Question

A bullet of mass 20g moving with a velocity of 200mpers gets embeded in a wooden block of mass 980g calculate the velocity acquired by a block

Answer 1

Hey!

Answering a question after such a long time!✌ So here it goes:

Mass of the bullet = 20gm
Mass of the Wooden block = 980gm
Velocity of the bullet = 200m/s
Velocity of the block = v1

According to the law of conservation of momentum :

m1u1 + m2u2 = m1v1 +m2v2

=> 0 = 20×200 +980v2

=> 0 = 4000 + 980v2

=> -4000 = 980v2

=> -4000/982 = v2

=> 4.07 = v2

Hence, the velocity acquired by the block is 4.1m/s [estimated value].

Thank you!

Answer 2

Given :-

» Mass of bullet = 20 gm = 20/1000 = 1/50 kg

» Velocity = 200 m/sec

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So

» Momentum of bullet = M × V

=> 1/50 × 200 = 4 kg.m/sec

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» Let's the velocity REQUARIED by block be "x"

» Mass of Block = 980 gm

= 980 gm = 980/1000 kg = 49/50 kg

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« Momentum of Block after bullet gets Embedded = Mass × Velocity

=> 4 = 49/50 × X = 49x/50

=> X = 200/49 = 4.08
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