Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:12

1 Answers

rohit rathi
47 Points
10 years ago

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)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free