Question

A bumper car with mass m1 = 103 kg is moving to the right with a velocity of v1 = 5 m/s. A second bumper car with mass m2 = 81 kg is moving to the left with a velocity of v2 = -3.3 m/s. The two cars have an elastic collision. Assume the surface is frictionless.What is the final velocity of car 1 in the ground (original) reference frame?

Answer 1

3.65 m/s

Step-by-step explanation:

`m_1` = Mass of first car = 103 kg

`m_2` = Mass of second car = 81 kg

`u_1` = Initial Velocity of first car = 5 m/s

`u_2` = Initial Velocity of second car = -3.3 m/s

`v_c` = Velocity of center of mass

For elastic collision

`m_1u_1 + m_2u_2 =(m_1 + m_2)v_c\\\Rightarrow v_c=(m_1u_1 + m_2u_2)/(m_1 + m_2)\\\Rightarrow v=(103* 5 + 81* -3.3)/(103 + 81)\\\Rightarrow v=1.35\ m/s`

Velocity of center of mass = 1.35 m/s

`v_1=u_1-v_c\\\Rightarrow v_1=5-1.35=3.65\\\Rightarrow v_1=3.65\ m/s`

The final velocity of car 1 is 3.65 m/s

Answer 2

The final velocity of car 1 in the center-of-mass reference frame is- 2.308 m/s

Step-by-step explanation:

For elastic collisions, the relative velocity of each mass in relation to the center of mass will be equal in magnitude but opposite in direction for each mass. (relative to the center of mass)

The velocity of the center of mass = [m1v1 + m2v2] / [m1 + m2] = [103*5 + 81(-3.3)] / 184 = 1.346 m/s (to the right)

The conservation law of momentum says that the centre of mass velocity remains the same after impact.

If we assume that Right is the positive direction , then the initial velocity of car 1 in the center-of-mass reference frame is:

= 5 - 1.346 = 3.654 m/s

this velocity would appear to be in the positive direction

The final velocity of car 1 in the center-of-mass reference frame is the initial velocityo of car 1, reversed.

= - 3.654 m/s )

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

-3.654 + 1.346 = -2.308 m/s