mm A bumper car with mass m 1 = 101 kg is moving to the right with a velocity of
ID: 1430525 • Letter: M
Question
mm
A bumper car with mass m1 = 101 kg is moving to the right with a velocity of v1 = 4.7 m/s. A second bumper car with mass m2 = 87 kg is moving to the left with a velocity of v2 = -3 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
1 )
the center of mass of the system is
CM = m1 v1 + m2 v2 / ( m1 + m2 )
= 101 X 4.7 - 87 X 3 / ( 101 + 87 )
= 474.7 - 261 / 188
CM = 1.13 m/s
2 )
the initial velocity of car 1 in the center-of-mass reference frame
= 4.7 - 1.13
= 3.57 m/s
and for the car = -3 - 1.136
= - 4 .136 m/s
3 )
the final velocity of car 1 in the center-of-mass reference frame
m1 v1 + m2 v2 = 0
the kinetic energy is conserved
0.5 X 101 X 3.572 + 0.5 X 87 X ( -4.136 )2 = 0.5 X
v1 = - 3.156 m/s
and v2 = 3.567 m/s
4 )
the final velocity of car 1 in the ground (original) reference frame is
v = 0.749 - 3.156
= - 2.407 m/s
5 )
the final velocity of car 2 in the ground (original) reference frame is
v = 4.136 + 1.13
v = 4.946 m/s
6 )
the final speed of the two bumper cars after the collision
101 X 4.7 + 87 X (-3) = 188 v
v = 101 X 4.7 + 87 X (-3 ) / 188
v = 1.13 m/s
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