a sphere of mass "m" and radius "R" , held between a wedge of same mass "m" and a rigid wall , is released from rest.

initially it is kept such that the centre of sphere is at a height of "2R" from ground

find the speed of both the bodies as sphere hits the ground, assuming all surfaces frictionless

To solve the problem of a sphere of mass "m" and radius "R" released from a height of "2R" between a wedge of the same mass "m" and a rigid wall, we need to analyze the motion of both the sphere and the wedge. Since all surfaces are frictionless, we can apply the principles of conservation of energy and momentum to find the speeds of both bodies when the sphere hits the ground.

Understanding the System

Initially, the sphere is at rest at a height of "2R." When it is released, it will fall under the influence of gravity, while the wedge will also move due to the reaction force exerted by the sphere. The key here is to recognize that the wedge will move horizontally as the sphere falls vertically.

Applying Conservation of Energy

As the sphere falls, it converts its gravitational potential energy into kinetic energy. The potential energy (PE) of the sphere at height "2R" is given by:

• PE = mgh = mg(2R) = 2mgR

When the sphere reaches the ground, all this potential energy will have converted into kinetic energy (KE). The kinetic energy of the sphere and the wedge can be expressed as:

• KE_sphere = (1/2)mv_s^2
• KE_wedge = (1/2)mv_w^2

Thus, the total kinetic energy when the sphere hits the ground is:

• Total KE = KE_sphere + KE_wedge = (1/2)mv_s^2 + (1/2)mv_w^2

Conservation of Momentum

Since there are no external horizontal forces acting on the system, we can apply the conservation of momentum in the horizontal direction. Initially, the total momentum is zero because both the sphere and the wedge are at rest. When the sphere falls, the wedge moves in the opposite direction to conserve momentum:

• 0 = mv_s - mv_w

This simplifies to:

• v_s = v_w

Combining the Equations

Now we can substitute \( v_w \) in terms of \( v_s \) into the kinetic energy equation. Since \( v_w = v_s \), we can write:

• Total KE = (1/2)mv_s^2 + (1/2)m(v_s)^2 = mv_s^2

Setting the total kinetic energy equal to the initial potential energy gives us:

• mv_s^2 = 2mgR

Dividing both sides by "m" (assuming m is not zero) leads to:

• v_s^2 = 2gR

Taking the square root of both sides, we find the speed of the sphere just before it hits the ground:

• v_s = √(2gR)

Finding the Speed of the Wedge

Since we established that \( v_w = v_s \), the speed of the wedge when the sphere hits the ground is also:

• v_w = √(2gR)

Final Results

In summary, both the sphere and the wedge will have the same speed when the sphere reaches the ground:

• Speed of the sphere, \( v_s = √(2gR) \)
• Speed of the wedge, \( v_w = √(2gR) \)

This elegant result arises from the conservation of energy and momentum principles, demonstrating the interconnected motion of the sphere and the wedge in a frictionless environment.