Question

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

Answer 1

Given Mass of bullet = 20gm
Mass of wooden block = 980gm
initial velocity = 20m/s
Final velocity = 0m/s (as it got embedded in wooden block)
Initial velocity of wooden block = 0 m/s
Final velocity = v
So from conservation of momentum
m1(0-20)/t = m2(v-0)/t
20(-20) = 980(v)
v = -400/980 = -20/49= -0.41m/s
so velocity of block is 0.41m/s

Answer 2

By using conservation of momentum equation

mu = (M + m)v

v = mu / (M + m)

= (20 × 10^-3 × 200) / [(980 + 20) × 10^-3]

= 4 m/s

Block will move with velocity equal to 4 m/s

OR

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/s

HOPE ANY OF TIS CAN HELP YOU