A body A of mass m=0.1 ; kg has an initial ve | vedclass

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(D) By the law of conservation of linear momentum:$m \vec{u}_A + m \vec{u}_B = m \vec{v}_A + m \vec{v}_B$Given $m = 0.1 \; kg$,$\vec{u}_A = 3 \hat{i}

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$1$

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(D) By the law of conservation of linear momentum:$m \vec{u}_A + m \vec{u}_B = m \vec{v}_A + m \vec{v}_B$Given $m = 0.1 \; kg$,$\vec{u}_A = 3 \hat{i} \; ms^{-1}$,$\vec{u}_B = 5 \hat{j} \; ms^{-1}$,and $\vec{v}_A = 4(\hat{i} + \hat{j}) \; ms^{-1}$.Substituting the values:$0.1(3 \hat{i}) + 0.1(5 \hat{j}) = 0.1(4 \hat{i} + 4 \hat{j}) + 0.1 \vec{v}_B$Dividing by $0.1$:$3 \hat{i} + 5 \hat{j} = 4 \hat{i} + 4 \hat{j} + \vec{v}_B$$\vec{v}_B = (3-4) \hat{i} + (5-4) \hat{j} = -\hat{i} + \hat{j} \; ms^{-1}$.The speed of $B$ after the collision is $|\vec{v}_B| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2} \; ms^{-1}$.The kinetic energy of $B$ after the collision is $K_B = \frac{1}{2} m |\vec{v}_B|^2$.$K_B = \frac{1}{2} (0.1) (\sqrt{2})^2 = \frac{1}{2} (0.1) (2) = 0.1 \; J$.Since $K_B = \frac{x}{10} \; J$,we have $0.1 = \frac{x}{10}$,which implies $x = 1$.

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