Question

A bumper car with mass m₁ = 122 kg is moving to the right with a velocity of v₁ = 4.5 m/s. A second bumper car with mass m₂ = 2m₁ = 244 kg is at rest. The two have an elastic collision and the first bumper car rebounds backwards at a speed that is 1/3 of its original speed (1.5 m/s). Assume the surface is frictionless.

1) What is the change in momentum of bumper car 1? (let the positive direction be to the right)

2) What is the change in momentum of BOTH bumper cars combined?

3) What is the change in momentum of bumper car 2?

4) What is the final velocity of car 2?

5) What is the change in energy of bumper car 1?

6) What is the change in energy of BOTH bumper cars combined?

7) What is the change in energy of bumper car 2?

Answer 1

Given that the collision is elastic,

the momentum and energy are conserved.

Given m1=122 kg, u1=+4.5m/s, m2=244kg, u2=0;

v1=-1.5m/s v2=? ui- velocity of bumper car 'i' before collision

vi- velocity of bumper car 'i' after collision

1. Change in momentum of bumper car 1

dp1 = m1v1-m1u1 = 122(-1.5-4.5) = -732 kgm/s   

2. Since momentum conserved total momentum change combined is zero

momentum of system before collision = momentum of system after collision

3. Conserving the momentum

m1u1+m2u2=m1v1+m2v2

122(4.5)+0 = 122(-1.5)+244*v2

v2 = 122*6/244 = +3m/s

change momentum of bumper car 2 = m2v2-m2u2 = 244*3 = 732m/s (Hence 1+4 = 0)

4 Final velocity of bumper car 2 = +3m/s

5. Change in energy of bumper car 1

dKE1 = 1/2m1v1^2-1/2m1u2^2

= 122*(2.25-20.25)/2 = -1098 J

6. Since energy is conserved total energy change of system (Combined) is zero.

7. Change of energy of first bumper car + second bumper car = 0;

Hence change of energy in second bumper car = 1098 J

or

chanfe of energy in bumper car 2 = 1/2m2v2^2-1/2m2u2^2

= 244*9/2 = 1098 J

. .