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# Answer to Question #8949 in Mechanics | Relativity for NGk

Question #8949
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.
Expert's answer
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.

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