Chapter 15, Problem 1FP

The 0.5kg ball strikes the rough ground and rebounds with the velocities shown. Determine the magnitude of the impulse the ground exerts on the ball. Assume that the ball does not slip when it strikes the ground, and neglect the size of the ball and the impulse produced by the weight of the ball.

To determine

The magnitude of the impulse the ground exerts on the ball.

Answer to Problem 1FP

The magnitude of the impulse the ground exerts on the ball is $\underset{\_}{\text{12}\text{.2}\text{N}\cdot \text{s}}$.

Explanation of Solution

Given:

The mass of a ball, $m=0.5\text{kg}$.

The initial velocity of the ball, $v_1=25\text{m/s}$.

The final velocity of the ball, $v_2=10\text{m/s}$.

The angle between the initial velocity and the ground, ${\theta }_1=45°$.

The angle between the final velocity and the ground, ${\theta }_2=30°$.

Show the impulse and momentum diagram as in Figure (1).

Express the principle of impulse and momentum for a system of particles.

$m{v}_1+I=m{v}_2$ (I)

Here, the impulsive force vector is I, the initial vector velocity is $v_1$, and the final vector velocity is $v_2$.

Use Figure (1) to find the value of $v_1$.

$v_1=v_1\cos {\theta }_{1}i-v_1\sin {\theta }_{1}j$ (II)

Here, i and j are the unit vectors.

Use Figure (1) to find the value of $v_2$.

$v_2=v_2\cos {\theta }_{2}i+v_2\sin {\theta }_{2}j$ (III)

Conclusion:

Substitute 25 m/s for $v_1$ and $45°$ for ${\theta }_1$ in Equation (II).

$\begin{array}{c}{v}_1=\left(\text{25}\text{m/s}\right)\cos 45°i-\left(\text{25}\text{m/s}\right)\sin 45°j\\ =\left(25\times 0.7071\right)i-\left(25\times 0.7071\right)j\\ =17.67i-17.67j\end{array}$

Substitute 10 m/s for $v_1$ and $30°$ for ${\theta }_2$ in Equation (III).

$\begin{array}{c}{v}_2=\left(10\text{m/s}\right)\cos 30°i+\left(10\text{m/s}\right)\sin 30°j\\ =\left(10\times 0.866\right)i+\left(10\times 0.5\right)j\\ =8.66i+5.0j\end{array}$

Substitute 0.5 kg for m, $17.67i-17.67j$ for $v_1$, and $8.66i+5.0j$ for $v_2$ in Equation (I).

$\begin{array}{l}\left(0.5\text{kg}\right)\left(17.67i-17.67j\right)+I=\left(0.5\text{kg}\right)\left(8.66i+5.0j\right)\\ 8.835i-8.835j+I=4.33i+2.5j\\ I=-4.505i+11.335j\end{array}$ (IV)

Calculate the magnitude of the impulse the ground exerts on the ball using Equation (IV).

$\begin{array}{c}I=\sqrt{{\left(-4.505\right)}^2+{\left(11.335\right)}^2}\\ =\sqrt{148.77}\\ =12.2\text{N}\cdot \text{s}\end{array}$

Thus, the magnitude of the impulse the ground exerts on the ball is $\underset{\_}{\text{12}\text{.2}\text{N}\cdot \text{s}}$.