Chapter 9: Q98P (page 254)

A 0.15kgball hits a wall with a velocity of $(5.00 \frac{m}{s}) \hat{i} + (6.50 m / s) \hat{j} + (4.00 m / s) \hat{k}$ . It rebounds from the wall with a velocity of $(2.00 \frac{m}{s}) \hat{i} + (3.50 m / s) \hat{j} + (- 3.20 m / s) \hat{k}$ . What are (a) the change in the ball’s momentum, (b) The impulse on the ball, and (c) the impulse on the wall?

Short Answer

Expert verified

The change in ball’s momentum is $- (0.45 k g . \frac{m}{s}) \hat{i} - (0.45 k g . \frac{m}{s}) \hat{j} - (1.1 k g . \frac{m}{s}) \hat{k}$ .
The impulse on the ball is $- (0.45 k g . \frac{m}{s}) \hat{i} - (0.45 k g . \frac{m}{s}) \hat{j} - (1.1 k g . \frac{m}{s}) \hat{k}$ .
The impulse on the concrete is $(0.45 k g . \frac{m}{s}) \hat{i} + (0.45 k g . \frac{m}{s}) \hat{j} + (1.1 k g . \frac{m}{s}) \hat{k}$ .

Step by step solution

Understanding the given information

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

Concept and formula used in the given question

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.

(a) Calculation for the change in the ball’s momentum

The change in the momentum can be calculated as,

$∆ p = m (v_{f} - v_{i})$

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

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

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

(b) Calculation for the impulse on the ball

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

$J = ∆ p$

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

$J = - (0.45 k g \cdot \frac{m}{s}) \hat{i} - (0.45 k g \cdot \frac{m}{s}) \hat{j} - (1.1 k g \cdot \frac{m}{s}) \hat{k}$

Therefore, the impulse on the ball is $- (0.45 k g \cdot \frac{m}{s}) \hat{i} - (0.45 k g \cdot \frac{m}{s}) \hat{j} - (1.1 k g \cdot \frac{m}{s}) \hat{k}$

(c) Calculation for the impulse on the wall

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 $(0.45 k g \cdot \frac{m}{s}) \hat{i} + (0.45 k g \cdot \frac{m}{s}) \hat{j} + (1.1 k g \cdot \frac{m}{s}) \hat{k}$ .

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