Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
2 masses of mass 10 kg and 20 kg are kept 1 meter apart and released assuming that only mutual gravitional forces are acting . Find the speed of particles at separation decreases 0.5 meter
Decrease in gravitational P.E of system=Increase in K.E
G.10.20/0.5=1/2.10.v^2+1/2.20.u^2 (v-speed of mass 10 kg.,u-speed of mass 20 kg.)
=>400G=5v^2+10u^2..................(1)
Now only internal conservative force is acting thus velocity of C.O.M=0.
v(com)=(10v+20u)/10+20
=>v=2u(taking magnitudes only)..........(2)
solving we get,v=6*10^(-5)m/s
u=3*10^(-5)m/s (approx. values)
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !