When are equipped with flexible bumpers, they will bounce off each other during

Question

When are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1815-kg car traveling to the right 1.51 m/s at collides with a 1365-kg car going to the left at 1.06 m/s. show that the heavier cars speed just after the collision was 0.210 m/s in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.

Explanation / Answer

a)Before collision:

Mass of the car moving to the right (m1) = 1815 Kg

Velocity of the car moving to the right (u1) = 1.51 m/s

Mass of the car moving to the left (m2) = 1365 Kg

Velocity of the car moving to the right (u2) =-1.06 m/s

Aftercollision:

Mass of the car moving to the right (m1) = 1815 Kg

Velocity of the car moving to the right (v1) = 0.210 m/s

Mass of the car moving to the left (m2) = 1365 Kg

Velocity of the car moving to the right (v2) =?

According to law of conservation of momentum, m1 u1+ m2 u2 = m1 v1+ m2 v2

From this equation, v2 = m1 u1+ m2 u2- m1 v1/m2

(1815)( 1.51)+(1365)(-1.06)- (1815)( 0.210) /1365 = 0.66 m/s

b) Change in Kinetic energy due to the combined system

K.E. before collision = ½ m1u12+ ½ m2u22

                                                     = ½ x 1815 x (1.51)2 + ½ x 1365 x (-1.06)2

= 2069.19+766.857 =2836.048 J

K.E. after collision = ½ m1v12+ ½ m2v2

= ½ x 1815 x (0.210)2 + ½ x 1365 x (0.66)2

= 297.297 J

So, change in Kinetic energy =2836.048 – 297.297 = 2538.75 J

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