Question

A bullet of mass 12.5 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.12 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 4.70 m/s.

(a) Find the initial speed of the bullet.

(b) How much kinetic energy is lost?

Answer 1

Answer 2

SOLUTION :

a.

Let the initial speed of the bullet be v1 m/s.

m1 = mass of bullet = 12.5 g = 0.0125 kg

m2 = mass of block = 1.12 kg

v2 = initial speed of the block = 0 

v’ = speed of block (with bullet) = 4.70 m/s 

So,

Momentum before collision = m1v1 + m2v2 = 0.0125 v1 + 1.12 * 0 = 0.0125 v1

Momentum after collision = (m1 + m2) v’ = (0.0125 + 1.12) * 4.70 = 5.32275

As per conservation of momentum :

Momentum before collision = Momentum after collision 

=> 0.0125 v1 = 5.32275

=> v1 = Initial speed of bullet =  5.32275 / 0.0125 = 425.82 m/s (ANSWER)

b.

K. E. lost

= Initial K.E. - Final K.E.

= 1/2 * 0.0125 * (425.82)^2 - 1/2 (0.0125+1.12) * (4.70)^2 

= 1120.76 J (ANSWER)