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

Two equal spheres of mass m are in contact on a smooth horizontal table. A third identical sphere impinges symmetrically on them and is reduced to rest. Find the loss of K.E


A

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B


Let u = velocity of sphere A before impact. As the sphere are identical, the triangle ABC formed by joining their centres is equilateral. The spheres B & C will move in direction AB and AC after impact making an angle of 30∘ with the original lines of motion of ball A.

Let v = speed of the other balls after impact

Momentum Conservation:

mu = mv cos 30∘ + mv cos 30∘

u = v√3 ........(i)

for an oblique collision, we have to take components along normal i.e., along AB for balls A and B

vB−vA = −e(uB−uA) ⇒ v−0 = −e(0−u cos 30∘)

v = eucos 30∘

combining (i) and (ii), we get e = 23.

Loss in KE = 12mu2−2(12mv2) = 12mu2−m(u√3)2 = 16mu2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Cut Shots in Carrom
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon