Question

A 50 kg man at rest on a frictionless horizontal floor throws along the floor a 5.0 kg ball with 2.0 m/s towards a wall. The ball rebounds elastically from the wall, the man catches the ball and throws it again with the same velocity towards the wall. How many times can he repeat this process? Also find the final velocities of the man and the ball.

Answer 1

5 times, Final velocities of man and ball are -2.2 m/s and -2.0 m/s respectively.

Answer 2

The problem involves a man and a ball on a frictionless horizontal floor, interacting with each other and a wall. We need to analyze the momentum conservation at each stage.

Let: $M = 50$ kg (mass of the man) $m = 5.0$ kg (mass of the ball) $v_0 = 2.0$ m/s (speed of the ball when thrown towards the wall)

We will use the convention that velocity towards the wall is positive and away from the wall is negative.

Interpretation of "throws along the floor a 5.0 kg ball with 2.0 m/s towards a wall":

This phrase is crucial. We interpret it to mean that the absolute velocity of the ball with respect to the ground is 2.0 m/s towards the wall.

• Initial State: Man and ball are at rest. Total momentum $P_0 = 0$.
• 1st Throw: The man throws the ball with $v_b = +2.0$ m/s. Let the man's velocity be $V_1$. By conservation of momentum: $0 = M V_1 + m v_b$ $0 = 50 V_1 + 5 \times 2.0 \implies 50 V_1 = -10 \implies V_1 = -0.2$ m/s. (Man moves away from the wall).
• Ball Rebounds: The ball hits the wall with $+2.0$ m/s and rebounds elastically with $v_b' = -2.0$ m/s. The man continues to move at $V_1 = -0.2$ m/s.
• 1st Catch: The man (moving at $V_1$) catches the ball (moving at $v_b'$). Let their combined velocity be $V_{1c}$. By conservation of momentum: $M V_1 + m v_b' = (M+m) V_{1c}$ $50(-0.2) + 5(-2.0) = (50+5) V_{1c}$ $-10 - 10 = 55 V_{1c} \implies -20 = 55 V_{1c} \implies V_{1c} = -\frac{20}{55} = -\frac{4}{11}$ m/s.
• 2nd Throw: The man+ball system is moving at $V_{1c}$. The man throws the ball again with $v_b = +2.0$ m/s. Let the man's new velocity be $V_2$. By conservation of momentum: $(M+m) V_{1c} = M V_2 + m v_b$ $55 \times (-\frac{4}{11}) = 50 V_2 + 5 \times 2.0$ $-20 = 50 V_2 + 10 \implies 50 V_2 = -30 \implies V_2 = -0.6$ m/s.
• Ball Rebounds (2nd time): $v_b' = -2.0$ m/s. Man at $V_2 = -0.6$ m/s.
• 2nd Catch: Let their combined velocity be $V_{2c}$. $M V_2 + m v_b' = (M+m) V_{2c}$ $50(-0.6) + 5(-2.0) = 55 V_{2c}$ $-30 - 10 = 55 V_{2c} \implies -40 = 55 V_{2c} \implies V_{2c} = -\frac{40}{55} = -\frac{8}{11}$ m/s.

Generalizing the process: Let $V_n$ be the man's velocity after the $n$-th throw. Let $V_{n-1,c}$ be the man+ball velocity before the $n$-th throw. From the $n$-th throw: $(M+m) V_{n-1,c} = M V_n + m v_b$. From the $n$-th catch: $M V_n + m (-v_b) = (M+m) V_{n,c}$.

Combining these, we can find a recurrence relation for $V_{n,c}$: $M V_n = (M+m) V_{n-1,c} - m v_b$ Substitute $M V_n$ into the catch equation: $(M+m) V_{n-1,c} - m v_b + m (-v_b) = (M+m) V_{n,c}$ $(M+m) V_{n,c} = (M+m) V_{n-1,c} - 2m v_b$ $V_{n,c} = V_{n-1,c} - \frac{2m v_b}{M+m}$

This is an arithmetic progression. With $V_{0,c}=0$: $V_{n,c} = n \left(-\frac{2m v_b}{M+m}\right) = n \left(-\frac{2 \times 5 \times 2}{50+5}\right) = n \left(-\frac{20}{55}\right) = -\frac{4n}{11}$ m/s.

