A particle of mass m travelling along the x-a | vedclass

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(A) According to the law of conservation of linear momentum,the initial momentum of the system must equal the final momentum of the system.Initial mom

Vedclass · Accepted Answer

$v_0 \left( \frac{3}{2} \hat{i} - \hat{j} \right)$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,the initial momentum of the system must equal the final momentum of the system.Initial momentum: $\vec{P}_i = m v_0 \hat{i}$.The particle splits into two parts: $m_1 = \frac{m}{3}$ with velocity $\vec{v}_1 = 2 v_0 \hat{j}$,and $m_2 = \frac{2m}{3}$ with unknown velocity $\vec{v}_2$.Applying conservation of momentum: $m v_0 \hat{i} = \left( \frac{m}{3} \right) (2 v_0 \hat{j}) + \left( \frac{2m}{3} \right) \vec{v}_2$.$m v_0 \hat{i} = \frac{2m v_0}{3} \hat{j} + \frac{2m}{3} \vec{v}_2$.Dividing by $m$ and rearranging: $v_0 \hat{i} - \frac{2}{3} v_0 \hat{j} = \frac{2}{3} \vec{v}_2$.Multiplying by $\frac{3}{2}$: $\vec{v}_2 = \frac{3}{2} v_0 \hat{i} - v_0 \hat{j} = \frac{v_0}{2} (3 \hat{i} - 2 \hat{j})$.Wait,checking the calculation: $\vec{v}_2 = \frac{3}{2} v_0 \hat{i} - v_0 \hat{j} = \frac{v_0}{2} (3 \hat{i} - 2 \hat{j})$.Re-evaluating the options: The expression $\frac{v_0}{2} (3 \hat{i} - 2 \hat{j})$ is equivalent to $v_0 (1.5 \hat{i} - \hat{j})$. This matches option $A$.

$A$ particle of mass $m$ travelling along the $x$-axis with speed $v_0$ shoots out $1/3$ of its mass with a speed $2v_0$ along the $y$-axis. The velocity of the remaining piece is:

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