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ID: 2179233 • Letter: H
Question
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A bumper car with mass m1 = 116.0 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 96.0 kg is moving to the left with a velocity of v2 = -3.0 m/s. The two cars have an elastic collision. Assume the surface is frictionless.
What is the velocity of the center of mass of the system?
What is the initial velocity of car 1 in the center-of-mass reference frame
What is the final velocity of car 1 in the center-of-mass reference frame
What is the final velocity of car 1 in the ground (original) reference frame?
What is the final velocity of car 2 in the ground (original) reference frame?
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?
PLEASE SHOW STEPS
Explanation / Answer
let the final velocity of car 1 be v1' towards left and of car2 be v2' towards right conserving momentum, 116*4.9-96*3=96*v2'-116*v1'.......(1) conserving kinetic energy 0.5*(116*4.9^2+96*3^2)=0.5*(116*v1'^2+96*v2'^2) solving both the equations, v1'=2.255 m/s,towards left v2'=5.644 m/s towards right a. velocity of center of mass=(m1*v1+m2*v2)/(m1+m2)=1.322 m/s towards right b. the initial velocity of car 1 in the center-of-mass reference frame =4.9-1.322=3.578 m/s towards right c. the final velocity of car 1 in the center-of-mass reference frame =2.255+1.322=3.577 m/s towards left d. the final velocity of car 1 in the ground (original) reference frame=2.255 m/s towards left e. the final velocity of car 2 in the ground (original) reference frame=5.644 m/s towards right f.In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide. let the final speed be v. then (116+96)*v=116*4.9-96*3 so v=1.322 m/s towards right
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