A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1

Question

A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 90 kg is moving to the left with a velocity of v2 = -3.2 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?

Explanation / Answer

1)
Velocity of the center of mass, Vcm = [m1v1 + m2v2] / (m1 + m2)
= [102 x 4.9 + 90 x (-3.2)] / (102 + 90]
= 211.8 / 192
= 1.1 m/s

2)
Initial velocity of car1 = 4.9 m/s
Initial velocity of the car1 with respect to the C.O.M = 4.9 - 1.1
= 3.8 m/s

In elastic collision, both momentum and kinetic energy is conserved.
Initial momentum = m1v1 + m2v2
= 211.8 kg.m/s [Calculated in earlier part]
Consider the new velocity of the car with mass m1 as v3 and car with mass m2 as v4
Final momentum = m1 v3 + m2 v4
= 102 v3 + 90 v4
Equating both momentum,
211.8 = 102 v3 + 90 v4
90 v4 = 211.8 - 102 v3
v4 = (211.8 - 102 v3) / 90       ...(1)

Initial kinetic energy = 1/2 m1(v1)2 + 1/2 m2 (v2)2
= 0.5 x 102 x (4.9)2 + 0.5 x 90 x (-3.2)2
= 1224.51 + 460.8
= 1685.31 J
Final kinetic energy = 1/2 m1(v3)2 + 1/2 m2 (v4)2.
Equating both energies,
1685.31 = 0.5 x 102 (v3)2 + 0.5 x 90 (v4)2.
Substituting equation (1),
1685.31 = 0.5 x 102 (v3)2 + 0.5 x 90 [(211.8 - 102 v3) / 90]2.
1685.31 = 51 (v3)2 + 45 x [(2.35 - 1.13 v3)]2.
1685.31 = 51 (v3)2 + 45 x [(5.538 - 5.334 v3 + 1.284 (v3)2.]
1685.31 = 51 (v3)2 + 45 x [(5.538 - 5.334 v3 + 1.284 (v3)2.]
1685.31 = 108.8 (v3)2 - 240.04 v3 + 249.22
108.8 (v3)2 - 240.04 v3 -1436.092
This is a quadratic equation, having solution
v3 = - 2.69 m/s
Substituting v3 in equation (1)
v4 = [211.8 - 102 x (-2.69)] / 90
v4 = 5.41 m/s

It can be seen that the center of mass velocity, Vcm = [m1v3 + m2v4] / (m1 + m2) = 1.1 remains the same as before.

3)
Final velocity of the car with respect ot C.O.M frame
= - 2.69 - 1.1
= -3.79 m/s

4)
Final velocity of the car 1 in the ground frame = - 2.69 m/s

5)
Final velocity of the car2 in the ground frame = 5.41 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.