Question

A 25.0 kg bumper car moving to the right at 5.00 m/s overtakes and collides elastically with a 35.0 kg bumper car moving to the right. After the collision, the 25.0 kg car slows to 1.50 m/a to the right, and the 35.0 pkg car moves 4.50 to the right. What was the velocity of the 35.0 pkg car before the collision?​

Answer 1

Step-by-step explanation:

To determine the velocity of the 35.0 kg car before the collision, we can use the principle of conservation of momentum. The total momentum of the system before the collision is equal to the total momentum after the collision.

Momentum before collision = Momentum after collision

m1v1 + m2v2 = (m1 + m2)vf

where:

m1 = 25.0 kg (mass of first car)

m2 = 35.0 kg (mass of second car)

v1 = velocity of first car before collision

v2 = velocity of second car before collision

vf = final velocity of the combined cars after collision

Since the first car slows down to 1.50 m/s to the right after the collision, and the second car moves 4.50 m/s to the right, we can write:

vf = (v1 + v2) / 2 = (1.50 + 4.50) / 2 = 3.00 m/s

Substituting the given values into the equation for conservation of momentum, we have:

25.0 kg x 5.00 m/s + m2 x v2 = 60.0 kg x 3.00 m/s

Solving for v2, the velocity of the second car before the collision, we get:

v2 = (60.0 kg x 3.00 m/s - 25.0 kg x 5.00 m/s) / 35.0 kg = 2.14 m/s

So the velocity of the 35.0 kg car before the collision was 2.14 m/s.