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

The position vector of three particles of mass m1=3 kg,m2=4 kg and m3=1 kg are r1=(2^i+^j+3^k) m, r2=(^i3^j+2^k) m and r3=(3^i2^j^k) m respectively. Find the position vector of centre of mass of the system of particles.

A
18(13^i12^j+15^k) m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
18(11^i13^j+16^k) m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
18(16^i+11^j13^k) m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
18(13^i11^j+16^k) m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 18(13^i11^j+16^k) m
The position vector of COM of the system of three particles will be given by:

rCM=m1r1+m2r2+m3r3m1+m2+m3...(i)

r1=(2^i+^j+3^k) m
r2=(^i3^j+2^k) m r3=(3^i2^j^k) m

Substituting the values in equation (i)

rCM=(3)(2^i+^j+3^k)+(4)(^i3^j+2^k)+(1)(3^i2^j^k)3+4+1

rCM=(6^i+3^j+9^k)+(4^i12^j+8^k)+(3^i2^j^k)8

rCM=18(13^i11^j+16^k) m

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Just an Average Point?
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon