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 1750kg car traveling to the right at 1.60m/s collides with a 1450kg car going to the left at 1.10m/s . Measurements show that the heavier car's speed just after the collision was 0.260m/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.

Answer 1

Step-by-step explanation:

We shall apply law of conservation of momentum .

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ , m₁ ,m₂ are masses of two bodies colliding with velocities u₁ and u₂ respectively . v₁ and v₂ are their velocities after collision.

m₁ = 1750 , m₂ = 1450 , u₁ = 1.6 m /s , u₂ = - 1.1 m /s , v₁ = .26 m /s

substituting the values

1750 x 1.6 + 1450 x - 1.1 = 1750 x .26 + 1450 v₂

2800 - 1595 = 455 + 1450v₂

1450v₂ = 750

v₂ = .517 m /s

B ) initial kinetic energy

= 1/2 x 1750 x 1.6²+ 1/2 x 1450 x -1.1²

= 2240+ 877.25

= 3117.25 J

final kinetic energy

= 1/2 x 1750 x .26²+ 1/2 x 1450 x .517²

= 59.15 + 193.78

= 252.93

loss of kinetic energy

= 3117.25 - 252.93

= 2864.32 J