a bullet of mass 20g moving with velocity of 200m/s strike and gets embedded into a stationary wooden block of mass 980g . find the velocity with which the block moves .

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Solution

M1(mass of bullet) = 20 g= 0.02kg
m2(mass of wooden block)=980 g= 0.98 kg
u1= 200 m/s (Initial velocity of the bullet)
u2= 0 m/s (Initial velocity of the block of wood)

After the collision, the block as well as the bullet move witha common welocity. Let this velocity be v.

By Law of Conservation of Momentum

m1u1 + m2u2 = m1v + m2v
= 0.02200 + 00.98 = v(m1+m2)
= 4 = v(0.98 + 0.02)
= 4 = v

Therefore the velocity with which both the wooden block and the bullet move is 4 m/s.

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