A spaceship of speed v_0 travelling along +y | vedclass

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(B) Let the total mass of the spaceship be $m$. The initial momentum is $\vec{P}_i = m v_0 \hat{j}$.After the explosion,the spaceship splits into two

Vedclass · Accepted Answer

$\frac{\sqrt{20}}{3}v_0$

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(B) Let the total mass of the spaceship be $m$. The initial momentum is $\vec{P}_i = m v_0 \hat{j}$.After the explosion,the spaceship splits into two parts: $m_1 = \frac{m}{4}$ and $m_2 = \frac{3m}{4}$.The velocity of the first part is $\vec{v}_1 = 2v_0 \hat{i}$.Let the velocity of the remaining part be $\vec{v}_2 = v_x \hat{i} + v_y \hat{j}$.By the law of conservation of linear momentum,$\vec{P}_i = \vec{P}_f$,so $m v_0 \hat{j} = \frac{m}{4}(2v_0 \hat{i}) + \frac{3m}{4}(v_x \hat{i} + v_y \hat{j})$.Equating the components:Along $x$-axis: $0 = \frac{2m v_0}{4} + \frac{3m v_x}{4} \Rightarrow 3v_x = -2v_0 \Rightarrow v_x = -\frac{2}{3}v_0$.Along $y$-axis: $m v_0 = \frac{3m v_y}{4} \Rightarrow v_y = \frac{4}{3}v_0$.The magnitude of the velocity of the remaining part is $v_2 = \sqrt{v_x^2 + v_y^2} = \sqrt{(-\frac{2}{3}v_0)^2 + (\frac{4}{3}v_0)^2} = \sqrt{\frac{4}{9}v_0^2 + \frac{16}{9}v_0^2} = \sqrt{\frac{20}{9}}v_0 = \frac{\sqrt{20}}{3}v_0$.

$A$ spaceship of speed $v_0$ travelling along $+y$ axis suddenly shoots out one-fourth of its part with speed $2v_0$ along $+x$ axis. The $xy$ axes are fixed with respect to the ground. The velocity of the remaining part is:

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