Momentum Problem: Find Speed of Block After Inelastic Collision

[dsptl]

* Block A of mass 5kg is released from the rest at height of 5m. Then it hits 10kg block B, which at ground level. The collision in inelastic. Find the speed of the block after the collision if the block B was intially at rest.


Attempt:

Ki+Ui=Kf+Uf, since the block was at rest, Ki=0, and it goes to ground level so, Uf=0, thus Ui=Kf.
mghi=1/2mvf2, 5*9.8*5=1/2(5)vf2, vf=9.9m/s


but the answer is wrong...please help someone

 

[alphysicist]

Hi dsptl,

* Block A of mass 5kg is released from the rest at height of 5m. Then it hits 10kg block B, which at ground level. The collision in inelastic. Find the speed of the block after the collision if the block B was intially at rest.


Attempt:

Ki+Ui=Kf+Uf, since the block was at rest, Ki=0, and it goes to ground level so, Uf=0, thus Ui=Kf.
mghi=1/2mvf2, 5*9.8*5=1/2(5)vf2, vf=9.9m/s


but the answer is wrong...please help someone

You've found the speed of block A right before the collision (assuming it slid freely down to ground level), but how do you take into account the effect of the collision itself? (By the way, since you mentioned that the collision was elastic, did you mean it was completely inelastic?)