Physics, asked by Anonymous, 1 year ago

A bullet of mass 0.01kg and moving with a velocity of 500 m/s strikes block of mass 2kg suspended feom a 5m long string.the centre of gravity of the block rises vertically upwards through a height of 0.1m the velocity of the bullet after emerging out of the block will be

Answers

Answered by Hiteshbehera74
3
h = v²/2g
v² = 2gh
v = √(2gh)
v = √(2×10×0.1)
v2 = √(2)

m1×u1 + m2×u2 = m1×v1 + m2×v2
0.01×500 + 2×0 = 0.01×v1 + 2×√2
5 + 0 = 0.01(v1)+2√2
0.01(v1) = 5-2√2
v1 = 2.12/0.01
v1 = 212 m/s

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