A bullet of mass m strikes an obstruction and deviates off at 60 ° to this original direction. If its speed…

1. $\text{m}\sqrt{{\text{u}}^2-\text{uv}+{\text{v}}^2}$
2. $\text{m}\sqrt{{\text{u}}^2+\text{uv}-{\text{v}}^2}$
3. $\text{m}\sqrt{{\text{u}}^4-\text{uv}+{\text{v}}^4}$
4. $\text{m}\sqrt{{\text{u}}^4+\text{uv}-{\text{v}}^4}$

Solution

$\text{impulse}=\text{Δ}{\text{P}}_x\widehat{\text{i}}+\text{Δ}{\text{P}}_y\widehat{\text{j}}$

$\text{Δ}{\text{P}}_x=\left({\text{mv cos 60}}^{\text{o}}-\text{mu}\right)\Rightarrow \text{Δ}{\text{P}}_{\text{X}}=(\frac{\text{mv}}{2}-\text{u})$

$\text{Δ}{\text{P}}_y=\left({\text{mv sin 60}}^{\text{o}}-0\right)\Rightarrow \text{Δ}{\text{P}}_y=\text{mv}\frac{\sqrt{3}}{2}$

$\left|\text{Δ}\overrightarrow{\text{P}}\right|=\sqrt{{\text{P}}_x^2+{\text{P}}_y^2}\Rightarrow \left|\text{Δ}\overrightarrow{\text{P}}\right|=\text{m}\sqrt{(\frac{\text{v}}{2}-\text{u}{)}^2+(\frac{\sqrt{3}\text{v}}{2}{)}^2}$

$\left|\text{Δ}\overrightarrow{\text{P}}\right|=\text{m}\sqrt{\frac{{\text{v}}^2}{4}+\text{u}{}^2-\text{uv}+\frac{3{\text{v}}^2}{4}}$

$\text{impulse}=\text{m}\sqrt{{\text{v}}^2+\text{u}{}^2-\text{uv}}$

Asked in: Practice

Practice more Center of Mass Momentum and Collision questions on Aicharya