Now, let's find $V_n$ (man's velocity after $n$-th throw): $M V_n = (M+m) V_{n-1,c} - m v_b$ $V_n = \frac{M+m}{M} V_{n-1,c} - \frac{m v_b}{M}$ Substitute $V_{n-1,c} = -\frac{4(n-1)}{11}$: $V_n = \frac{55}{50} \left(-\frac{4(n-1)}{11}\right) - \frac{5 \times 2}{50}$ $V_n = \frac{11}{10} \left(-\frac{4(n-1)}{11}\right) - \frac{10}{50}$ $V_n = -\frac{4(n-1)}{10} - \frac{1}{5} = -\frac{2(n-1)}{5} - \frac{1}{5} = -\frac{2n-2+1}{5} = -\frac{2n-1}{5}$ m/s.

How many times can he repeat this process?

The process involves the man catching the ball. The man is moving away from the wall (negative velocity $V_n$). The ball returns from the wall with velocity $-v_b = -2.0$ m/s. For the man to catch the ball, the ball's speed must be greater than the man's speed in the same direction (away from the wall). So, $|V_n| < |-2.0|$ m/s. $|-\frac{2n-1}{5}| < 2.0$ $\frac{2n-1}{5} < 2.0$ $2n-1 < 10$ $2n < 11$ $n < 5.5$

This means the man can successfully complete the process (throw, rebound, catch) for $n=1, 2, 3, 4, 5$ times. For the 6th throw, the man's velocity after throwing would be $V_6 = -\frac{2(6)-1}{5} = -\frac{11}{5} = -2.2$ m/s. The ball would go to the wall and return with $-2.0$ m/s. Since $|V_6| = 2.2$ m/s, which is greater than $2.0$ m/s, the man is moving faster away from the wall than the ball returning from the wall. Thus, the ball will never reach the man. Therefore, the man can repeat the process 5 times.

Final velocities of the man and the ball:

After 5 repetitions, the man makes the 5th throw. His velocity becomes $V_5 = -1.8$ m/s. The ball goes to the wall with $+2.0$ m/s, rebounds with $-2.0$ m/s. The man (at $-1.8$ m/s) catches the ball (at $-2.0$ m/s). This is the 5th catch. Their combined velocity is $V_{5c} = -\frac{4 \times 5}{11} = -\frac{20}{11}$ m/s $\approx -1.818$ m/s. At this point, the process ends because he cannot perform the 6th throw and subsequent catch.

After the 5th successful cycle. The man then attempts the 6th throw.

For the 6th throw: $(M+m) V_{5c} = M V_6 + m(2.0)$ $55 \times (-\frac{20}{11}) = 50 V_6 + 5(2.0)$ $-100 = 50 V_6 + 10$ $50 V_6 = -110 \implies V_6 = -2.2$ m/s.

The ball goes to the wall at $+2.0$ m/s and rebounds at $-2.0$ m/s. At this point, the man's velocity is $-2.2$ m/s, and the ball's velocity is $-2.0$ m/s. Since $|-2.2| > |-2.0|$, the man is moving away from the wall faster than the ball is returning. The man cannot catch the ball. Therefore, the process stops here.

The number of times he can repeat the process (complete a full cycle of throw-rebound-catch) is 5 times. The final velocities are the velocities of the man and the ball when the process ends, i.e., after the 6th throw and the ball's rebound.

Final Answer: Number of times he can repeat the process = 5 times. Final velocity of the man = $-2.2$ m/s (away from the wall). Final velocity of the ball = $-2.0$ m/s (away from the wall).