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 into the block, causing the block, with the bullet embedded, to swing upwards to a height of 50.0cm above its original position.

A) What type of collisions best describes the bullet striking the block?

B) Do a walk-through of the energy of the system, starting just before the bullet strikes the block and continue until the block and bullet reach maximum height.

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

Please answer all three parts fully and clearly to get the points! Thanks!!

Explanation / Answer

A)THIS IS AN inelastic collision

B) energy of the system = mgh=2.002*9.8*0.5=9.8098J

C)kinetic energy of the bullet=potential energy of the block

1/2*0.002*v^2=9.8098

v=99.04m/s

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