How Does Momentum Affect the Velocity of a Caught Ball?

[devilz_krypt]

sorry but i have one last question momentum and impulse...

A boy with mass of 60 kg standing on a frozen lake throws a ball with mass of 10 kg at 10 m/s towards his worst enemy. His enemy with a mass of 50 kg catches the ball. What velocity does the ball have after being caught? My personal guess is 0 m/s but I am pretty sure that is incorrect because 1. That is to eas of an answer, 2. This is sort of like a test grade, 3. There would be no work required. So can someone smart trll me how to work this question out, I would really appreciate it...

 

[Mentallic]

The point of the frozen lake implies there are no frictional forces involved. You need to calculate this question by using the law of conservation of momentum.

Think about it logically for a sec. You catch a 10kg mass traveling at 10ms^-1 (I think we assume it is horizontally) and at the same time you are standing on ice. Once you catch the rock, because of its momentum, you will begin to move in the direction the rock was travelling. Obviously not at 10ms^-1 since you're heavier, but whatever velocity you are traveling at after the catch, is the velocity the rock will be traveling at too.

 

[kevtimc]

Mentallic is correct, but I would just like to add that the first boy should be removed from your thought process. You have a ball traveling at you (50 kg person) with a certain momentum (mass * velocity). It doesn't matter how it got its momentum at this point. The momentum when the ball is traveling at you (10 kg * 10 m/s) is the same as when you catch the ball. (Hint: think of you and the ball as an inelastic equation, your masses combine)

 

[kevtimc]

You catch a 10kg mass traveling at 10ms^-1 (I think we assume it is horizontally)

I don't know that it will be traveling at 10 m/s at the instance of the catch . . .

Actually, if it follows a perfectly horizontal path, I guess it is safe to assume.

 

[Mentallic]

I don't know that it will be traveling at 10 m/s at the instance of the catch . . .

Actually, if it follows a perfectly horizontal path, I guess it is safe to assume.

Of course there must be gravity present, so a parabolic projection of the rock's path will be travelled. However, because no further info is given about how far apart or how much time the rock was in flight for, we can only assume the simplest case scenario which is that it makes an inelastic collision with the enemy parallel to the frozen lake (ground).

A lot of assumptions are required in this question. and damn a rock that size thrown at that speed... can we assume attempted murder?