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

Find the velocity of separation of the spheres 1 and 2 if the sphere 3 comes down with a velocity of 3 m/s. Assume all spheres to be identical and frictionless.


A
23 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
43 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
63 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
83 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 63 m/s
The spheres are identical, the centre to centre distance between the spheres will be equal and thus, they will form an equilateral triangle.

So, the value of θ will be 30 in the above figure shown.

By condition of wedge constraint, acceleration along normal to the common interface should be same.
So, V3cosθ=V2sinθ
V2=V3cotθ=3cot30
θ is 30
V2=33
Also, due to similarity V1 will be 33
Now, the spheres 1 and 2 will move apart in the opposite direction
velocity of separation will be
V12=V1(V2)=33+33=63 m/s

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Problem Solving Using Newton's Laws
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon