A bumper car with mass m 1 = 118 kg is moving to the right with a velocity of v
ID: 1339985 • Letter: A
Question
A bumper car with mass m1 = 118 kg is moving to the right with a velocity of v1 = 4.3 m/s. A second bumper car with mass m2 = 82 kg is moving to the left with a velocity of v2 = -3.4 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?
m/s
Your submissions:
2)
What is the initial velocity of car 1 in the center-of-mass reference frame?
m/s
Your submissions:
3)
What is the final velocity of car 1 in the center-of-mass reference frame?
m/s
Your submissions:
4)
What is the final velocity of car 1 in the ground (original) reference frame?
m/s
Your submissions:
5)
What is the final velocity of car 2 in the ground (original) reference frame?
m/s
Your submissions:
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?
m/s
Explanation / Answer
1)
vcom=(m1u1 + m2u2)/(m1+m2)=(118*4.3 - 82*3.4)/200 =1.143m/s
2)initial velocity of car 1 in the center-of-mass reference frame is v1-vcom= 4.3-1.143=3.157m/s
3) by linear momentum consevation
intial momentum = final momentum
118*4.3 - 82*3.4=118v1+82v2 ................(1)
due to elastic collision
v2 - v1 =u1 - u2 = 4.3+3.4=7.7 ..........(2)
from equation (1) and (2).......
v1= -2.014 m/s and v2=5.686 m/s
final velocity of car 1 in the center-of-mass reference frame = v1-vcom= -2.014 -1.143 = - 3.157
4) final velocity of car 1 in the ground (original) reference frame is v1= -2.014 m/s
5) final velocity of car 2 in the ground (original) reference frame is v2=5.686 m/s
6) new (inelastic) collision
by linear momentum consevation
intial momentum = final momentum
118*4.3 - 82*3.4=118v1+82v2 ................(1)
and
due to inelastic collision
v2 - v1 = 0...........(2)
from equation (1) and (2).......
v2 = v1 = 1.143 m/s
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