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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 :
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Verified Answer
The correct answer is:
8
$\mathbf{r}_1=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m}$
$\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{m}$
$\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ 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
$\therefore \quad 2=\frac{8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{(a-4) \hat{\mathbf{j}}+3 \hat{\mathbf{j}}} \quad[\because t=2 \mathrm{~s}]$
$(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}=4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
$a-4=4$
$a=8$
$\mathbf{r}_2=(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mathrm{m}$
$\mathbf{v}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ 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
$\therefore \quad 2=\frac{8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{(a-4) \hat{\mathbf{j}}+3 \hat{\mathbf{j}}} \quad[\because t=2 \mathrm{~s}]$
$(a-4) \hat{\mathbf{i}}+4 \hat{\mathbf{j}}=4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
$a-4=4$
$a=8$
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