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

give solutin

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 as the sphere falls. Since all surfaces are frictionless, we can apply the principles of conservation of momentum and energy 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" above the ground. When it is released, it will fall under the influence of gravity. As the sphere falls, it will also exert a force on the wedge, causing the wedge to move horizontally. We can break this problem down into two main parts: the motion of the sphere and the motion of the wedge.

Applying Conservation of Energy

First, let's consider the potential energy of the sphere at the initial height. The potential energy (PE) when the sphere is at height "2R" is given by:

• PE_initial = m * g * (2R)

As the sphere falls to the ground, this potential energy will convert into kinetic energy (KE). The kinetic energy of the sphere when it hits the ground can be expressed as:

• KE_sphere = (1/2) * m * v_s^2

Here, "v_s" is the speed of the sphere just before it hits the ground. At the moment the sphere hits the ground, all the potential energy will have converted into kinetic energy:

• m * g * (2R) = (1/2) * m * v_s^2

We can simplify this equation by canceling "m" from both sides:

• g * (2R) = (1/2) * v_s^2

Multiplying both sides by 2 gives:

• 2g * (2R) = v_s^2

Thus, we find:

• v_s^2 = 4gR

Taking the square root, we have:

• v_s = 2√(gR)

Analyzing the Wedge's Motion

Now, we need to determine the speed of the wedge as the sphere falls. Since the system is frictionless, we can use the conservation of momentum in the horizontal direction. Initially, both the sphere and the wedge are at rest, so the total momentum is zero. When the sphere falls, it moves downward, and the wedge moves horizontally. Let "v_w" be the speed of the wedge when the sphere hits the ground.

Using conservation of momentum:

• m * v_s = m * v_w

From this, we can simplify to:

• v_s = v_w

Substituting the expression we found for "v_s":

• v_w = 2√(gR)

Final Speeds

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

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

This result illustrates the interplay between gravitational potential energy and kinetic energy, as well as the conservation of momentum in a frictionless system. Both bodies move with the same speed due to their equal masses and the nature of their interaction during the fall of the sphere.