Q152 / 160 · AP EAMCET 2003 · Physics

Two particles having position vectors $\mathbf{r}_1=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})$ metres and $\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})$ metres are moving with velocities $\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ and $\mathbf{v}_2=(a \hat{\mathbf{i}}+7 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. If they collide after $2 \mathrm{~s}$, then the value of $a$ is :

  • A
    2
  • B
    4
  • C
    6
  • Correct answer
    D
    8

Solution

$\mathbf{r}_1=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m}$ $\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{m}$ $\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ and $\mathbf{v}_2=(a \hat{\mathbf{i}}+7 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ Time taken in collision, $t=\frac{r}{v}$ ...(i) where, $r=$ resultant position vector $=\mathbf{r}_1-\mathbf{r}_2$ $=8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}$ $v=$ resultant velocity $\mathbf{v}_2-\mathbf{v}_1=(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ From Eqs. (i), we get $\therefore \quad 2=\frac{8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{(a-4) \hat{\mathbf{j}}+3 \hat{\mathbf{j}}} \quad[\because t=2 \mathrm{~s}]$ $(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}=4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ $a-4=4$ $a=8$

Hence, the correct answer is option: 'D'.

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