# Easy collision question, momentum and energy

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1. Mar 29, 2016

### PoohBah716

1. The problem statement, all variables and given/known data

A spring (k=4200N/m) and box A (mA=120kg) are on a frictionless incline, as shown below . Box A is pressed against the spring such that it is compressed 1.0m, and then released. Box A then hits, and sticks to box B, 1.0m farther up the ramp from the uncompressed position of the spring (massB=80kg) (i.e. the collision happens 2.0m total distance from A's initial position). Box B is held at rest until it is struck by A, then it is free to move. How far up the ramp will the two of them travel up the ramp before starting to slide back down?

2. Relevant equations
MaVa+MbVb=MaVaf+MbVbf

3. The attempt at a solution
h=2sin30
h=1
v=1/2*(m+m)=(M+m)gh

2. Mar 29, 2016

### drvrm

what is m and M and how you are trying it
v=1/2*(m+m)=(M+m)gh .... your this equation has dimensional problem
try to do conservation of momentum to find out the velocity of A+B after collision. then you can find the distance as energy conservation will hold after the collision.

3. Mar 29, 2016

### PoohBah716

P1=p2
M1(Vf-Vi)=-M2(Vf-Vi)
KE1f+KE2f=KE1i+KE2i
1/2M1f+V1f^2+1/2M2fV2f^2=1/2M1iV1i^2+1/2M2V2f^2

Could you walk me through this problem, Im getting stuck.

4. Mar 29, 2016

### drvrm

actually we can only provide hints and you have to do the work.
initially the spring will give push when released - one can calculate the velocity acquired by this push by mass A- you have k given so thats not a problem- the spring energy will be spent in providing kinetic energy as well as potential energy of massA
now this massA is hitting B at rest so equate momentum of A and B before the hit with momentum of two body combined with the unknown velocity.
this equality can give you the final velocity of the combined mass.
but one must take care when equating energy of this mass going uphill to the velocity zero as it is climbing up also against gravitational pull.