Two identical particles of mass m carry a charge Q each. Initially, one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first particle from a large distance, with a speed V. The closest distance of approach is 
.Find the value of x
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(4)
Sol.
The particle is projected from infinity towards other particle. As the 2 nd particle gets closer to 1 st particle, force of repulsion is acting on both of them, which decreases one’s speed and increases other's speed. At minimum separation, both particles have same velocity (v 1 ). Let closest distance of approach = r;
So, by energy conservation:
O + 
mv 2 = 
mv 1 2 + 
mv 1 2 + 
⇒ 
= 
mv 2 – mv 1 2 .......
Also, by momentum conservation: mv = mv 1 + mv 1
⇒ v 1 = 
So by eq 
= 
mv 2 – 
mv 2 = 
mv 2
⇒ r = 
= 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems