A 0.30 kgsoftball has a velocity of 15 m/sat an angle of$\mathbf{35}\mathbf{°}$below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) Velocity of 20 m/s , vertically downward, and (b) Velocity of 20 m/s, horizontally back toward the pitcher?

The mass of a softball, m = 0.30 kg .

The initial velocity of a softball, ${\mathrm{v}}_1=15\mathrm{m}/\mathrm{s}$.

The initial velocity is directed at an angle $35°$below the horizontal.

The final velocity is, ${\mathrm{v}}_{\mathrm{f}}=20\mathrm{m}/\mathrm{s}$.

We need to consider the vector nature of velocity in this case. So, we calculate the components of velocity before using the equation of momentum. The magnitude of change in momentum can be calculated using the equation of magnitude of the vector from its components.

$$\begin{gathered} \mathrm{P}=\mathrm{m}\times \mathrm{v} \\ \mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_{\mathrm{f}}-{\mathrm{v}}_{\mathrm{i}}\right) \end{gathered}$$

As velocity is a vector quantity, we consider the components of velocity.

So, the initial velocity components are:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{ix}}={\mathrm{v}}_{\mathrm{i}}\cos \left(-35°\right) \\ =\left(15\mathrm{m}/\mathrm{s}\right)\mathrm{xcos}\left(35°\right) \\ =12.29\mathrm{m}./\mathrm{s} \\ {\mathrm{v}}_{\mathrm{iy}}={\mathrm{v}}_{\mathrm{i}}\sin \left(-35°\right) \\ =-\left(15\mathrm{m}/\mathrm{s}\right)\times \sin \left(35°\right) \\ =-8.60\mathrm{m}/\mathrm{s} \end{gathered}$$

The final velocity is vertically downward, so the components of final velocity are,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{fx}}={\mathrm{v}}_{\mathrm{f}}\cos \left(-90°\right) \\ =\left(20\mathrm{m}/\mathrm{s}\right)\times \cos \left(90°\right) \\ =0\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{fy}}={\mathrm{v}}_{\mathrm{f}}\sin \left(-90°\right) \\ =-\left(20\mathrm{m}/\mathrm{s}\right)\times \sin \left(90°\right) \\ =-20\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the change in momentum along x is

$$\begin{gathered} {\mathrm{P}}_{\mathrm{x}}\mathrm{m}\times \left({\mathrm{v}}_{\mathrm{fx}}-{\mathrm{v}}_{\mathrm{ix}}\right) \\ {\mathrm{P}}_{\mathrm{x}}=0.30\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-12.29\mathrm{m}/\mathrm{s}\right) \\ =-3.69\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Similarly, the change in momentum along y is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{y}}=\mathrm{m}\times \left({\mathrm{v}}_{\mathrm{fy}}-{\mathrm{v}}_{\mathrm{iy}}\right) \\ {\mathrm{P}}_{\mathrm{y}}=0.30\mathrm{kg}\times \left(-20\mathrm{m}/\mathrm{s}-\left(-8.60\mathrm{m}/\mathrm{s}\right)\right) \\ =-3.42\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the magnitude of change in momentum can be calculated using the equation of magnitude of a vector. So, we get

$$\begin{gathered} \mathrm{P}=\sqrt{{\mathrm{P}}_{\mathrm{x}}^2+{\mathrm{P}}_{\mathrm{y}}^2} \\ =\sqrt{{\left(-3.69\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2+{\left(-3.42\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2} \\ =5.0\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The final velocity is horizontally back toward the pitcher, so the components of the final velocity are,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{fx}}={\mathrm{v}}_{\mathrm{f}}\cos \left(180°\right) \\ =\left(20\mathrm{m}/\mathrm{s}\right)\times -1 \\ =-20\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{fx}}={\mathrm{v}}_{\mathrm{f}}\sin \left(180°\right) \\ =\left(20\mathrm{m}/\mathrm{s}\right)\times 0 \\ =0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the change in momentum along x is

role="math" localid="1661492302946" $$\begin{gathered} {\mathrm{P}}_{\mathrm{x}}=\mathrm{m}\times \left({\mathrm{v}}_{\mathrm{fx}}={\mathrm{v}}_{\mathrm{ix}}\right) \\ {\mathrm{P}}_{\mathrm{x}}=0.30\mathrm{kg}\times \left(-20\mathrm{m}/\mathrm{s}-12.29\mathrm{m}/\mathrm{s}\right) \\ =-9.7\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Similarly, the change in momentum along y is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{x}}=\mathrm{m}\times \left({\mathrm{v}}_{\mathrm{fy}}={\mathrm{v}}_{\mathrm{iy}}\right) \\ {\mathrm{P}}_{\mathrm{x}}=0.30\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-(-8.60\mathrm{m}/\mathrm{s}\right) \\ =2.58\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the magnitude of change in momentum can be calculated using the equation of magnitude of a vector. So, we get

$$\begin{gathered} \mathrm{P}=\sqrt{{\mathrm{P}}_{\mathrm{x}}^2+{\mathrm{P}}_{\mathrm{y}}^2} \\ =\sqrt{{\left(-97\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2+\left(2.58\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)} \\ =10\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$