A 2.00g bullet strikes a 2.00kg block hanging from a long cord. The bullet imbed

Question

A 2.00g bullet strikes a 2.00kg block hanging from a long cord. The bullet imbeds itself in the block causing the block to swing upwards to a height of 75cm above its original position.
A. What type of collision best describes the bullet striking the block?

* I believe this is a complete inelastic collision (Please correct me if I'm wrong).


B. Do a walk-through of the energy of the system starting just before the bullet strikes the block.


*Not sure how to answer this one

C. Find the velocity of the bullet before it collides with the block.

*Need help here

Help with these please and I will rate as Awesome!

Explanation / Answer

a) It is a completely inelastic collision

b)The bullet has some kinetic energy before it strikes the block. On striking the block, the bullet loses some of the kinetic energy and embeds in the block which gains some kinetic energy which converts to potential energy when the block rises with the cord.

c)Potential energy gain=mgh=2.002*9.8*0.75=14.715J

velocity of the system at lowermost point=v

0.5*2.002*v2=14.715J

v=3.834m/s

Now conserving momentum,

0.002*vi=2.002*3.834

vi=3838m/s

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