1. Bumper cars A bumper car with mass m 1 = 119 kg is moving to the right with a
ID: 1389747 • Letter: 1
Question
1. Bumper cars
A bumper car with mass m1 = 119 kg is moving to the right with a velocity of v1 = 4.4 m/s. A second bumper car with mass m2 = 84 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? 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
Lit us take u1 =4.4m/s and u2 = -3.6 m/s so that u refers initial. And final is denoted by v.
1 and 2 represent car 1 and car 2.
Momentum lost by car 1 = momentum gained by car 2.
119(4.4-v1) = 84(v2+3.6)
v1= 1.858-0.7058v2
K.E lost by car 1 = K.E gained by car 2.
119(4.4^2-v1^2) = 84(v2^2 - 3.6 ^2) (after canceling 0.5 on both sides)
27.42-1.416v12=v22-12.96
1.416v12+v22=40.38
(4.4 + v1) = (v2- 3.6)
v1 = v2-8
From 3 and 1A
v2-8= 1.858-0.7058v2
1.882 v2= 10.461
v2 =5.77 m/s
v1 = (5.77-8) = -2.3 m/s
Momentum before collision
(m1u1 +m2u2) = (119*4.4-84*3.6) = 173.6
Momentum before collision
(m1v1 +m2v2) = (-119*2.3+ 84*5.77) =210.98
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CENTER OF MASS REFERENCE SYSTEM
V of center of mass = total momentum / (m1+m2) = 173.6/ (203)
= 0.85 which agrees with your answer.
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In the center of mass reference system u1 of car 1 is 4.4-0.85 = 3.55 m/s
Initial Momentum of car 1 is m1u1 = 119*3.55= 422.45
The initial momentum of car 2 is - 373.8
(it is because the total momentum of the system is zero in the center of mass reference system)
In the center of mass reference system u2 of car 2 is = - 373.8/84= -4.45m/s
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It can be shown that the final velocity of car 1 is -3.55 m/s opposite to the initial value of u1.
And v2 = +4.45 opposite to the initial value of u1
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