Abullet with a mass of 50g is shot out from a gun having a mass of 1kg. if the kinetic energy at the moment when the bullet was shot is 1050j, find the velocity of the bullet when it was shot out and the velocity of the gun when it gets the opposite reaction.
(answer - 200 & 10)
pls show all the steps

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.

answer: electrons