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 with bullet embedded to swing upwards to a height of 50.0 cm above its original position.

A. What type of collision 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 max height.

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

**Remember to convert to MKS and show all work**

Explanation / Answer

A.

completey inelastic

b. No, not just brfore bullet strike.

Yes , after bullet strike.

c. using energy conservation after bullet hit to till max. heright

mgh = mv^2 /2

9.8 x 0.50 = v^2 /2

v = 3.13 m/s

now momentum conservation .

0.002 x u = (0.002 +2) x 3.13

u = 3133.63 m/s

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