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

Two bodies A and B of masses 5 kg and 10 kg moving in free space in opposite directions with velocity 4 m/s and 0.5 m/s undergo head on collision. The force F of their mutual interaction varies with time t according to given graph. We can conclude

A
Period of deformation is 0.2 sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
At the time of maximum deformation speed of 5 kg is 1 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Coefficient of restitution is 0.5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Speed of 10 kg after collision is 1.75 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A Period of deformation is 0.2 sec
B At the time of maximum deformation speed of 5 kg is 1 m/s
C Coefficient of restitution is 0.5
D Speed of 10 kg after collision is 1.75 m/s
Area of F-t graph is impulse and impulse J=ΔP
At the time of maximum deformation, the velocities of the blocks are same. By applying conservation of momentum at instant of maximum deformation,

5×410×0.5=15×vv=1m/s

For 5 kg J=ΔP (For0<t<t0)

12×t0×150=5[41]t0=2×15150=0.2sec

After collision,

By conservation of momentum, 5×410×0.5=10V15V2(1)
By J=ΔP for 5 kg mass,
12×0.3×150=5(4+V2)(2)

By (1) and (2),

V2=0.5 and
V1=1.75

and e=V1+V24.5=2.254.5=0.5

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Centre of Mass in Galileo's World
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon