wiz-icon
MyQuestionIcon
MyQuestionIcon
16
You visited us 16 times! Enjoying our articles? Unlock Full Access!
Question

A man of mass 60 kg is standing on a platform of mass 140 kg kept on a smooth horizontal surface. The man starts moving on the platform with a velocity 2 m/s relative to the platform. Find the recoil velocity of the platform.

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

The correct option is C 0.6 m/s
Given, mass of man, (m1)=60 kg
mass of platform, (m2)=140 kg


Let the man move at a speed u towards right and the platform recoil at a speed v towards left, relative to the smooth surface.

Thus, speed of the man relative to platform is
u+v=vr
u=vrv ...(1)

Here, man and platform form a system. There is no external horizontal force on the system. So, linear momentum of the system remains constant in horizontal direction.

Initially, both man and platform were at rest. So, applying conservation of linear momentum (taking rightward +ve)
0=m2v+m1u
m2v=m1u
m2v=m1(vrv) (from (i))
v=m1vrm1+m2
v=60×260+140=0.6 m/s

So, recoil velocity of the platform is 0.6 m/s.

flag
Suggest Corrections
thumbs-up
2
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