Question

A 900-kg car traveling east at 20.0 m/s collides with a 750-kg car traveling north at 15.0 m/s. The cars stick together. Assume that any other unbalanced forces are negligible. (Draw Diagrams)

(a) What is the speed of the wreckage just after the collision?

(b) In what direction does the wreckage move just after the collision?

(c) What is the total Kinetic Energy before the collision?

(d) What is the total Kinetic Energy after?

Answer 1

Using conservation of momentum, we have

X : m₁ v₁ = (m₁ + m₂) v_f cos                                                            { eq.1 }

Y : m₂ v₂ = (m₁ + m₂) v_f sin                                                              { eq.2 }

dividing eq.2 by eq.1 & we get -

(m₂ v₂ / m₁ v₁) = (m₁ + m₂) v_f sin / (m₁ + m₂) v_f cos

(m₂ v₂ / m₁ v₁) = tan

tan = [(750 kg) (15 m/s)] / [(900 kg) (20 m/s)]

= tan^-1 (0.625)

= 32 degree

(a) What is the speed of the wreckage just after the collision?

From eq.1 & we get -

m₁ v₁ = (m₁ + m₂) v_f cos   

v_f = m₁ v₁ / (m₁ + m₂) cos   

v_f = [(900 kg) (20 m/s)] / [(900 kg) + (750 kg)] cos 32⁰

v_f = (18000 kg.m/s) / (1399.2 kg)

v_f = 12.8 m/s

(b) In what direction does the wreckage move just after the collision?

tan = (m₂ v₂ / m₁ v₁)

tan = [(750 kg) (15 m/s)] / [(900 kg) (20 m/s)]

= tan^-1 (0.625)

= 32 degree