JEE Advanced 2013 · Paper 1 · Physics Q20

← JEE Advanced 2013 · Paper 1

PhysicsInteger+4 / -1Question 20 of 60

A bob of mass $m$ , suspended by a string of length $l_1$ , is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass $m$ suspended by a string of length $l_2$ , which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $\dfrac{l_1}{l_2}$ is

Your answer

Answer and worked solution

Official answer: 5

First bob at the top. The minimum speed to complete the circle gives $v=\sqrt{gl_1}$ at the highest point.

Collision. Equal masses, elastic: the second bob (at rest at its lowest point) takes all of this speed.

Second bob needs $\sqrt{5gl_2}$ at its lowest point: $gl_1=5gl_2$ , so $\dfrac{l_1}{l_2}=5$ .

Answer: 5

Topics:Motion in a vertical circle Elastic and inelastic collisions

← Previous question Next question →

JEE Advanced 2013 · Paper 1, Physics question 20