A bullet of mass 20g strikes a block of mass 980g with a velocity u and gets embedded in it. The block is in contact with a spring whose force constant is 100N/m. After the collision the spring is compressed to 10cm. Find: 1. Velocity of the block after the collision. 2. Magnitude of the velocity of the bullet. 3. Loss in kinetic energy due to the collision.

Refer to the diagram shown below.

m = 20g = 0.02 kg, mass of bullet

M = 980 g = 0.98 kg, mass of block

u = velocity of bullet before impact

k = 100 N/m, spring constant

x = 10 cm = 0.1 m, compression of spring

Before impact,

The bullet travels at velocity u m/s, and the block is stationary.

The momentum of the system is

P = mu = (0.02 kg) * (u m/s) = 0.02u (kg-m) / s

The initial kinetic energy, KE, of the system is

KE₁ = (1/2) mu² = (1/2) (0.02 kg) * (u m/s) ² = 0.01u² J

At impact, the momentum is preserved. The bullet is embedded in the block, and they move together with velocity v on a frictionless surface.

Therefore

(M+m) v = mu

(0.02+0.98) v = 0.02u

v = 0.02u

The KE of the system becomes

KE₂ = (1/2) * (0.02+0.98 kg) * (v m/s) ² = 0.5v² J

The KE₂ of the system will be stored in the spring as strain energy (if energy losses are ignored).

The strain energy stored in the spring is

SE = (1/2) kx² = (1/2) * (100 N/m) * (0.1 m) ² = 0.5 J

Equate KE₂ to SE to obtain

0.5v² = 0.5

v = 1 m/s

Because v = 0.02u, therefore

u = 1/0.02 = 50 m/s

The loss in KE due to the collision is

KE₁ - KE₂ = 0.01 (50²) - 0.5 (1²) = 24.5 J

Answer:

The velocity of the block after the collision is 1.0 m/s

The velocity of the bullet is 50 m/s

Loss in KE due to the collision is 24.5 J