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

Question

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750 kg car traveling to the right at 1.40 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.270 m/s in its original direction. You can ignore any road friction during the collision.

Part A

What was the speed of the lighter car just after the collision?

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Part B

Calculate the change in the combined kinetic energy of the two-car system during this collision.

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750 kg car traveling to the right at 1.40 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.270 m/s in its original direction. You can ignore any road friction during the collision.

Part A

What was the speed of the lighter car just after the collision?

v =   m/s  

SubmitMy AnswersGive Up

Part B

Calculate the change in the combined kinetic energy of the two-car system during this collision.

Explanation / Answer

m1 = mass of heavy car = 1750 kg

m2 = mass of light car = 1450 kg

V1i = initial speed of heavy car before collision = 1.40 m/s                   going towards right

V2i = initial speed of light car before collision = - 1.10 m/s             towards left

V1f = final speed of heavy car after collision = 0.270 m/s                   going towards right

V2f = final speed of heavy car after collision

Using conservation of momentum

m1 V1i + m2 V2i = m1 V1f + m2 V2f

1750 x 1.40 + 1450 (-1.10) = 1750 x 0.270 + 1450 V2f

V2f = 0,264 m/s    towards right

Part B)

initial kinetic energy of the system is given as

KEi = (0.5) (m1 V21i + m2 V22i )

KEi = (0.5) (1750(1.40)2 + 1450 (-1.10)2)

KEi = 2592.25 J

final kinetic energy of the system is given as

KEf = (0.5) (m1 V21f + m2 V22f )

KEf = (0.5) (1750(0.270)2 + 1450 (0.264)2)

KEf = 114.32 J

Change in KE = KEf - KEi = 114.32 - 2592.25 = - 2477.93

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