A09 and A10. In these questions you will use the \"center of mass frame\", that

Question

A09 and A10. In these questions you will use the "center of mass frame", that is, a reference frame moving at the velocity of the center of mass of the colliding objects, to solve the problem. You must understand that momentum of the center of mass remains unchanged in the collision

Two spheres of equal radius and masses 5.00 kg and 20.0 kg are moving towards each other along the same straight line along across a frictionless, horizontal surface as illustrated below. Vi1 = +10.0m/s and Vi2 = ?5.00m/s respectively. During the collision, 25% of the total initial kinetic energy is lost.

A09 (i)  Find the velocity of the center of mass of this system.
(ii) What are the velocities of each sphere in the center of mass reference frame (i.e., from the point of view of someone moving with the velocity of the center of mass)?
(iii) What is total initial momentum in the center of mass reference frame? (You should find that this is zero!)

A10 (i) What are the initial and final total KEs in the center of mass frame?
(ii) Find the final velocities of the two objects in the center of mass reference frame.
(iii) Now find the final velocities of the two objects in the laboratory frame

Explanation / Answer

A09)
i)
Vi1 = 10 m/s, Vi2 = -5 m/s
m1 = 5 kg/ m2 = 20 kg
Velocity of the center of mass, Vcm = (m1 Vi1 + m2Vi2) / (m1 + m2)
Vcm = (5 x 10 - 20 x 5)/(5 + 20)
= -2 m/s

ii)
To find the relative velocity, we must subtract the Vcm from the initial velocities
For mass m1, V1 = Vi1 - Vcm = 10 - (-2) = 12 m/s
For mass m2, V2 = Vi2 - Vcm = - 5 - (-2) = -3 m/s.

iii)
Total initial momentum
= m1V1 + m2V2
= 5 x 12 - 20 x 3 = 0

A10)
i)
In the center of mass frame
V1 = 12 m/s and V2 = -3 m/s
Initial KE = 1/2 m1V12 + 1/2 m2V22
= 0.5 [ 5 x 122 + 20 x 32]
= 450 J

ii)
Take V1f and V2f be the velocities of m1 and m2 after collision in the center of mass frame
Total KE = 75% of the initial KE (25% is lost)
1/2 (m1V1f2 + m2 V2f2) = 0.75 x 450
5 V1f2 + 20V2f2 = 675

From conservation of momentum, final momentum in center of mass reference frame must be zero.
m1V1f + m2V2f = 0
Vif = -4V2f

Substituting,
5 x 16 V2f2 +  20V2f2 = 675
V2f = 2.6 m/s
V1f = - 4 x 2.6 = -10.4 m/s

iii)
To find this, we need to add Vcm with V1f and V2f
For mass m1, V = -10.4 + (-2) = -12.4 m/s
For mass m2, V = 2.6 - 2 = 0.6 m/s

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