wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in Fig. Ball 1 moves with a velocity v­1 towards balls 2 and 3. All collisions are assumed to be elastic. If M <m, the number of collisions between the balls will be

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

The correct option is B Two
The first collision will be between balls 1 and 2. Since both have the same mass, after the collision ball 1 will come to rest and ball 2 will move with speed v1. This ball will collide with the stationary ball 3. After this second collision, let v2 and v3 be the speeds of balls 2 and 3 respectively. Since the collisions are elastic,v2 and v3 are given by (see Sec. 9)
v2=(mMm+M)v1 (i)
and v3=(2mm+M)v1 (ii)
If M < m, it follows from (i) and (ii) that v2 < v3 and both have the same direction.
Therefore, ball 2 cannot collide with ball 3 again. Hence there are only two collisions.
Thus the correct choice is (b).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
A No-Loss Collision
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon