Question

A 940 kg car traveling east at 12.3 m/s collides with a 670 kg car traveling north at 20.3 m/s . The cars stick together. Assume that any other unbalanced forces are negligible.

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

B) What is the speed of the wreckage just after the collision?

Answer 1

A)
Let East be +x axis

let vx and vy are components of final velocity of wreckage.

Apply conservation of momentum in x-direction

940*12.3 = (940 + 670)*vx

==> vx = 940*12.3/(940 + 670)

= 7.18 m/s

Apply conservation of momentum in y-direction

670*20.3 = (940 + 670)*vy

==> vy = 670*20.3/(940 + 670)

= 8.45 m/s

direction of motion: theta = tan^-1(vy/vx)

= tan^-1(8.45/7.18)

= 49.6 degrees North of East <<<<<<<<----------------Answer

b) final speed of wreckage, v = sqrt(vx^2 + vy^2)

= sqrt(7.18^2 + 8.45^2)

= 11.1 m/s <<<<<<<<----------------Answer