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A bullet of mass 20g moving with a velocity of 200m/s gets embedded in a wooden block of mass 980g. Calculate the velocity acquired by the block.
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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
= 0.02*200 + 0*0.98 = 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.
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
= 0.02*200 + 0*0.98 = 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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Answer: 4m/sec
Explanation: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
= 0.02*200 + 0*0.98 = 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 4m/sec
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