1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A man of mass 75 kg is standing on a platform of mass 15 kg kept on a smooth horizontal surface. The man starts moving on the platform with a velocity of 20 m/s relative to the platform. Then, magnitude of the recoil velocity of the platform is

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

## The correct option is D 16.66 m/sTaking the platform and the man as a system, Fext=0 on the system in horizontal direction. ∴ The linear momentum of the system remains constant and system was at rest intitially. Pi=Pf=0 ...(i) Mass of man (m1)=75 kg Mass of platform (m2)=15 kg Given, velocity of man is 20 m/s be w.r.t platform (along +ve x direction) Let the recoil velocity of platform be −v m/s (along −ve x axis). ∴ Velocity of man w.r.t ground is (20−v) m/s From Eq (i): Pf=0 75(20−v)+15(−v)=0 90v=1500 ∴v=150090=16.66 m/s

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program