CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two masses m and 2m are placed in fixed horizontal circular smooth hollow tube of radius r as shown. The mass m is moving with speed u and the mass 2m is stationary. After their first collision, the time elapsed for next collision. (coefficient of restitution e = 1/2)
1079930_c644429635c0402296e7e762cc941579.png

A
2πru
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4πru
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3πru
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12πru
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2πru

Let the velocity of the particle of mass m, after the collision is u1 and the velocity of the particle of mass 2m is u2.

From conservation of momentum, it can be written as,

mu=mu1+2mu2

u=u1+2u2

Since, the collision is elastic, the kinetic energy is conserved in the collision and it can be written as,

12mu2=12mu12+122mu22

u2=u12+2u22

u2u12=2u22

(uu1)(u+u1)=2u22

(u+u1)=u2

u=u2u1

The time taken is given as,

t=2πru2u1

t=2πru

Thus, the time elapsed for the next collision is 2πru.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
A Sticky Situation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon