Question

Three interacting particles of masses $100g,200g$ and $400g$ each have a velocity of $20m/s$ magnitude along the positive direction of $x-$axis, $y-$axis and $z-$axis. Due to force of interaction the third part stops moving. The velocity of the second particle is $(10\hat{j}+5\hat{k})$. What is the velocity of the first particle?

• A. $20\hat{i}+20\hat{j}+70\hat{k}$
• B. $10\hat{i}+20\hat{j}+8\hat{k}$
• C. $30\hat{i}+10\hat{j}+7\hat{k}$
• D. $15\hat{i}+5\hat{j}+60\hat{k}$

Answer 1

The correct option is A $20\hat{i}+20\hat{j}+70\hat{k}$

By conservation of momentum

$m_1{v}_1+m_2{v}_2+m_3{v}_3=finalmomentum$

Moment will conserved along all axis

And along $x-axis$

$100\times 20=100\times x$

As $3rd$ particle stops and $2nd$ have no velocity along $x-axis$

$x=20m{s}^{-1}$

Along $y-axis$

$200\times 20=200\times 10+100\times y$

$y=20m{s}^{-1}$

Along $z-axis$

$400\times 20=200\times 5+100\times z$

$z=70m{s}^{-1}$

So, velocity of particle in vector form

$20\hat{i}+20\hat{j}+70\hat{k}$

Hence option $A$ is correct.