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

A man of mass 40 kg is standing on a platform of 12 kg. If the man starts running at 6 m/s relative to the platform, the velocity of the platform relative to the ground is

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

The correct option is B 4.6 m/s

Mass of the platform m1=12 kg
Mass of the man m2=40 kg
Taking rightward direction as +ve,
Let velocity of platform relative to the ground be v towards left.
v1=v
Then, velocity of man relative to the ground is (6v)
v2=6v
Using momentum conservation,
0=m1v1+m2v2 (initially, both at rest)
40(6v)=12v
240=52v
v=24052
v=4.6 m/s (towards left)

flag
Suggest Corrections
thumbs-up
28
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