the 6th round trip.

The 1st, 2nd, 3rd, 4th, 5th round trips are completed. A rolls the ball for the 6th time, B catches it, B rolls it back, but A cannot catch it. So A rolls the ball 6 times.

(e) Where is the centre of mass of the system "A + B + ball" at the end of the nth trip?

As established, the total momentum of the system is conserved and is zero. Therefore, the velocity of the center of mass is zero. Since the initial velocity of the CM is zero, the CM remains at its initial position. Let A's initial position be $x_A(0) = 0$ and B's initial position be $x_B(0) = d$. The ball is initially with A, so $x_b(0) = x_A(0) = 0$. The initial position of the center of mass is: $X_{CM}(0) = \frac{m_A x_A(0) + m_B x_B(0) + m_b x_b(0)}{m_A + m_B + m_b} = \frac{40(0) + 40(d) + 4(0)}{40 + 40 + 4} = \frac{40d}{84} = \frac{10d}{21}$. At the end of the $n$-th trip, the ball is held by A. The CM position is constant. The position of the center of mass at the end of the $n$-th trip is: $\frac{10d}{21}$ from A's initial position.