Question

A wooden block of mass 1190 g is initially at rest on a frictionless floor. A bullet of mass 10 g moving with a velocity of 300 m/s strikes the block and gets embedded in it. Find the speed with which the block moves after the bullet hits it.

Answer 1

Mass of the block (M) = 1190 g = 1.19 kg

Initial velocity of the block = 0 m/s

Mass of the bullet (m) = 0.01 kg

Initial velocity of bullet (u) = 300 m/s

Let the common velocity after the collision be v.

Initial momentum = M × 0 + 0.01 × 300 = 3 kg m/s

Final momentum = (M + m) × v

Since, the forces between block and bullet are internal forces. Hence, the momentum of the system will be conserved.

Initial momentum = Final momentum

(1.19 + 0.01) × v = 3

1.2 × v = 3

$v = \frac{3}{1.2}$

v = 2.5 m/s

Hence, the common velocity of block and bullet is 2.5 m/s after the collision.