Question

An Escort and a Camaro traveling at right angles collide and stick together. The Escort has a mass of 1300 kg and a speed of 25 km/h in the positive x direction before the collision. The Camaro has a mass of 1700 kg and was traveling in the positive y direction. After the collision, the two move off at an angle of 55 ^o to the x axis.

What was the speed of the Camaro?(in km/h)

Please Show Work.

Answer 1

Escort
p1i = m1*v1 i + m1*v1 j = 1200*25 i + 1200*0 j = 30000 i + 0 j

Camero
p2i = m2*v2 i + m2*v2 j = 1700*0 i + 1700*v2 j = 0 i + 1700*v2 j

Conservation of Momentum
p1i + p2i = p1f + p2f

momentum in the x direction
m1*v1i + m2*v2i = m1*v1f +m2*v2f

Since they stick together v1f = v2f =vfi and v2i = 0
m1*v1i = (m1+m2)vfi

vfi = (m1*v1i)/(m1+m2)

momentum in the y direction
m1*v1j + m2*v2j = m1*v1f +m2*v2f

Again, since they stick together v1f = v2f =vf and v1j = 0
m2*v2j = (m1+m2)*vfj

vfj =m2*v2j / (m1+m2)

vfj = 1700 x 13.38/(1700 + 1300) = 5.68 km/h

Now, since you have the angle, you can find the velocity of the camero, v2j

tan(56) = o/a = y/x = vfj/vfi = ((m1*v1i)/(m1+m2))/(m2*v2j / (m1+m2))
tan(56) = (m1*v1i)/(m2*v2j)

Solve for v2j

v2j = (m1*v1i)/(m2*tan(56))

v2j = (1300*25) / (1700*tan(55))

V2j = 13.39 km/h