Step-by-step explanation:
Momentum is conserved:
Momentum before firing = momentum after firing
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Before the bullet is fired, both the bullet and gun have zero velocity, so u₁ and u₂ are 0.
0 = m₁v₁ + m₂v₂
0 = (1 kg) v₁ + (0.05 kg) v₂
0 = 20v₁ + v₂
After being fired, the kinetic energy of the system is 1050 J:
KE = ½m₁v₁² + ½m₂v₂²
1050 J = ½ (1 kg) v₁² + ½ (0.05 kg) v₂²
42000 = 20v₁² + v₂²
Two equations, two variables. Solve with substitution:
v₂ = -20v₁
42000 = 20v₁² + (-20v₁)²
42000 = 20v₁² + 400v₁²
42000 = 420 v₁²
v₁ = 10 m/s
v₂ = -200 m/s
The velocity of the gun is 10 m/s, and the velocity of the bullet is -200 m/s.