A bullet of mass 0.01 kg travelling at 500 m/s strikes a wood block of mass . 5

Question

A bullet of mass 0.01 kg travelling at 500 m/s strikes a wood block of mass . 5 kg from below. The bullet bullet emerges from the block travelling at 200m/s (a) what is the velocity of the block right after the bullet leaves it. (b) how high does that block rise (c) if the bullet was in contact with the block for . 001 s what was the average force exerted by the bullet on the block?

A bullet of mass 0.01 kg travelling at 500 m/s strikes a wood block of mass . 5 kg from below. The bullet bullet emerges from the block travelling at 200m/s (a) what is the velocity of the block right after the bullet leaves it. (b) how high does that block rise (c) if the bullet was in contact with the block for . 001 s what was the average force exerted by the bullet on the block?

Explanation / Answer

a) From momentum conservation

Mass of bullet X Initial velocity of bullet = Mass of bullet X Final velocity of bullet + Mass of block X Final velocity of block (v)

0.01 X 500 = 0.01 X 200 + 0.5 X v

v = 6 m/s

b) From energy conservation

1/2mv2 = mgh

h = v2/2g = 62/2X9.8 = 1.84 m

c) Impulse = Change in momentum

F t = m ( vf - vi )

F X 0.001 = 0.5 ( 6-0)

F = 3000 N

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