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

A man of mass M=50 kg stands at one end of a boat (of length L) which is floating in still water. The man walks to the other end of the boat. Length of the boat is L=10 m and mass M=20 kg. Then the distance moved by the boat relative to the ground is

A
507 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7.5 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
257m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12.57m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 507 m
Here, no external force is acting on the system. Hence Xcom will not change its position.



Assume that when the man reaches the other end, boat moves x distance left, relative to the ground.

Hence, distance travelled by the man with respect to the ground, x1=(Lx)=(10x) m
Distance travelled by the boat with respect to the ground (x2)=x m

As we know, Xcom=m1x1+m2x2m1+m2
Here Xcom=0
0=50×(10x)+(20)(x)50+20
50050x20x=0
70x=500
x=507 m towards left

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon