A ballistic pendulum has a bullet shot into a block in which it will get buried

Question

A ballistic pendulum has a bullet shot into a block in which it will get buried such that the block and bullet rise to a certain height in order to determine the velocity of the bullet when it left the muzzle. If the bullet's mass is 0.015kg, the block has a mass of 2.000kg and it rises to a height of 0.500m: a) What is the initial velocity of the block once the bullet got buried in it? b) What was the initial velocity of the bullet when it left the barrel of the gun? c) What type of collision is this between the bullet and the block? d) How much work did the block do on the bullet? e) If the bullet is in contact with the block for 0.005 seconds before being coming to rest relative to the block, what average force must the block have applied to the bullet?

Explanation / Answer

We will use conservation of energy after the bullet is embedded inside the block,

a) ( m+M)V^2 / 2 = ( m+M) gh

V= sqroot ( 2gH) =sqroot *(2 x 9.8 x 0.5) = 3.13 m/s

b) conserving the momentum

0.015 u = ( 0.015 + 2) 3.13

u = 420.463 m/s apprx

c) Inelastic collision

d)work done by block in bullet = chang in KE of bullet = iNitial KE of bullet - final KKE = 1/2 ( 0.015) (420.463 )^2 - 0 = 1326 J apprx

e) force applied = change in momentum / time = 0.015 (420.463 )/ 0.005 = 1261.389 N apprx

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