Question

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the collision is 14 m/s, what will be the final speed of the coupled carts after the collision?

Answer 1

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$