In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s (Fig. P8.87). Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37° from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

The height of package from bottom of cart is$h=4\text{\hspace{0.17em}m}$

The initial speed of the cart is $u_c=5\text{\hspace{0.17em}m}/\text{s}$

The initial speed of package at incline is $u_p=3\text{\hspace{0.17em}m}/\text{s}$

The mass of cart is $M=50\text{\hspace{0.17em}kg}$

The mass of package is: $m=15\text{\hspace{0.17em}kg}$

Thespeed of the package is calculated by using third equation of motion and final speed of cart is found by momentum conservation.

The speed of the package just before landing in the cart is given as is given as:

$v_p=\sqrt{u_p^2+2gh}$

Here, $g$ is the gravitational acceleration.

Substitute all the values in the above equation.

$\begin{array}{l}{v}_p=\sqrt{{(3\text{\hspace{0.17em}m}/\text{s})}^2+2(9.8\text{\hspace{0.17em}m}/\text{s})\left(4\text{\hspace{0.17em}m}\right)}\\ v_p=9.35\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, the speed of the package just before landing in the cart is $9.35\text{\hspace{0.17em}m}/\text{s}$.

Apply the momentum conservation to find the final speed of cart.

$m{u}_c+M{u}_p=m{v}_c+M{v}_p$

Here, $v_c$is the final speed of cart.

Substitute all the values in the above equation.

$\begin{array}{c}\left(50\text{\hspace{0.17em}kg}\right)(5\text{\hspace{0.17em}m}/\text{s})+\left(15\text{\hspace{0.17em}kg}\right)(3\text{\hspace{0.17em}m}/\text{s})=\left(50\text{\hspace{0.17em}kg}\right)v_c+\left(15\text{\hspace{0.17em}kg}\right)(9.35\text{\hspace{0.17em}m}/\text{s})\\ v_c=3.29\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, the final speed of cartis $3.29\text{\hspace{0.17em}m}/\text{s}$.