First slide

Gravitaional potential energy

Question

Two spherical bodies of the mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

Easy

A

2R
B

4.5R
C

7.5R
D

1.5R

Solution

Let the collision takes place at a point O. Suppose the smaller sphere moves to a distance x. Now the bigger mass moves to a distance (9R - x). Therefore
$F = G \frac{M \times 5 M}{(12 R - x)^{2}}$
Now $a_{small} = \frac{F}{M} = G \times \frac{5 M}{(12 R - x)^{2}}$
$a_{big} = \frac{F}{5 M} = G \times \frac{M}{(12 R - x)^{2}}$
$∴ x = \frac{1}{2} \cdot a_{small} \cdot t^{2} = \frac{1}{2} \cdot G \cdot \frac{M}{(12 R - x)^{2}} \cdot t^{2}$ ………….(1)
and $(9 R - x) = \frac{1}{2} \cdot a_{big} \cdot t^{2} = \frac{1}{2} \cdot G \cdot \frac{M}{(12 R - x)^{2}} \cdot t^{2}$ …….(2)
Dividing eq. (1) by eq. (2), we get
$\frac{X}{(9 R - x)} = 5 or X = 45 R - 5 X$
Solving we get, $x = 7 \cdot 5 R$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App