91Ó°ÊÓ

1. Mass of the ball, m is 0.15 kg.
2. Speed of the ball before hitting the wall$v_{i}is\left(5.00\frac{m}{s}\right)\hat{i}+\left(6.50\frac{m}{s}\right)\hat{j}+\left(4.00\frac{m}{s}\right)\hat{k}$.
3. Speed of the ball after hitting the wall is, $v_i=\left(2.00\frac{m}{s}\right)\hat{i}+\left(3.50\frac{m}{s}\right)\hat{j}+\left(-3.20\frac{m}{s}\right)\hat{k}$ .

Using the formula of momentum, you can find the initial and final momentum of the ball. Then using this momentum, you can find the change in the ball’s momentum and the impulse on the ball and on the wall.

The change in the momentum can be calculated as,

$∆p=m\left(v_f-v_i\right)$

Substitute the values in the above expression, and we get,

role="math" localid="1661322378994" $$\begin{gathered} ∆p=\left(0.15kg\right)\left[\left(2.00\frac{m}{s}\right)\hat{i}+\left(3.50\frac{m}{s}\right)\hat{j}+\left(-3.20\frac{m}{s}\right)\hat{k}-\left(5.00\frac{m}{s}\right)\hat{i}+\left(6.50\frac{m}{s}\right)\hat{j}+\left(4.00\frac{m}{s}\right)\hat{k}\right] \\ =\left(0.15kg\right)\left[\left(-3.00\frac{m}{s}\right)\hat{i}+\left(-3.00\frac{m}{s}\right)\hat{j}+\left(-7.20\frac{m}{s}\right)\hat{k}\right] \\ =-\left(0.45kg·\frac{m}{s}\right)\hat{i}-\left(0.45kg·\frac{m}{s}\right)\hat{j}-\left(1.1kg·\frac{m}{s}\right)\hat{k} \end{gathered}$$

Therefore, the change in the ball’s momentum is $-\left(0.45kg·\frac{m}{s}\right)\hat{i}-\left(0.45kg·\frac{m}{s}\right)\hat{j}-\left(1.1kg·\frac{m}{s}\right)\hat{k}$

As the change in the momentum is nothing but the impulse. Therefore,

$J=∆p$

Substitute the values in the above expression, and we get,

$J=-\left(0.45kg·\frac{m}{s}\right)\hat{i}-\left(0.45kg·\frac{m}{s}\right)\hat{j}-\left(1.1kg·\frac{m}{s}\right)\hat{k}$

Therefore, the impulse on the ball is $-\left(0.45kg·\frac{m}{s}\right)\hat{i}-\left(0.45kg·\frac{m}{s}\right)\hat{j}-\left(1.1kg·\frac{m}{s}\right)\hat{k}$

According to Newton’s third law, for every action, thereis an equaland opposite reaction.

The impulse on the ball will be equal to the impulse on the wall, but the direction will be opposite. From sub part b we can say that the impulse on the wall will be$\left(0.45kg·\frac{m}{s}\right)\hat{i}+\left(0.45kg·\frac{m}{s}\right)\hat{j}+\left(1.1kg·\frac{m}{s}\right)\hat{k}$.

Therefore, impulse on the wall is $\left(0.45kg·\frac{m}{s}\right)\hat{i}+\left(0.45kg·\frac{m}{s}\right)\hat{j}+\left(1.1kg·\frac{m}{s}\right)\hat{k}$.