A particle of mass m moving with velocity (3h | vedclass

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(B) The impulse $J$ received by a body is equal to the change in its linear momentum,given by $J = \Delta p = m(v_f - v_i)$.For the particle of mass $

Vedclass · Accepted Answer

impulse received by $m$ is $m[-5\hat{i} - \hat{j}]$

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(B) The impulse $J$ received by a body is equal to the change in its linear momentum,given by $J = \Delta p = m(v_f - v_i)$.For the particle of mass $m$:Initial velocity $v_i = (3\hat{i} + 2\hat{j}) \ m/s$.Final velocity $v_f = (-2\hat{i} + \hat{j}) \ m/s$.Impulse received by $m = m(v_f - v_i) = m[(-2\hat{i} + \hat{j}) - (3\hat{i} + 2\hat{j})] = m[-5\hat{i} - \hat{j}]$.According to Newton's third law,the impulse received by mass $M$ is equal and opposite to the impulse received by mass $m$:Impulse received by $M = -(Impulse \ received \ by \ m) = -m[-5\hat{i} - \hat{j}] = m[5\hat{i} + \hat{j}]$.Thus,the correct statement is that the impulse received by $m$ is $m[-5\hat{i} - \hat{j}]$.

$A$ particle of mass $m$ moving with velocity $(3\hat{i} + 2\hat{j}) \ m/s$ collides with another body of mass $M$ and finally moves with velocity $(-2\hat{i} + \hat{j}) \ m/s$. Then during the collision:

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