A point mass m moving with velocity v0 hits a ball of same mass m at rest elastically. After the collision, point mass moves with velocity v0/2 at an angle θ from initial direction of motion. Then value of θ is
A
300
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
450
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
600
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
900
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C600
Let v=Velocity of ball Applying conservation of linear momentum in the direction of motion, mv0=m.v02cosθ+m.vcosα ...(1)....Applying conservation of linear momentum in the direction perpendicular to the direction of motion, mVo2sin(θ)=mVsin(α)....(2) Applying conservation of kinetic energy, 12mV2o=12m(Vo2)2+12mV234V2o=V2⇒V=√32V.....(3) solving (1), (2) and (3), we get θ=600