A 2.0 kg block hangs from a string of length 1.0 m. A 5.0 gram bullet moving at

Question

A 2.0 kg block hangs from a string of length 1.0 m. A 5.0 gram bullet moving at 500 m/s is fired into the 2.0 kg block. The bullet rebounds with some unknown speed,and the block swings to s height of 20 cm above the original positions. 1) what is the speed of the block immediately after the collision? 2) what is the rebound speed of the bullet? 3) is this colliosion elastic or inelastic? 4) if the bullet gets embedded in the block instead of rebounding what new height will the block reach? A 2.0 kg block hangs from a string of length 1.0 m. A 5.0 gram bullet moving at 500 m/s is fired into the 2.0 kg block. The bullet rebounds with some unknown speed,and the block swings to s height of 20 cm above the original positions. 1) what is the speed of the block immediately after the collision? 2) what is the rebound speed of the bullet? 3) is this colliosion elastic or inelastic? 4) if the bullet gets embedded in the block instead of rebounding what new height will the block reach? 1) what is the speed of the block immediately after the collision? 2) what is the rebound speed of the bullet? 3) is this colliosion elastic or inelastic? 4) if the bullet gets embedded in the block instead of rebounding what new height will the block reach?

Explanation / Answer

1) rebound velocity of the bullet

initial momentum = final momentum

mbvb = mbvb'+mblock vblock

block raises to a height of 20 cm

so its velocity 1) vblock = (2gh)^0.5= 1.97 m/sec

so the bullet recoil velociy vb' can be found by the above eqaution

2) vb' = (mbvb-mblockvblock )/(mb) = -288m/sec

3) the collision is elastic collision

4) if the bullet gets embedded the raises to which the block raises

is with a common velocity Vc

mbvb = (mblock+mb)Vc

Vc = 1.24668 m/sec

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