Two putty balls of equal mass m moving with e | vedclass

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(B) Let the two balls have mass $m$ and move along the $x$ and $y$ axes respectively.Initial velocity of ball $1$: $\vec{v}_1 = 45\sqrt{2} \hat{i} \,

Vedclass · Accepted Answer

$45$

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(B) Let the two balls have mass $m$ and move along the $x$ and $y$ axes respectively.Initial velocity of ball $1$: $\vec{v}_1 = 45\sqrt{2} \hat{i} \, m/s$.Initial velocity of ball $2$: $\vec{v}_2 = 45\sqrt{2} \hat{j} \, m/s$.Initial total momentum: $\vec{P}_i = m\vec{v}_1 + m\vec{v}_2 = m(45\sqrt{2} \hat{i} + 45\sqrt{2} \hat{j})$.The magnitude of initial momentum is: $|\vec{P}_i| = m \sqrt{(45\sqrt{2})^2 + (45\sqrt{2})^2} = m \sqrt{4050 + 4050} = m \sqrt{8100} = 90m \, kg \cdot m/s$.After collision,the balls stick together to form a mass $2m$ moving with velocity $\vec{V}$.By the law of conservation of linear momentum: $\vec{P}_i = \vec{P}_f$.$90m = (2m)V$.$V = \frac{90m}{2m} = 45 \, m/s$.

Two putty balls of equal mass $m$ moving with equal velocity $v = 45\sqrt{2} \, m/s$ in mutually perpendicular directions,stick together after collision. The velocity of their combined mass after collision is .................. $m/s$.

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