Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of $\left(20.0m/s\right)\hat{\mathbf{i}}\mathbf{-}\left(40.0m/s\right)\hat{\mathbf{j}}$. During the 3.00 ms the racket and ball are in contact, the net force on the ball is constant and equal to$\mathbf{-}\left(380N\right)\hat{\mathbf{i}}\mathbf{+}\left(110N\right)\hat{\mathbf{j}}$. What are the x- and y- components?

1. Of the impulse of the net force applied to the ball;
2. Of the final velocity of the ball?

Given that just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of $\left(20.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}-\left(4.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$. During the 3.00 ms the racket and ball are in contact, and the net force on the ball is constant and equal to $-\left(380\mathrm{N}\right)\hat{\mathrm{i}}+\left(110\mathrm{N}\right)\hat{\mathrm{j}}$.

Weight of racket, $\mathrm{w}=0.560\mathrm{N}$

Mass of racket,

$\begin{array}{rcl}\mathrm{m}& =& \raisebox{1ex}{$\mathrm{w}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{g}$}\right.\\ & =& \raisebox{1ex}{$0.560\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{$9.8\mathrm{m}/{\mathrm{s}}^2$}\right.\\ & =& 0.0571\mathrm{kg}\end{array}$

Net force, $\mathrm{F}=-\left(380\mathrm{N}\right)\hat{\mathrm{i}}+\left(110\mathrm{N}\right)\hat{\mathrm{j}}$

Initial velocity of ball, $\mathrm{u}=\left(20.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}-\left(4.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

Time, $∆\mathrm{t}=3.00\mathrm{ms}=0.003\mathrm{s}$

Impulse is product of Force applied for a change in time.

$\mathbf{J}\mathbf{=}\mathbf{F}\mathbf{∆}\mathbf{t}$

Here, J is impulse, F is the applied force, and $\mathbf{∆}\mathbf{t}$is the change in time.

The mpulse can be write as a product of mass and change in velocity.

$\mathbf{J}\mathbf{=}\mathbf{m}\left(v-u\right)$

Here, m is the mass of the object, u is the initial velocity, and v is the final velocity.

Let impulse be $\mathrm{J}={\mathrm{J}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{J}}_{\mathrm{y}}\hat{\mathrm{j}}$

Since, $\mathrm{J}=\mathrm{F}∆\mathrm{t}$

Therefore,

$$\begin{gathered} {\mathrm{J}}_{\mathrm{x}}={\mathrm{F}}_{\mathrm{x}}∆\mathrm{t} \\ {\mathrm{J}}_{\mathrm{y}}={\mathrm{F}}_{\mathrm{y}}∆\mathrm{t} \end{gathered}$$

So

Hence, x- component of the impulse is -1.14 N s and y – component is 0.33 N s.

Let final velocity be $\mathrm{v}={\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}$

Since $\mathrm{v}-\mathrm{u}=\frac{\mathrm{J}}{\mathrm{m}}$

Therefore, you can write,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\frac{{\mathrm{J}}_{\mathrm{x}}}{\mathrm{m}}+{\mathrm{u}}_{\mathrm{x}} \\ {\mathrm{v}}_{\mathrm{y}}=\frac{{\mathrm{J}}_{\mathrm{y}}}{\mathrm{m}}+{\mathrm{u}}_{\mathrm{y}} \end{gathered}$$

Substitute known values in the both above equation.

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{x}}& =& \frac{-1.14}{0.0571}+20\\ & =& 0.04\mathrm{m}/\mathrm{s}\\ {\mathrm{v}}_{\mathrm{y}}& =& \frac{0.33}{0.0571}-4\\ & =& 1.8\mathrm{m}/\mathrm{s}\end{array}$

Hence, x-component of final velocity is 0.04 m/s and y- component is 1.8 m/s.