Please so work! A bumper car with mass m1 = 104 kg is moving to the right with a
ID: 1474931 • Letter: P
Question
Please so work! A bumper car with mass m1 = 104 kg is moving to the right with a velocity of v1 = 4.2 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.
I found that the velocity of the center of mass of the system to be .496
1) What is the final velocity of car 1 in the center-of-mass reference frame?
2) What is the final velocity of car 1 in the ground (original) reference frame?
3) What is the final velocity of car 2 in the ground (original) reference frame?
4) 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
mass m1 = 104 kg
velocity v1 = 4.2 m/s
mass m2 = 94 kg
velocity v2 = -3.6 m/s
velocity of Center of Mass = [m1u1 + m2u2] / [m1+ m2] = [116(4) + 98(-3.1)] / 214 = 0.496 m/s
2) since due to conservation of momentum, the centre of mass velocity remains same after impact.
the initial velocity of car 1 in the center-of-mass reference frame:
vf= 4.2 - 0.496 = 3.7 m/s
(and of car2 = -3.6- 0.496 = - 4.096 m/s )
3) the final velocity of car 1 in the center-of-mass reference frame:
in the Center of Mass reference frame total momentum is 0 (cause the Center of Mass is relatively at rest)
m1v1 + m2v2 = 0
total kinetic energy is conserved since it is an elastic collision:
(1/2)*104*(3.7)2 + (1/2)*94*(-4.02)2 = (1/2)*104*(v1)2 + (1/2)*94(v2)2
solving gives us v1 = - u1 and v2' = - u2
=> v1 = -3.7m/s and v2 = 4.02 m/s
iv) the final velocity of car 1 in the ground (original) reference frame:
since relative to Center of Mass v1 = -3.7 =v1 - velcoity of Center of Mass = v1 - 0.496
=> v1 = 0.496 - 3.7 = -3.204 m/s
v) the final velocity of car 2 in the ground (original) reference frame:
since relative to Center of Mass v2 = 4.02 = v2 - velcoity of Center of Mass = v2 - 0.496
=> v2 = 0.496 + 4.02 = 4.516 m/s
6) in a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.
the final speed of the two bumper cars after the collision:
equation is: 104*(4.2) + 94(-3.6) = 98.4 v
=> v = 98.4 / 202 = 0.319 m/s = velocity of Center of Mass since its velocity is unchanged when momentum is conserved.