A bumper car with mass m1 = 123 kg is moving to the right with a velocity of v1
ID: 1786873 • Letter: A
Question
A bumper car with mass m1 = 123 kg is moving to the right with a velocity of v1 = 4.8 m/s. A second bumper car with mass m2 = 2m1 = 246 kg is at rest. The two have an elastic collision and the first bumper car rebounds backwards at a speed that is 1/3 of its original speed (1.6 m/s). Assume the surface is frictionless.
1. What is the change in momentum of bumper car 1? (let the positive direction be to the right)
2. What is the change in momentum of BOTH bumper cars combined?
3. What is the change in momentum of bumper car 2?
4. What is the final velocity of car 2?
5. What is the change in energy of bumper car 1?
6. What is the change in energy of BOTH bumper cars combined?
7. What is the change in energy of bumper car 2?
Explanation / Answer
m1 = 123 kg
m2 = 246 kg
v1 = 4.8 m/s
v2 = 0
after collision
v1' = v1/3 = -1.6 m/s
v2' = ?
1)
change in momentum dP1 = m*(v1' - v1)
dP1 = 123*(-1.6-4.8)
dP1 = -787.2 kg m/s
(2)
In elastic collision total momentum remains same
momentum before collision = momentum after collision
Pbefore = Pafter
change in momentum dP = Pafter - Pbefore = 0
===============================
(3)
dP = Pafter - Pbefore = 0
(P1' + P2') - (P1 + P2) = 0
P2' - P2 = -(P1'-P1) = -dP1
dP2 = -dP1 = +787.2 kg m/s
==================
(4)
dPtotal = 0
Pafter = m1*v1' + m2*v2'
Pbefore = m1*v1 + m2*v2
m1*v1' + m2*v2' = m1*v1 + m2*v2
-123*1.6 + 246*v2' = 123*4.8 + 246*0
v2' = 3.2 m/s
==================================
5)
dE1 = (1/2)*m1*(v1'^2 - v1^2)
dE1 = (1/2)*123*(1.6^2-4.8^2)
dE1 = -1259.52 J
=================
6)
since the colliison is elastic
total energy remains same before and after collision
Eafer = Ebefore
dE = Eafter - Ebefore = 0
==================
7)
Eafter = Ebefore
E1' + E2' = E1 + E1'
E2' - E2 = E1' - E1
dE1 = dE1
dE2 = 1259.52 J
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