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A pair of bumper cars in an amusement park ride collide elastically as one approaches the...

A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear, as seen in part (a) of the figure below. ((a) before collision, (b) after collision) One has a mass of m1 = 441 kg and the other m2 = 562 kg, owing to differences in passenger mass. If the lighter one approaches at v1 = 4.30 m/s and the other is moving at v2 = 3.70 m/s, calculate the velocity of the lighter car after the collision. Submit Answer Tries 0/10 Calculate the velocity of the heavier car after the collision. Submit Answer Tries 0/10 Calculate the change in momentum of the lighter car. Submit Answer Tries 0/10 Calculate the change in momentum of the heavier car.

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Answer #1

From the conservation of momentum we can write as

Sum of the moments before the collision = sum of the moments after the collision

m1 v1 + m2 v2 = m1 (v1)f + m2 (v2 )f  

{ m1 = 441 kg , m2 =562 kg , v1 =4.30 m/s , v2 =3.70 m/s }

(441 kg )(4.30 m/s ) + (562 kg ) (3.70 m/s ) = (441 kg ) (v1)f + (562 kg )(v2 )f  

(441 kg ) (v1)f + (562 kg )(v2 )f = 3975.7 kg m/s

441 (v1)f + 562 (v2 )f = 3975.7 m/s ------------------(1)

We know that in eleastic head on collision the Velocity of approach equals velocity of recession.

Velocity of approach = velocity of recession.

v1 - v2 = - ( (v1)f - (v2 )f )

4.30 m/s - 3.70 m/s =   (v2 )f -  (v1)f

(v2 )f -  (v1)f  = 0.6 m/s

(v1)f - (v2 )f = -0.6 m/s -------------(2)

By solving equations (1) and (2) we get

a) (v1)f =3.628 m/s (velocity of the lighter car )

b) (v2 )f =4.228 m/s (Velocity of heavier car )

c) Change in the momentum of the lighter car is

\Delta P = ( Momentum before collision ) - (Momentum after the collision)

\Delta P= m1 v1 -  m1 (v1)f = m1 (v1 - (v1)f )

\Delta P = (441 kg ) ( 4.30 m/s - 3.628 m/s ) = 296.35 kg m/s

d) Change in the momentum of the heavier car is

\Delta P = ( Momentum before collision ) - (Momentum after the collision)

\Delta P=  m2 v2 -  m2 (v2 )f   =m2 (v2 -  (v2 )f )

\Delta P = (562 kg) (3.70 m/s - 4.228 m/s ) = - 296.74 kg m/s

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