A bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s. if the thickness of the block reduces to one half of the previous value
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Is this block of wood stationary ?
If so
loss of energy in block = F of friction time thickness
energy loss = (1/2)m (100)^2 - (1/2)m(10^2)
= (m/2)(9900)
If we only lost half as much then
(m/2)(100)^2 - (m/2)v^2 = (m/2)(4950)
v^2 = 5050
v = 71.1 m/s
Explanation:
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