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?)

 

[baba89]

Hi there,

I can see that you have attempted to use the conservation of energy equation to solve this problem. However, since this is an inelastic collision, some of the kinetic energy will be lost in the form of heat, sound, and deformation. This means that the final kinetic energy will not be equal to the initial potential energy.

Instead, we can use the conservation of momentum equation to solve this problem. In an inelastic collision, the total momentum before the collision will be equal to the total momentum after the collision. So, we can set up the following equation:

m1v1 + m2v2 = (m1 + m2)vf

Where m1 and v1 are the mass and velocity of block A before the collision, m2 and v2 are the mass and velocity of block B before the collision, and vf is the final velocity of the combined blocks after the collision.

We know that block A is initially at rest, so v1 = 0. We also know that block B is initially at rest, so v2 = 0. Substituting these values into the equation, we get:

0 + 0 = (5+10)vf

Solving for vf, we get:

vf = 0 m/s

This means that the two blocks will stick together and have a final velocity of 0 m/s after the collision.

I hope this helps! Keep up the good work in your scientific problem-solving.