Determine the velocity of their center of mass before the collision?

Question

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65 103 kg and moves at a velocity of v01 = +0.79 m/s. Car 2, with a mass of m2 = 92 103 kg and a velocity of v02 = +1.2 m/s, overtakes car 1 and couples to it. Neglect the effects of friction in your answer.
(a) Determine the velocity of their center of mass before the collision
______m/s

(b) Determine the velocity of their center of mass after the collision
______m/s

(c) Should your answer in part (b) be less than, greater than, or equal to the common velocity vf of the two coupled cars after the collision?
?less than
?greater than
? equal to
Justify your answer.

Answer 1

The center of mass speed is actually the sum of the momentum of the two divided by the total mass

=(65103*0.79+92103*1.2)/(65103+92103)
=1.03 m/s

This will be the same after the collision since the conservation of momentum equation for an inelastic collision such as this is was used to calculate.

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Answer 2

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The velocity of center of mass,

Vcm=[m1v1+m2v2] /(m1+m2)

(a) m1=65*10^3 kg

m2=92*10^3 kg

v1=+0.79 m/s

v2=+1.2 m/s

Vcm=[m1v1+m2v2] /(m1+m2)

Vcm=[ 65*10^3*0.79 +92*10^3*1.2 ] / ( 65*10^3 +92*10^3*)

Vcm=161.75 / 157 =1.03 m/s
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(b)The momentum of the system remains conserved.

The coupled boxcars move with same velocity V.

(m1+m2)V=m1v1+m2v2

V = [ m1v1+m2v2] / (m1+m2)

m1=65*10^3 kg

m2=92*10^3 kg

v1=+0.79 m/s

v2=+1.2 m/s

V=[m1v1+m2v2] /(m1+m2)

V=[ 65*10^3*0.79 +92*10^3*1.2 ] / ( 65*10^3 +92*10^3*)

V=161.75 / 157 =1.03 m/s

As the two boxcars are coupled, the velocity of center of mass Vcm =V=1.03 m/s
_______________________________________________________________________________________
(c)The momentum remains conserved in perfectly INELASTIC collision

As no external force has acted,the velocity of center of mass will not change.

Vcm before collision IS EQUAL TO Vcm after the collision
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Answer 3

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