Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
At a given instant of time two particles are having the position vectors $4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+57 \hat{\mathbf{k}}$ metres and $2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ respectively. If the velocity of the first particle be $0.4 \hat{\mathbf{i}} \mathrm{ms}^{-1}$, the velocity of second particle in metre per second if they collide after $10 \mathrm{sec}$ is
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A $6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
  • B $0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
  • C $6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
  • D $0.6\left(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\frac{1}{3} \hat{\mathbf{k}}\right)$
Solution:
2271 Upvotes Verified Answer
The correct answer is: $0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
$\begin{aligned} s_1+\left(u_1\right) t & =s_2+\left(u_2\right) t \\ & =(4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+(0.4 \hat{\mathbf{i}}) 10 \\ & =(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\left(u_2\right) 10 \\ \Rightarrow \quad u_2 & =\frac{(6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{10} \\ & =0.6\left[\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right]\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.