A bumper car with mass m 1 = 114 kg is moving to the right with a velocity of v
ID: 2029856 • Letter: A
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.|KEelastic| = |KEinelastic|
b.|KEelastic| > |KEinelastic|
c.|KEelastic| < |KEinelastic|
Explanation / Answer
1)velocity of CoM
= [mu + mu] / [m + m]
= [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)
mv + mv = 0
total kinetic energy is conserved since it is an elastic collision:
½114(3.796)² + ½94(-4.6038)² = ½114(v')² + ½94(v')²
solving gives us v' = - u' and v' = - u'
=> v' = - 3.796 m/s and v' =4.6038 m/s
4) the final velocity of car 1 in the ground (original) reference frame:
since relative to CoM
v ' = -3.796
= v - velcoity of CoM
= v - 0.9038
v = 9038 - 3.796
= -2.8922 m/s
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