Question

A 900-kg car traveling east at 20.0 m/s collides with a 750-kg car traveling north at 15.o m/s. The cars stick together. Assume that any otherunbalanced forces are negligible. (Draw Diagrams)

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

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

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

(d) What is the total Kinetic Energy after?

Answer 1

let m1 = 900 kg, v1 = 20.0 m/s

m2 = 750 kg, v2 = 15 m/s

let East +x axis

a) let vx and vy are the component of velocity of wreckage.

Apply conservation of momentum in x-direction

m1*v1 = (m1 + m2)*vx

==> vx = m1*v1/(m1 + m2)

= 900*20/(900 + 750)

= 10.91 m/s

Apply conservation of momentum in y-direction

m2*v2 = (m1 + m2)*vy

==> vy = m2*v2/(m1 + m2)

= 750*15/(900 + 750)

= 6.82 m/s

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

= sqrt(10.91^2 + 6.82^2)

= 12.9 m/s <<<<<<<<<----------------Answer

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

= tan^-1(6.82/10.91)

= 32.0 degrees North of East <<<<<<<<<----------------Answer

c) KE_before = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= (1/2)*900*20^2 + (1/2)*750*15^2

= 2.64*10^5 J <<<<<<<<<----------------Answer

d) KE_after = (1/2)*(m1+m2)*v^2

= (1/2)*(900 + 750)*12.9^2

= 1.37*10^5 J <<<<<<<<<----------------Answer