A 0.550 kgball falls directly down onto concrete, hitting it with a speed of$\mathbf{12}\mathbf{.}\mathbf{0}\frac{\mathbf{m}}{\mathbf{s}}$and rebounding directly upward with a speed of$\mathbf{3}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}$. Extend a yaxis upward. In unit-vector notation, what are (a) the change in the ball’s momentum, (b) the impulse on the ball, and (c) the impulse on the concrete?

1. Mass of the ball is,$M=0.550\mathrm{kg}$
2. Speed of the ball before hitting the concrete is,$v_i=12.0\frac{\mathrm{m}}{\mathrm{s}}$
3. Speed of the ball after hitting the concrete is,$v_f=3.00\frac{\mathrm{m}}{\mathrm{s}}$

Using the formula of momentum, find the initial and final momentum of the ball. Then using this momentum, find the change in the momentum. Then, using this momentum, find the impulse on the ball and on the concrete.

Formulae are as follow:

$p=mv$

Here, m is mass, v is velocity, p is momentum and J is impulse.

Consider the formula for momentum as:

$p=mv$

As, speed of the ball before hitting the concrete is $v_i=12.0\frac{\mathrm{m}}{\mathrm{s}}$,

$$\begin{gathered} p_i=m{v}_i \\ {p}_i=\left(0.550\mathrm{kg}\right)\left(12.0\frac{\mathrm{m}}{\mathrm{s}}\right) \\ {p}_i=6.6\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

And, speed of the ball after hitting the concrete is, $v_f=3.00\frac{\mathrm{m}}{\mathrm{s}}$

$$\begin{gathered} p_f=m{v}_f \\ {p}_f=\left(0.550\mathrm{kg}\right)\left(-3.0\frac{\mathrm{m}}{\mathrm{s}}\right)) \\ {p}_f=-1.65\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

So,

role="math" localid="1661310893949" $$\begin{gathered} ∆p=p_i-p_f \\ ∆p=6.6\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}-\left(-1.65\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right) \\ ∆p=\left(8.25\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j} \end{gathered}$$

Therefore, the change in ball’s momentum is$\left(8.25\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$.

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

$$\begin{gathered} J=∆p \\ J=\left(8.25\mathrm{kg}\mathrm{m}/\mathrm{s}\right)\hat{j} \end{gathered}$$

Therefore, the impulse on the ball is $\left(8.25\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$.

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

Therefore, impulse on the concrete is $-\left(8.25\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$

The change in momentum of the ball $\left(8.25\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)j$has been found using the formula of momentum. As change in momentum is impulse, the impulse on the ball is the same.