**Answer:**

**Given:**

Radius of the sphere, R

Mass of the sphere, m

The sphere strikes the wall elastically

**To find:**

Magnitude of angular momentum of the sphere just after the strike about the point of contact on the ground

**Solution:**

**Step 1:**

Let the velocity of the center of mass of the sphere be V and its angular velocity be ω just before the collision.

Since the sphere is rolling, the velocity of its point of contact with the ground will be zero.

Hence, V = Rω

**Step 2:**

Let the sphere rebound elastically. Since the wall is vertical, the direction of the velocity of the center of mass will be reversed. Hence the velocity after the collision will be -V.

**Step 3:**

The point of contact on the ground will have a velocity of -2V.

**Step 4:**

The angular velocity of the sphere just after the collision can be found using conservation of angular momentum.

The angular momentum of the sphere just before the collision about the point of contact on the ground is given by:

L = Iω

where I is the moment of inertia of the sphere about an axis passing through its center of mass and perpendicular to the plane of motion.

For a hollow sphere, I = 2/3mr^2

Hence, L = 2/3mr^2ω

**Step 5:**

Just after the collision, the angular momentum of the sphere about the point of contact on the ground is given by:

L' = Iω'

where ω' is the angular velocity of the sphere just after the collision.

Since the sphere rebounds elastically, the kinetic energy of the sphere is conserved. Hence,

1/2mv^2 + 1/2Iω^2 = 1/2mv'^2 + 1/2Iω'^2

Substituting I = 2/3mr^2 and v' = -2V, we get:

1/2mV^2 + 1/2(2/3mr^2)ω^2 = 1/2m(2V)^2 + 1/2(2/3mr^2)ω'^2

Simplifying, we get:

ω' = ω(3/2)

**Step 6:**

Substituting the value of ω' in the expression for angular momentum, we get:

L' = 2/3mr^2(ω(3/2))

Simplifying, we get:

L' = 2/3mr^2(3/2)V/R

Hence, the magnitude of angular momentum of the sphere just after the strike about the point of contact on the ground is given by:

|L'| = (2/3)mVR

Therefore, the magnitude of angular momentum of the sphere just after the strike about the point of contact on the ground is (2/3)mVR.