Question

A body of mass $3kg$ moving with a velocity $(2\hat{i}+3\hat{j}+3\hat{k})m/s$ collides with another body of mass $4kg$ 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

• A. $\frac{1}{7}(4\hat{i}+6\hat{j}-3\hat{k})$
• B. $\frac{1}{7}(18\hat{i}+17\hat{j}-3\hat{k})$
• C. $\frac{1}{7}(6\hat{i}+4\hat{j}-6\hat{k})$
• D. $\frac{1}{7}(9\hat{i}+8\hat{j}-6\hat{k})$

Answer 1

Answer: (B)

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

Given - $m_1=3kg,m_2=4kg,v_1=2\hat{i}+3\hat{j}+3\hat{k},v_2=3\hat{i}+2\hat{j}-3\hat{k}$

Let $v$ be the velocity of composite body .

As the bodies stick together , hence mass of composite body

$m=m_1+m_2=3+4=7kg$

Now , by law of conservation of linear momentum

total momentum before collision= total momentum after collision

$m_1{v}_1+m_2{v}_2=mv$

$3(2\hat{i}+3\hat{j}+3\hat{k})+4(3\hat{i}+2\hat{j}-3\hat{k})=7v$

or $6\hat{i}+9\hat{j}+9\hat{k}+12\hat{i}+8\hat{j}-12\hat{k}=7v$

or $18\hat{i}+17\hat{j}-3\hat{k}=7v$

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