Question

A body of mass $2kg$ moving with a velocity $\left(\hat{i}+2\hat{j}-3\hat{k}\right)m/s$ collides with another body of mass $3kg$ moving with a velocity $\left(2\hat{i}+\hat{j}+\hat{k}\right)m/s$ . If they stick together, the velocity in $m/s$ of the composite body is

• A. $\frac{1}{5}\left(8\hat{i}+7\hat{j}-3\hat{k}\right)$
• B. $\frac{1}{5}\left(-4\hat{i}+\hat{j}-3\hat{k}\right)$
• C. $\frac{1}{5}\left(5\hat{i}+\hat{j}-\hat{k}\right)$
• D. $\frac{1}{5}\left(-4\hat{i}+8\hat{j}-3\hat{k}\right)$

Answer 1

Answer: (A)

We know that $p=mv$
So, $p_1=2(\hat{i}+2\hat{j}-3\hat{k})kgm/s$
$=(2\hat{i}+4\hat{j}-6\hat{k})kgm/s$
and $p_2=3(2\hat{i}+\hat{j}+\hat{k})m/s$
$=(6\hat{i}+3\hat{j}+3\hat{k})kgm/s$
Resultant momentum is
$p=p_1+p_2$
$=(2\hat{i}+4\hat{j}-6\hat{k})+(6\hat{i}+3\hat{j}+3\hat{k})$
$=(8\hat{i}+7\hat{j}-3\hat{k})kgm/s$
Total mass $(m)=2+3=5kg$
Therefore, the velocity of body is
$v=\frac{p}{m}=\frac{1}{5}(8\hat{i}+7\hat{j}-3\hat{k})m/s$
$=\frac{1}{5}(8\hat{i}+7\hat{j}-3\hat{k})m/s$