Question

A 700 kg car is moving at 10 m/s west rear ends a 500 kg car at rest. They both stuck together after the collision and continue to move. How fast are they both moving together after the collision?

Answer 1

5.83 m/s

Step-by-step explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

mu+m'u' = V(m+m')..................... Equation 1

Where m = mass of the first car, m' = mass of the second car, u = initial velocity of the first car, u' = initial velocity of the second car, V = common velocity of both cars after collision

make V the subject of the equation,

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 700 kg, m' = 500 kg, u = 10 m/s, u' = 0 m/s ( at rest)

Substitute into equation 2

V = (700×10+500×0)/(700+500)

V = 7000/1200

V = 5.83 m/s

Answer 2

`v = 5.833\,(m)/(s)`

Step-by-step explanation:

The collision is inelastic and can be described by the Principle of Momentum Conservation:

`(700\,kg)\cdot (10\,(m)/(s) ) + (500\,kg)\cdot (0\,(m)/(s) ) = (1200\,kg)\cdot v`

The speed after the collision is:

`v = 5.833\,(m)/(s)`