Question

A car of mass 1689-kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is essentially elastic. If the velocity of the truck is 17 km/h (in the same direction as the car's initial velocity) after the collision, what was the initial speed of the car?

Answer 1

Answer:

5.16 m/s

Explanation:

mass of car, m1 = 1689 kg

mass of truck, m2 = 2000 kg

Velocity of truck after collision, v2 = 17 km/h = 4.72 m/s

Let the initial velocity of car is u1.

initial velocity of truck, v1 = 0

velocity of car after collision, v1 = ?

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

1689 x u1 + 2000 x 0 = 1689 x v1 + 2000 x 4.72

1689 u1 = 1689 v1 + 9444.4      .... (1)

As the collision is elastic, so coefficient of restitution is 1.

Use the formula for the coefficient of restitution.

$e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}$

e = 1

v1 - 4.72 = 0 - u1

v1 = 4.72 - u1

Substitute the value of v1 in equation (1)

1689 u1 = 1689 (4.72 - u1) + 9444.44

1689 u1 = 7972.08 - 1689 u1 + 9444.44

3378 u1 = 17416.52

u1 = 5.16 m/s

Thus, the speed of car before collision is 5.16 m/s.

Answer 2

Answer:

The initial velocity of the car is V1= 5.58 m/s = 20.12 km/h

Explanation:

m1= 1689 kg

v1= ?

m2= 2000kg

v2= 17 km/h = 4.72 m/s

by the conservation of quantity of movement :

m1*v1 = m2*v2

v1= m2*v2 / m1

v1= 5.58 m/s = 20.12 km/h