a 2.0 kg mass moving to the east at a speed of 4.0 m/s collides head-on in a perfectly inelastic collision with a stationary 2.0 kg mass. how much kinetic energy is lost during

12J

Explanation:

Kinetic Energy before collision = 1/2mv1^2 = 1/2*2*4^2 = 16J

Velocity after collision (v2) = m1v1/m1+m2 = 2*4/2+2 = 8/4 = 2m/s

Kinetic Energy after collision = 1/2mv2^2 = 1/2*2*2^2 = 4J

Kinetic Energy lost = 16J - 4J = 12J

Lost in kinetic energy = 12 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V (m+m') ... Equation 1

Where m = mass of first body, u = initial of the first body, m' = mass of the second body, u' = initial velocity of the second body, V = common velocity.

Making V the subject of the equation,

V = mu+m'u' / (m+m') ... Equation 2

Where m = 2.0 kg, m' = 2.0 kg, u = 4.0 m/s, u' = 0 m/s (stationary).

Substitute into equation 2

V = (2*4 + 2*0) / (2+2)

V = 8/4

V = 2 m/s.

Total kinetic energy before collision = 1/2mu² = 1/2 (2) (2) ² = 16 J.

Total Kinetic energy after collision = 1/2V² (m+m') = 1/2 (2²) (4) = 4 J.

Thus

Lost in kinetic energy = 16 - 4

Lost in kinetic energy = 12 J