A ball of mass 10, kg moving with a velocity | vedclass

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(C) Let the mass of the first ball be $m_1 = 10 \, kg$ and its initial velocity be $\vec{u}_1 = 10 \sqrt{3} \hat{i} \, m/s$. The second ball has mass

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Let the mass of the first ball be $m_1 = 10 \, kg$ and its initial velocity be $\vec{u}_1 = 10 \sqrt{3} \hat{i} \, m/s$. The second ball has mass $m_2 = 20 \, kg$ and is at rest $(\vec{u}_2 = 0)$.After the collision,the first ball comes to rest. The second ball disintegrates into two equal pieces of mass $m_p = 10 \, kg$ each.Let the velocity of the first piece be $\vec{v}_1 = 10 \hat{j} \, m/s$.Let the velocity of the second piece be $\vec{v}_2 = v_x \hat{i} - v_y \hat{j}$ (based on the figure,it moves at $30^{\circ}$ below the $x$-axis).By the law of conservation of linear momentum:$m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_{1,final} + m_p \vec{v}_1 + m_p \vec{v}_2$$10(10 \sqrt{3} \hat{i}) + 0 = 0 + 10(10 \hat{j}) + 10 \vec{v}_2$$100 \sqrt{3} \hat{i} = 100 \hat{j} + 10 \vec{v}_2$$10 \vec{v}_2 = 100 \sqrt{3} \hat{i} - 100 \hat{j}$$\vec{v}_2 = 10 \sqrt{3} \hat{i} - 10 \hat{j}$The magnitude of velocity $\vec{v}_2$ is $x = |\vec{v}_2| = \sqrt{(10 \sqrt{3})^2 + (-10)^2} = \sqrt{300 + 100} = \sqrt{400} = 20 \, m/s$.

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