Posted by Sandi on .
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.

Physics 
bobpursley,
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 
Physics 
bobpursley,
oops...
distancealongplane=h/sin37 
Physics 
Sandi,
where are you getting 6kg from?