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Question

A bullet of mass 20 g moving with a velocity of 100 ms1​ strikes a wooden block of mass 80 g and gets embedded into it. Calculate velocity of the combined system.


A

100 ms1

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B

30 ms1

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C

20 ms1

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D

5 ms1

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Solution

The correct option is C

20 ms1


Let mass of bullet be m1 =20g
Let mass of the wooden block be m2 =80g
Let initial velocity of the bullet be u1 = 100 ms1 and final velocity of the bullet be v1.
Let the initial velocity of wooden block be u2, since it was at rest, the initial velocity is zero, i.e u2=0 ms1
Let final velocity of the wooden block be v2.
By applying law of conservation of momentum,
intial momentum of the system = final momentum of the system
i.e, m1u1+m2u2=m1v1+m2v2
After the bullet is fired, the bullet gets embedded on the block.
Therefore, it acts as a system
Hence , the final velocity of the bullet and wooden block after collision will be equal as it works as a system
i.e v1=v2=v where v is the velocity of the system after collision.
Substituting on the equation, we get
(0.020×100)+(0.080×0)=v×(0.020+0.080)
2=0.1×v
v = 20 ms1

Velocity of system after collision = 20 ms1


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