A m = 1.4 kg object moving at v = 15 m/s collides with a stationary 2.0 kg object. If the collision is perfectly inelastic, how far along the inclined plane (37degrees) will the combined system travel? Neglect friction.

Answer 1

I have never know what a perfectly inelastic collision is. I do know what an inelastic collision is: Energy is NOT conserved. Ask your teacher this question, see if they know.

Momentum applies:
Now on this statement, conservation ofm momentum applies
momentum before=momentum after
along the path of the moving object..

1.4*15m/s+2*0=(6kg*V) assuming the objects are stuck together, the problem did not state that.

Now, you can consider energy:
initialKE after collision= change PEnergy

1/2 (6)V^2= (6)g height

where distancealongplane=hSin37

Answer 2

oops...

distancealongplane=h/sin37

Answer 3

where are you getting 6kg from?