Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The total mass of the sled, man, and rock is 96.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 14.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving).
Also determine the speed of the sled if he throws the rock directly backwards.

Answer 1

you can do this at least 2 ways.

first and most obvious is conservation of momentum. no external force has acted on the system, ie the man, the sled and the rock. thus the momentum of the man the rock and the sled should remain constant throughout.

By telling you that "Both speeds are relative to the ground" the question is telling you to do all your calculations with respect to ground speed.

momentum before = 96 x 0.480 kgm/s.

******if the man throws the rock forward (i.e. in the direction the sled is moving).
momentum after = (96 - 0.31) * v + 0.31 * 14.5
momentums before and after should be equal
THUS solve for v

******if he throws the rock directly backwards THEN velocity has -ve sign as velocity is a VECTOR, ie its direction really matters
momentum after = (96 - 0.31) * v + 0.31 * (-14.5)
momentums before and after should be equal
THUS solve for v again

OK?