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Two uniform solid spheres of equal radii R,but mass M and 4M have a centre to centre separation 6R,the two spheres are held fixed on a horizontal floor.A projectile of mass m is projeced from the surface of the sphere of mass M directly towards the centre of the second sphere.Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere?
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Two uniform solid spheres of equal radii R,but mass M and 4M have a ce...
In this example, we define a neutral point (N) where the gravitational forces of the two bodies cancel each other out. Now, the position of this neutral point would depend upon the distribution of masses. It will always be placed closer to the lighter body, as shown in the figure.

The position of the point can be found by equation the gravitational forces due to the two masses (which what a neutral point represents), as done in the example. It was found out to be that r= 2R 

A body is projected from mass M to 4M it will have certain mechanical energy, which is the sum of potential and kinetic energy.

The kinetic energy of the body is simply (1/2)mv^2, where v is the velocity of the object and m is its mass.

the potential energy would of the form of Gravitational Potential energy of the form - (GMm/r).

So, the mechanical energy of the particle at the surface of mass M is the kinetic energy of the particle + gravitational potential energy due to M and 4M as well.

this comes out to be, Ei =  (1/2)mv^2 - GMm/r - G4Mm/r
similarly we can find out the mechanical energy of the particle at point N, which would be

here we will consider the distance of N from the two masses and as v = 0, KE = 1/2mv^2 = 0, here

so,
En = -GMm/2R - G4Mm/4R

Now, finally we can use the law of conservation of energy, so

Ei = En
or (1/2)mv^2 - GMm/r - G4Mm/r = -GMm/2R - G4Mm/4R

we can thus calculate v, which would be

v = (3GM/5R)^1/2

which is the minimum speed required by the particle to escape the gravitational field of M and go to 4M.
This question is part of UPSC exam. View all NEET courses
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Two uniform solid spheres of equal radii R,but mass M and 4M have a ce...
Problem Analysis:
We are given two solid spheres with equal radii R, but with masses M and 4M. The spheres are fixed on a horizontal floor and are separated by a center-to-center distance of 6R. A projectile of mass m is projected from the surface of the sphere with mass M towards the center of the second sphere. We need to find the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Solution:

Step 1: Analyzing the Forces:
When the projectile is projected from the surface of the first sphere, it experiences two types of forces:
1. Gravitational Force (mg): This force acts vertically downward due to the gravitational pull.
2. Centripetal Force (Fc): This force acts towards the center of the second sphere, as the projectile moves in a circular path.

Step 2: Equating the Forces:
To find the minimum speed v of the projectile, we need to equate the gravitational force and the centripetal force.

1. Gravitational Force (mg): The gravitational force acting on the projectile is given by mg, where m is the mass of the projectile and g is the acceleration due to gravity.
2. Centripetal Force (Fc): The centripetal force acting on the projectile is given by Fc = mv^2 / R, where m is the mass of the projectile, v is its velocity, and R is the radius of the circular path.

Setting mg equal to mv^2 / R, we have:
mg = mv^2 / R

Step 3: Solving for v:
Canceling out the mass m from both sides of the equation, we get:
g = v^2 / R

Solving for v, we have:
v^2 = gR

Taking the square root of both sides, we get:
v = sqrt(gR)

Step 4: Substituting values:
Since the radii of both spheres are equal (R), we can substitute the value of R into the equation:
v = sqrt(gR) = sqrt(g * 6R) = sqrt(6gR)

Hence, the minimum speed v of the projectile so that it reaches the surface of the second sphere is given by v = sqrt(6gR).

Conclusion:
The minimum speed v of the projectile required to reach the surface of the second sphere is given by v = sqrt(6gR), where g is the acceleration due to gravity and R is the radius of the spheres.
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Two uniform solid spheres of equal radii R,but mass M and 4M have a centre to centre separation 6R,the two spheres are held fixed on a horizontal floor.A projectile of mass m is projeced from the surface of the sphere of mass M directly towards the centre of the second sphere.Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere?
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Two uniform solid spheres of equal radii R,but mass M and 4M have a centre to centre separation 6R,the two spheres are held fixed on a horizontal floor.A projectile of mass m is projeced from the surface of the sphere of mass M directly towards the centre of the second sphere.Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two uniform solid spheres of equal radii R,but mass M and 4M have a centre to centre separation 6R,the two spheres are held fixed on a horizontal floor.A projectile of mass m is projeced from the surface of the sphere of mass M directly towards the centre of the second sphere.Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two uniform solid spheres of equal radii R,but mass M and 4M have a centre to centre separation 6R,the two spheres are held fixed on a horizontal floor.A projectile of mass m is projeced from the surface of the sphere of mass M directly towards the centre of the second sphere.Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere?.
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