How Does Catching a Ball Illustrate the Conservation of Momentum?

[erik-the-red]

Somebody throws me a ball with mass .400 $$kg$$ at velocity 10.0 $$m/s$$. My mass is 65.5 $$kg$$.

1. If I catch the ball, with what speed do the ball and I move afterwards?

Shouldn't the answer be $$(.400)(10.0=(65.9)(v_2)$$, where $$v_2=.0607$$$$m/s$$?

 

[Doc Al]

That's correct. (Assuming no other forces act on you.)

 

[quicknote]

Yes, that is correct. According to the law of conservation of momentum, the total momentum of the system before and after the ball is thrown should be the same. This means that the momentum of the ball before it is thrown (0.400 kg * 10.0 m/s) should be equal to the combined momentum of the ball and you after you catch it (0.400 kg * v2 + 65.5 kg * v2). Solving for v2, we get v2 = 0.0607 m/s. This means that you and the ball will move at a speed of 0.0607 m/s after you catch it. This is a result of the transfer of momentum from the ball to you, as you absorb its motion.