Problem: A bomb of mass m is projected from the ground with speed v at an angle theta with the horizontal. At the maximum height from the ground, it explodes into two fragments of equal mass. If one fragment comes to rest immediately after the explosion, the horizontal range of the center of mass is:

Solution:

Concepts:
- Conservation of momentum
- Conservation of energy
- Center of mass

Explanation:

When the bomb explodes into two fragments of equal mass, each fragment will have a mass of m/2. One of the fragments comes to rest immediately after the explosion, which means that its final velocity is zero.

Step 1: Calculate the initial velocity of the bomb before the explosion.

Let's assume that the initial velocity of the bomb is v0 and the angle with the horizontal is θ. The horizontal and vertical components of the initial velocity can be calculated as:

vx = v0 cos θ
vy = v0 sin θ

Step 2: Calculate the velocity of each fragment after the explosion.

Since the fragments have equal mass, the conservation of momentum can be used to calculate their final velocities:

m(v0 cos θ) = (m/2)v1 + (m/2)v2
0 = (m/2)v1 - (m/2)v2

Solving these equations, we get:

v1 = v0 cos θ
v2 = -v0 cos θ

We know that one of the fragments comes to rest, which means that v2 = 0.

Step 3: Calculate the horizontal range of the center of mass.

The horizontal range of the center of mass can be calculated using the conservation of energy. The potential energy at the maximum height is equal to the kinetic energy at the horizontal range.

At the maximum height, the kinetic energy of the bomb is:

K1 = (1/2)m(v0 sin θ)^2

After the explosion, the kinetic energy of the fragments is:

K2 = (1/2)(m/2)(v0 cos θ)^2 + (1/2)(m/2)(0)^2 = (1/4)m(v0 cos θ)^2

Since the total kinetic energy is conserved, we have:

K1 = K2
(1/2)m(v0 sin θ)^2 = (1/4)m(v0 cos θ)^2
v0^2 = 4(v0 sin θ)^2/(cos θ)^2

The horizontal range of the center of mass can be calculated using the time of flight:

t = 2(v0 sin θ)/g
R = (v0 cos θ)t = (2v0^2 sin θ cos θ)/g = (2v0^2 sin 2θ)/g

Substituting v0^2 = 4(v0 sin θ)^2/(cos θ)^2, we get:

R = (8v0^2 sin θ cos θ)/g = (4v0^2 sin 2θ)/g

Therefore, the horizontal range of the center of mass is (4v0^2 sin 2θ)/g.

Conclusion:

The horizontal