A train car with mass = 609 kg is moving to the right with a speed of = 7.7 m/s and collides with a second train car. The two cars latch together during the collision and then move off to the right at = 4.6 m/s.
a. What is the initial momentum of the first train car?
b. What is the mass of the second train car?
c. What is the change in kinetic energy of the two train system during the collision?

a. 4689.3 kgm/s b 410.41 kg c. -7268.45 J

Explanation:

a. The initial momentum of the first train car is given by p₁ = m₁v₁ where m₁ = 609 kg and v₁ = 7.7 m/s. So p₁ = m₁v₁ = 609 kg × 7.7 m/s = 4689.3 kgm/s

b. From the law of conservation of momentum,

m₁v₁ + m₂v₂ = (m₁ + m₂)v where m₂ = mass of second train car, v₂ = speed of second train car = 0 m/s (since it is stationary relative to the first train car )and v = velocity during collision = 4.6 m/s

m₁v₁ + m₂v₂ = (m₁ + m₂)v

609 kg × 7.7 m/s + m₂ × 0 = (609 + m₂)4.6

4689.3 + 0 = 2801.4 + 4.6m₂

4689.3 - 2801.4 = 4.6m₂

1887.9 = 4.6m₂

m₂ = 1887.9/4.6 = 410.41 kg

c. The kinetic energy change of the two train system ΔK = K₂ - K₁ where K₂ = final kinetic energy of two train system and K₁ = initial kinetic energy of two train system.

K₁ = 1/2m₁v₁² + 1/2m₂v₂² = 1/2 × 609 × 7.7² + 1/2 × 410.41 × 0² = 18053.81 J + 0 J = 18053.81 J

K₂ = 1/2(m₁ + m₂)v² = 1/2 (609 + 410.41)4.6² =1/2 × 1019.41 × 4.6² = 10785.36 J.

So, ΔK = K₂ - K₁ = 10785.36 J - 18053.81 J = -7268.45 J

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg