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

- 23 m
- 0 m
- 53 m
- 103 m

Solution

The correct option is **D** 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=(L−x)=(10−x) m

Distance travelled by the boat w.r.t ground is x m

As we know,

Δxcom=m1Δx1+m2Δx2m1+m2

Δxcom=0

0=50×(10−x)+(100)(−x)50+100

(−x because we assumed rightward movement as positive)

⇒500−50x−100x=0

⇒x=500150

∴x=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=(L−x)=(10−x) m

Distance travelled by the boat w.r.t ground is x m

As we know,

Δxcom=m1Δx1+m2Δx2m1+m2

Δxcom=0

0=50×(10−x)+(100)(−x)50+100

(−x because we assumed rightward movement as positive)

⇒500−50x−100x=0

⇒x=500150

∴x=103 m

Suggest corrections

0 Upvotes

Similar questions

View More...

People also searched for

View More...

- About Us
- Contact Us
- Investors
- Careers
- BYJU'S in Media
- Students Stories - The Learning Tree
- Faces of BYJU'S – Life at BYJU'S
- Social Initiative - Education for All
- BYJU'S APP
- FAQ
- Support

© 2021, BYJU'S. All rights reserved.