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? v =   m/s   SubmitMy AnswersGive Up Part B Calculate the change in the combined kinetic energy of the two-car system during this collision.

Answer 1

m₁ = mass of heavy car = 1750 kg

m₂ = mass of light car = 1450 kg

V_1i = initial speed of heavy car before collision = 1.40 m/s                   going towards right

V_2i = initial speed of light car before collision = - 1.10 m/s             towards left

V_1f = final speed of heavy car after collision = 0.270 m/s                   going towards right

V_2f = final speed of heavy car after collision

Using conservation of momentum

m₁ V_1i + m₂ V_2i = m₁ V_1f + m₂ V_2f

1750 x 1.40 + 1450 (-1.10) = 1750 x 0.270 + 1450 V_2f

V_2f = 0,264 m/s    towards right

Part B)

initial kinetic energy of the system is given as

KE_i = (0.5) (m₁ V²_1i + m₂ V²_2i )

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

KE_i = 2592.25 J

final kinetic energy of the system is given as

KE_f = (0.5) (m₁ V²_1f + m₂ V²_2f )

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

KE_f = 114.32 J

Change in KE = KE_f - KE_i = 114.32 - 2592.25 = - 2477.93