Twobodies, A and B, collide. The velocities before the collision are ${\overrightarrow{\mathbf{v}}}_{\mathbf{A}}\mathbf{=}\left(15\stackrel{⏜}{i}+30\stackrel{⏜}{j}\right)\mathbf{m}\mathbf{/}\mathbf{s}$and ${\overrightarrow{\mathbf{v}}}_{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{10}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{5}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$ . After the collision, ${\stackrel{\mathbf{\to }\mathbf{\text{'}}}{\mathbf{v}}}_{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{20}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$ . What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?

Mass of each body is m=2.0 kg

Velocity of body A before collision is${\overrightarrow{\mathrm{v}}}_{\mathrm{A}}=\left(15\stackrel{⏜}{\mathrm{i}}+30\stackrel{⏜}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$

Velocity of body B before collision is${\overrightarrow{\mathrm{v}}}_{\mathrm{B}}=\left(-10\stackrel{⏜}{\mathrm{i}}+5\stackrel{⏜}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$

Velocity of body A before collision is ${\overrightarrow{\mathrm{v}}}_{\mathrm{A}}=\left(-5\stackrel{⏜}{\mathrm{i}}+20\stackrel{⏜}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$

We can get two final velocities of body B after collision by applying the law conservation of momentum to the given system. We can easily find the magnitudes of the velocities of bodies before and after collision. Inserting these values in the formula for K.E we can find the initial and final K.E of the system. From the difference between them, we can get the change in the total K.E of the system.

Formula:

$$\begin{gathered} P_i=P_f \\ K.E=\frac{1}{2}m{v}^2 \end{gathered}$$

Conservation of linear momentum gives,

${\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}}$

${\mathrm{m}}_{\mathrm{A}}{\overrightarrow{\mathrm{v}}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\overrightarrow{\mathrm{v}}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}\overrightarrow{\mathrm{v}}{\text{'}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\overrightarrow{\mathrm{v}}{\text{'}}_{\mathrm{B}}$

Inserting given values,

role="math" localid="1661249930011" $$\begin{gathered} 2\left(15\stackrel{⏜}{\mathrm{i}}+30\stackrel{⏜}{\mathrm{j}}\right)+2\left(-10\stackrel{⏜}{\mathrm{i}}+5\stackrel{⏜}{\mathrm{j}}\right)=2\left(-5\stackrel{⏜}{\mathrm{i}}+20\stackrel{⏜}{\mathrm{j}}\right)+\overrightarrow{{\mathrm{v}}_{\mathrm{B}}} \\ {\overrightarrow{\mathrm{v}}}_{\mathrm{B}}^{\text{'}}=\left(15\stackrel{⏜}{\mathrm{i}}+30\stackrel{⏜}{\mathrm{j}}\right)+\left(-10\stackrel{⏜}{\mathrm{i}}+5\stackrel{⏜}{\mathrm{j}}\right)-\left(-5\stackrel{⏜}{\mathrm{i}}+20\stackrel{⏜}{\mathrm{j}}\right) \\ {\overrightarrow{\mathrm{v}}}_{\mathrm{B}}^{\text{'}}=\left(10\stackrel{⏜}{\mathrm{i}}+15\stackrel{⏜}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

Final velocity of body B after collision is$(10\stackrel{⏜}{\mathrm{i}}+15\stackrel{⏜}{\mathrm{j}})\mathrm{m}/\mathrm{s}$.

Therefore, Final velocity of body B after collision is$(10\stackrel{⏜}{\mathrm{i}}+15\stackrel{⏜}{\mathrm{j}})\mathrm{m}/\mathrm{s}$

Let’s find the magnitude of velocities of bodies,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{A}}=\sqrt{{\left(15\right)}^2+{\left(30\right)}^2}=33.54\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{B}}=\sqrt{{\left(-10\right)}^2+{\left(5\right)}^2}=11.18\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{A}}^{\text{'}}=\sqrt{{\left(-5\right)}^2+{\left(20\right)}^2}=20.62\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{B}}^{\text{'}}=\sqrt{{\left(10\right)}^2+{\left(15\right)}^2}=18.03\mathrm{m}/\mathrm{s} \end{gathered}$$

Initial K.E of the system is,

$$\begin{gathered} K.E_i=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_A{v}_B^2 \\ =\frac{1}{2}\times 2{\left(33.54\right)}^2+\frac{1}{2}\times 2{\left(11.18\right)}^2 \\ =1250\mathrm{J} \\ ~1.3\times {10}^3\mathrm{J} \end{gathered}$$

$$\begin{gathered} K.E_f=\frac{1}{2}{m}_A{v}_A^{\text{'}2}+\frac{1}{2}{m}_B{v}_B^{\text{'}2} \\ =\frac{1}{2}\times 2{\left(22.62\right)}^2+\frac{1}{2}\times 2{\left(18.03\right)}^2 \\ =836.75 \\ ~8.0\times {10}^2\mathrm{J} \end{gathered}$$

Change in the K.E is,

$$\begin{gathered} K.E=K.E_f-K.E_i \\ =8.0\times {10}^2-1.3\times {10}^3 \\ =-5.0\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the Change in the total K.E of the system is$-5.0\times {10}^2\mathrm{J}$