Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

The fig shows the position and velocities of two particles.If the particles move under the mutual attraction of each other, then the position of centre of mass at t=1s is

qsnImg
Asked by m.nilu 17th September 2018, 7:34 PM
Answered by Expert
Answer:
Initial position X of Centre of Mass (CM) is given by,  X = begin mathsize 12px style fraction numerator m subscript 1 X subscript 1 space end subscript plus space m subscript 2 X subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction space equals space fraction numerator 1 cross times 2 space plus space 1 cross times 8 over denominator 1 plus 1 end fraction space equals space 5 space m end style
velocity V of CM is given by, V = begin mathsize 12px style fraction numerator m subscript 1 cross times v subscript 1 space plus space m subscript 2 cross times v subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction space equals space fraction numerator 1 cross times 5 space minus space 1 cross times 3 over denominator 1 plus 1 end fraction space equals space 1 space m divided by s end style
hence position of CM after 1 second  = 5+1×1 = 6 m
Answered by Expert 18th September 2018, 3:07 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp