Question

A 2.45 kg brick is added onto a 14.3 kg cart that is moving with an initial horizontal speed of 5.4m/s. Find the horizontal speed of the cart and brick after the brick is in the cart.

Answer 1

Given,

The mass of the brick, m=2.45 kg

The mass of the cart, M=14.3 kg

The initial horizontal speed of the cart, u=5.4 m/s

From the law of conservation of momentum, the total momentum of a system always remains constant. That is, the total momentum of the cart and the brick before the brick was put in the cart is equal to the total momentum of the cart and the brick after that.

Thus,

`Mu=(M+m)v`

Where v is the velocity of the cart and the brick after the brick was put in the cart.

On rearranging the above equation,

`v=(Mu)/((M+m))`

On substituting the known values,

`\begin{gathered} v=(14.3*5.4)/((14.3+2.45)) \\ =4.61\text{ m/s} \end{gathered}`

Thus the horizontal speed of the cart and the brick after the brick is in the cart is 4.61 m/s