Problem 65 - A Complete Resource Book in Physics for JEE Main

Question

Two spherical bodies of mass \(M\) and \(5 M\) and radii \(R\) and \(2 R\) 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 (A) \(2.5 R\) (B) \(4.5 R\) (C) \(7.5 R\) (D) \(1.5 R\)

Step-by-Step Solution

Answer

The distance covered by the smaller body just before the collision is \(1.5R\). Therefore, the correct answer is (D) \(1.5 R\).

Step 1: Setup the initial conditions

The initial separation between the centers of the two bodies is given as 12R. The mass of the smaller body is M and the mass of the larger body is 5M. The radii of the bodies are R and 2R, respectively.

Step 2: Apply the conservation of linear momentum

The principle of conservation of linear momentum states that the total momentum of an isolated system (i.e., the two bodies) remains constant. Since the bodies attract each other due to gravitational force and there is no external force acting on the system, the total linear momentum is conserved. Let's denote the distances traveled by the smaller and larger bodies as \(x\) and \(y\), respectively. Then, the conservation of linear momentum can be expressed as: \[M\frac{dx}{dt} = 5M\frac{dy}{dt}\]

Step 3: Apply the formula for gravitational force

According to Newton's law of universal gravitation, the force between two masses is given by: \[F = G\frac{Mm}{r^2}\] where F is the gravitational force, G is the gravitational constant, M and m are the masses of bodies and r is the distance between their centers. In this case, m = 5M and r = 12R initially. The force equation becomes: \[F = G\frac{M(5M)}{(12R)^2}\]

Step 4: Setup a relation for relative motion

We know that force is equal to mass times acceleration. Since the smaller body has mass M, its acceleration due to the gravitational force can be expressed as: \[a = \frac{F}{M}\] Using the gravitational force expression obtained in step 3: \[a = \frac{G(5M)}{(12R)^2}\] and substituting the value of 'a' in the linear momentum conservation equation obtained in step 2, we get: \[\frac{dx}{dt} = 5\frac{dy}{dt}\] \[dx = 5dy\]

Step 5: Determine the distance covered by the smaller body

When the two bodies collide, their centers will be separated by a distance equal to the sum of their radii, which is R + 2R = 3R. So, we can express the sum of the distances covered by both bodies (\(x\) and \(y\)) as: \[x + y = 12R - 3R\] Now, since we know that \(dx = 5dy\), we can write the distance covered by the smaller body as \(x = 5y\). Substituting this expression back into the equation for the sum of the distances gives us: \[5y + y = 9R\] \[6y = 9R\] \[y = 1.5R\] So, the distance covered by the smaller body just before the collision is \(1.5R\). Therefore, the correct answer is (D) \(1.5 R\).