Question

A grocery shopper tosses a 6 kg bag of rice into a stationary 27 kg grocery cart. The bag hits the cart with a horizontal speed of 8.0 m/s toward the front of the cart. What is the final speed of the cart and the bag? Round to the hundredths place

Answer 1

Given,

The mass of the bag, m=6 kg

The mass of the cart, M=27 kg

The initial velocity of the cart, u=0 m/s

The initial velocity of the bag, v=8.0 m/s

From the law of conservation of momentum, the sum momentum of the bag and the momentum of the cart before the bag hits the cart must be equal to the momentum of the bag and the cart after the bag hits the cart.

Thus,

$mv+Mu=(M+m)V$

Where V is the velocity of the bag and the cart after the bag hits the cart.

On rearranging,

$V=\frac{mv+Mu}{(M+m)}$

On substituting the known values,

$\begin{gathered} V=\frac{6\times8.0+0}{(27+6)} \\ =\frac{48}{33} \\ =1.45\text{ m/s} \end{gathered}$

Thus the final speed of the cart and the bag is 1.45 m/s