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

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be:

A
0.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.25
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 0.25
Coefficient of Restitution (e) -

e = v2v1u2u1

Ratio of relative velocity after collision to relative velocity before collision.

Let velocity of 4m mass is v after collision

then mv = 4mv

v = v4v

e=velocity of separationvelocity of approach=v4v

=14

e=0.25



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