Question

a bullet of mass 20g moving with a velocity of 200m s get embedded in a wooden block of mass 980g . calculate the velocity acquired by the block.​

Answer 1

by the law of conservation of momentum,

20/1000 *200 + 980/1000 *0= 20/1000*0 + 980/1000*v

4=0.98v

v= 4.081 m per s

Answer 2

Answer:

4m/s

Explanation:

Total momentum before collision

Mass of bullet (M) =20g=0.02kg

Mass of block (m) =980g=0.98kg

Velocity of bullet (V) =200m/s

Velocity of block (v) =0m/s

Total momentum=MV+mv

=0.02×200+0.98×0

=4

Total momentum after collision

Mass of bullet (M)=0.02kg

Mass of block (m)=0.98kg

[velocity of bullet and block are same after collision]

Velocity=?

Total momentum=Mv+mv

=v(M+m)

=v(0.02+0.98)=v×1

According to Law of Conservation of Momentum

Total momentum(before collision =after collision)

4=v×1

therefore v=4m/s