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