Question

A 40 kilogram child moving at 2.0 meters per second jumps onto a 60 kilogram sled that is initially at rest on a long, frictionless, horizontal sheet of ice.
Speed = .8 m/s
KE = 32 J

After coasting at constant speed for a short time, the child jumps off the sled in such a way that she is at rest with respect to the ice.

a. Determine the speed of the sled after the child jumps off it.

b. Determine the kinetic energy of the child-sled system when the child is at rest on the ice.

Answer 1

Well if the kid and sled are moving at v1i=0.8m/s and v2i=o.8m/s and the kid jumps off and has a final velocity of v1f=0m/s then just use the momentum equation.
m1v1i + m2v2i = m1v1f + m2v2f
m1=40kg v1i=0.8m/s m2=60kg v2i=0.8m/s v1f=0m/s v2f=?
40(0.8) + 60(0.8) = 40(0) + 60(?)
40(0.8) + 60(0.8) =80 \[\therefore\]
v2f=80/60\[\therefore\]
v2f=1.333...m/s

Then just use the kinetic energy equation for the second question.
1/2m1v1^2 + 1/2m2v2^2
So
1/2x40x0^2 + 1/2x60x(1.333...)^2
Ek=53.33333331

Answer 2

Thank You! This is so much easier than it looks! I understand it much better now! Thank You Again! : D