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

An object of m kg with a speed of v m/s strikes a wall at an angle θ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

A
2mvcosθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2mvsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2mv
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2mvcosθ

Initial momentum P1=mvsinθ^imvcosθ^j
and final momentum P2=mvsinθ^i+mvcosθ^j
So change in momentum
ΔP=P2P1=2mvcosθ^j
|ΔP|=2mvcosθ

flag
Suggest Corrections
thumbs-up
22
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