CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A man of mass M=50 kg stands at one end of the boat which is floating on still water. The man walks to the other end of the boat. Length of the boat is L=10 m and mass M=100 kg. Then find the distance moved by the boat relative to the ground

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

The correct option is A 103 m
Here no external force is acting on the system. Hence xcom=0.
COM will not change its position.

Assume


When man reaches the other end boat moved x distance relative to the ground.

Hence,
Distance travelled by the man w.r.t ground,
x1=(Lx)=(10x) m
Distance travelled by the boat w.r.t ground is x m
As we know,
Δxcom=m1Δx1+m2Δx2m1+m2
Δxcom=0
0=50×(10x)+(100)(x)50+100
(x because we assumed rightward movement as positive)
50050x100x=0
x=500150
x=103 m

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image