Answer :
Final answer:
The 93 kg man will be moving at a speed of 0.161 m/s after catching the ball that he threw at a wall and rebounded back to him, based on the principle of conservation of momentum.
Explanation:
This question involves the principle of conservation of momentum. According to the law of conservation of momentum, the total momentum of the system (man and the ball) is conserved before and after the collision with the wall. The total initial momentum, before the ball is thrown, is zero as everything is at rest.
When the man throws the ball, the ball gains momentum in one direction, and to keep the total momentum zero, the man must gain an equal amount of momentum in the opposite direction. On returning, the ball has changed its direction, so it has lost the initial momentum and gained the same amount in the opposite direction.
The man catches the ball, meaning he gains the momentum that the ball brought back, causing him to move. To determine how fast he is moving, we set the final momentum of the system (which should still be zero) equal to the initial momentum.
To calculate the speed (v) of the man we'll use the equation: v = 2 × ball_mass × ball_speed / man_mass = 2 × 0.653 kg × 11.3 m/s / 93 kg = 0.161 m/s
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Final answer:
The man will be moving at approximately 0.078 m/s after he catches the ball. This is found using the conservation of momentum, considering the ball's momentum before and after it collides with the wall and bounces back to the man.
Explanation:
The student's question involves the conservation of momentum, a fundamental concept in physics. When the man throws the ball and catches it after it bounces off the wall, both the ball and the man are involved in an isolated system where external forces are negligible. As momentum is conserved, the velocity of the man after catching the ball can be calculated using the conservation of momentum principle.
To solve this problem, we first consider the momentum before and after the collision with the wall. The man throws the ball at 11.3 m/s. The direction of the ball's momentum changes after it bounces back, but the speed remains the same due to the 'elastic' nature of the collision as mentioned in the problem statement (no energy loss).
The initial momentum of the system (man and ball) is zero because they're both at rest initially. When the man throws the ball, the ball gains momentum in the forward direction, and the man gains momentum in the opposite direction. This momentum for the man (in the opposite direction) is what will be calculated.
The initial momentum of the ball and man is:
- Initial Ball: 0 kg·m/s (at rest)
- Initial Man: 0 kg·m/s (at rest)
The final momentum after throwing the ball is:
- Final Ball: -0.653 kg * 11.3 m/s (the negative sign indicates the direction change after bouncing)
- Final Man: 93 kg * v (where v is the velocity we want to calculate)
By conservation of momentum:
Initial momentum = Final momentum
0 = -0.653 kg * 11.3 m/s + 93 kg * v
Solving for v gives:
v = (0.653 kg * 11.3 m/s) / 93 kg
v ≈ 0.078 m/s
This is the speed of the man after catching the ball, moving opposite to the direction in which he threw the ball.
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