Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

Two small particle of equal masses start moving in opposite direction from A in a horizontal circular path of radius r Their tangential velocity at A are v and 2v, respectively as shown. Where will be the first collision in the path of the circle.

a) at distance $\pi \;r$  anticlockwise from A

b) at distance $\large\frac{2\pi}{3} $$r$  anticlockwise from A

c) at distance $\large\frac{\pi}{2} $$r$  clockwise from A

d) at distance $\large\frac{\pi \;r}{2}$  anticlockwise from A

Can you answer this question?

1 Answer

0 votes
Let the two make the first collision in time t,
distance travelled by particle 1 with velocity 'v' is $vt$
and distance travelled by particle 2 with velocity $2v=2vt$
$vt + 2vt=2 \pi r$
$\therefore t=\large\frac{2 \pi r}{3v}$
Distance travelled by particle with velocity $v =v \times \large\frac{2 \pi r}{3 v}$
$\qquad=\large\frac{2 \pi r}{3}$
So the first collision will take place at distance $\large\frac{2 \pi r}{3}$ in anticlockwise from A
Hence b is the correct answer.


answered Aug 8, 2013 by meena.p
edited Feb 17, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App