A body of mass 3 kg moving with a velocity ( | vedclass

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(B) Given,mass of the first body $m_1 = 3 \ kg$ and its velocity $v_1 = (2 \hat{i}+3 \hat{j}+3 \hat{k}) \ m/s$.Mass of the second body $m_2 = 4 \ kg$

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$\frac{1}{7}(18 \hat{i}+17 \hat{j}-3 \hat{k}) \ m/s$

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(B) Given,mass of the first body $m_1 = 3 \ kg$ and its velocity $v_1 = (2 \hat{i}+3 \hat{j}+3 \hat{k}) \ m/s$.Mass of the second body $m_2 = 4 \ kg$ and its velocity $v_2 = (3 \hat{i}+2 \hat{j}-3 \hat{k}) \ m/s$.According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$Substituting the values:$3(2 \hat{i}+3 \hat{j}+3 \hat{k}) + 4(3 \hat{i}+2 \hat{j}-3 \hat{k}) = (3 + 4) v$$(6 \hat{i}+9 \hat{j}+9 \hat{k}) + (12 \hat{i}+8 \hat{j}-12 \hat{k}) = 7 v$$(6+12) \hat{i} + (9+8) \hat{j} + (9-12) \hat{k} = 7 v$$18 \hat{i} + 17 \hat{j} - 3 \hat{k} = 7 v$$v = \frac{1}{7}(18 \hat{i}+17 \hat{j}-3 \hat{k}) \ m/s$.

$A$ body of mass $3 \ kg$ moving with a velocity $(2 \hat{i}+3 \hat{j}+3 \hat{k}) \ m/s$ collides with another body of mass $4 \ kg$ moving with a velocity $(3 \hat{i}+2 \hat{j}-3 \hat{k}) \ m/s$. The two bodies stick together after collision. The velocity of the composite body is

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