A bumper car with mass m 1 = 114 kg is moving to the right with a velocity of v

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 = -3.6 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?

m/s

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

m/s

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

m/s

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

m/s

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

m/s

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?

Explanation / Answer

1)
Vcm = m1 v1 + m2v2 / (m1 + m2)
= (114 x 4.7 - 94 x 3.6) / 114 + 94 = 0.95 m/s

2)
Initial velocity of car 1 in the center of mass reference frame
Vi = 4.7 - 0.95 = 3.75 m/s.

3)
Final velocity will be same as the initial velocity in the center of mass reference frame
Vf = 3.75 m/s

4)
Take the final velocity of car 1 as v3
Take the final velocity of car 2 as v4
Using the conservation of momentum
m1 v1 + m2v2 = m1 v3 + m2v4 ...(1)
Using the conservation of kinetic energy,
m1 v12 + m2v22 = m1 v32 + m2v42 ...(2)
Solving (1) and (2), we get
v3 = -2.8 m/s.
5)
v4 = 5.5 m/s

6)
m1 v1 + m2v2 = v5 [m1 + m2]
(114 x 4.7 - 94 x 3.6) = (114 + 94)v5
v5 = 0.95 m/s, which is the velocity of their center of mass.

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