1) 2) 3) 4) 5) 6) 7) A bumper car with mass m1 = 105 kg is moving to the right w
ID: 1374113 • Letter: 1
Question
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A bumper car with mass m1 = 105 kg is moving to the right with a velocity of v1 = 4.4 m/s. A second bumper car with mass m2 = 96 kg is moving to the left with a velocity of v2 = -3.8 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 2) What is the initial velocity of car 1 in the center-of-mass reference frame? m/s 3) What is the final velocity of car 1 in the center-of-mass reference frame? m/s 4) What is the final velocity of car 1 in the ground (original) reference frame? m/s 5) What is the final velocity of car 2 in the ground (original) reference frame? m/s 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 7) Compare the loss in energy in the two collisions:Explanation / Answer
1) velocity of center of mass
Vcm = m1*V1 + m2*V2 / (m1 + m2) = 105*4.4 - 96*3.8 / (105 + 96) = 0.4836 m/s
2) initial velocity of car 1 in center of mass frame
V1/cm = V1 - Vcm = 4.4 - 0.4836 = 3.9164 m/s
3) Since it is 'perfectly' elastic, the velocity in the reference frame would be the same magnitude, opposite direction so = -3.9164 m/s
4) & 5) let final velocities of car 1 and car 2 be V1f and V2f respectively
Conservation of momentum
m1*V1 - m2*V2 = m1*V1f + m2*V2f
97.2 = 105*V1f + 96*V2f
Conservation of kinetic energy
1/2 * m1 *V1^2 + 1/2*m2*V2^2 = 1/2 * m1 *V1f^2 + 1/2*m2*V2f^2
3419.04 = 105*V1f^2 + 96*V2f^2
solve both equations
349551.36 = 19334.44*V2f^2 - 18662.4 V2f
V1f = -3.7966 m/s
V2f = 4.762 m/s
6) Conservation of momentum
m1*V1 - m*V2 = (m1 + m2)*Vc
Vc = 0.4836 m/s
7) In elastic collison loss of kinetic energy is zero
In inelastic collision loss in KE is
1/2 * 201 * 0.4836 ^2 - 1/2 * 105*4.4 ^2 - 1/2 * 96 * 3.8^2 = 1686 J
KE elastic < KE inelastic
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