Answer to Question #8949 in Mechanics | Relativity for NGk

Question

A 10000 kg truck travelling at 5.5 m/s along a straight level road rear ends a 1300 kg
passenger vehicle travelling at 2.2 m/s in the same direction. If the coefficient of restitution
between the two vehicles is 0.4 determine the velocity of each vehicle immediately after
impact.

Answer 1

This problem involves the conservation of momentum.
Momentum before collision = Momentum after collision.
And the momentum = mass X velocity.

So;
mass of truck = m1
mass of car = m2
Velocity of truck before collision = u1
Velocity of car before collision = u2
Velocity of truck after collision = v1
Velcoity of car after collision = v2

You want to find v1 and v2.
The conservation of momentum then states
Momentum before collision = momentum after collision
m1u1 + m2u2 = m1v1 + m2v2 (1)

Now, the coefficient of restitution is 0.4 and is given by

C = (v2 - v1) / (u1 - u2) (2)

Rearrange equation (1) in terms of v1 to get:

v1 = ( m1u1 + m2u2 - m2v2 ) / m1 (3)

Rearrange equation (2) in terms of v2 to get:

v2 = C( u1 - u2 ) + v1 (4)

Substitute equation (4) in to equation (3)

v1 = ( m1u1 + m2u2 - m2*C*( u1-u2 ) - v1*m2 ) / m1 (5)

Now you want to bring the v1 all to one side:

v1 = (( m1u1 + m2u2 + m2*C*( u2-u1 ) ) / m1 ) - (v1*m2 ) / m1

v1 + (v1*m2 ) / m1 = (( m1u1 + m2u2 + m2*C*( u2-u1 ) ) / m1 )

v1*(1 + (m2 / m1) ) = (( m1u1 + m2u2 + m2*C*( u2-u1 ) ) / m1 )

v1 = (( m1u1 + m2u2 + m2*C*( u2-u1 ) ) / m1 ) / (m1 + m2) (6)

So there you have an expression for v1 the final velocity of the truck and all the unknowns are given in the question.

v1 = (10000*(5.5) + 1300*(2.2) + 1300(0.4)(2.2-5.5)) / (10000 + 1300)
= 4.97 m/s

You can do a similar substitution to get v2 or you can you can just use equation (4) above, now that you know v1.
v2 = 0.4 (5.5 - 2.2) + 4.97 = 6.29 m/s.