91Ó°ÊÓ

The height of package from bottom of cart is$h=4\text{″¾}$

The initial speed of the cart is $u_c=5\text{″¾}/\text{s}$

The initial speed of package at incline is $u_p=3\text{″¾}/\text{s}$

The mass of cart is $M=50\text{ k²µ}$

The mass of package is: $m=15\text{ k²µ}$

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.

$\begin{array}{l}{v}_p=\sqrt{{(3\text{″¾}/\text{s})}^2+2(9.8\text{″¾}/\text{s})\left(4\text{″¾}\right)}\\ v_p=9.35\text{″¾}/\text{s}\end{array}$

Therefore, the speed of the package just before landing in the cart is $9.35\text{″¾}/\text{s}$.

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.

$\begin{array}{c}\left(50\text{ k²µ}\right)(5\text{″¾}/\text{s})+\left(15\text{ k²µ}\right)(3\text{″¾}/\text{s})=\left(50\text{ k²µ}\right)v_c+\left(15\text{ k²µ}\right)(9.35\text{″¾}/\text{s})\\ v_c=3.29\text{″¾}/\text{s}\end{array}$

Therefore, the final speed of cartis $3.29\text{″¾}/\text{s}$.