wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

Two men jump simultaneously from one buggy to another on the same track, towards left direction with a velocity of u . Initially, both the buggies are stationary. Find the final velocities (in m/s) of the buggies after they jump. Given: masses of the men =50 kg each, mass of buggy =200 kg and relative velocity of jumping u=10 m/s w.r.t. buggy.

A
103, 209
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
503, 259
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2513, 2518
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
503, 209
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 103, 209
Given:
Mass of men, m=50 kg
Mass of buggies, M=200 kg
Velocity of man w.r.t. buggy, u=10 m/s

There is no external force is acting on the system so momentum will be conserved.
(pi=pf)

Take +ve direction leftward.
When both men jumped from the buggy. Let v be the velocity of the buggy (towards right)


0=m(uv)+m(uv)Mv
2mu=2mv+Mv
v=2mu(2m+M)=2×50×10(2×50+200) v=103 m/s

(2) When both men land on the other buggy. Let v1 be the velocity of the other buggy (left)


pi=pf
2m(uv)=(2m+M)v1
2×50(10103)=(2×50+200)×v1
v1=209 m/s

Velocity of buggy (1) is v=103 m/s & buggy (2) is v1=209 m/s

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