Question

A bumper car with mass m₁ = 114 kg is moving to the right with a velocity of v₁ = 4.7 m/s. A second bumper car with mass m₂ = 94 kg is moving to the left with a velocity of v₂ = -3.7 m/s. The two cars have an elastic collision. Assume the surface is frictionless.

1)What is the velocity of the center of mass of the system?

2)What is the initial velocity of car 1 in the center-of-mass reference frame?

3)What is the final velocity of car 1 in the center-of-mass reference frame?

4)What is the final velocity of car 1 in the ground (original) reference frame?

5)What is the final velocity of car 2 in the ground (original) reference frame?

6)In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide. What is the final speed of the two bumper cars after the collision?

7)Compare the loss in energy in the two collisions:

a.|ΔKE_elastic| = |ΔKE_inelastic|

b.|ΔKE_elastic| > |ΔKE_inelastic|

c.|ΔKE_elastic| < |ΔKE_inelastic|

Answer 1

1)velocity of CoM

= [mᵢuᵢ + mᵢᵢuᵢᵢ] / [mᵢ + mᵢᵢ]

= [114(4.7) + 94(-3.7)] / 208

= 0.9038 m/s


2)the initial velocity of car 1 in the center-of-mass reference frame:

= 4.7 - 0.9038

= 3.796 m/s

3) since due to conservation of momentum, the center of mass velocity remains same after impact.

the final velocity of car 1 in the center-of-mass reference frame:

in the CoM reference frame total momentum is 0 (cause the CoM is relatively at rest)

mᵢvᵢ + mᵢᵢvᵢᵢ = 0

total kinetic energy is conserved since it is an elastic collision:

½114(3.796)² + ½94(-4.6038)² = ½114(vᵢ')² + ½94(vᵢᵢ')²

solving gives us vᵢ' = - uᵢ' and vᵢᵢ' = - uᵢᵢ'

=> vᵢ' = - 3.796 m/s and vᵢᵢ' =4.6038 m/s

4) the final velocity of car 1 in the ground (original) reference frame:

since relative to CoM

vᵢ ᵢ' = -3.796

= vᵢ - velcoity of CoM

= vᵢ - 0.9038

vᵢ = 9038 - 3.796

= -2.8922 m/s