A shell is fired from a gun with a muzzle velocity um/sec at an angle  with the horizontal. At the topof the trajectory the shell explodes into twofragments P and Q of equal mass. If the speed ofthe fragment P immediately after explosionbecomes zero, where does the centre of mass ofthe fragments hit the ground ?

To solve the problem of where the center of mass of the fragments hits the ground after the shell explodes, we need to analyze the motion of the shell and the subsequent behavior of the fragments. Let's break this down step by step.

Understanding the Initial Conditions

When the shell is fired from the gun, it follows a projectile motion path. The initial velocity of the shell is given as u m/s at an angle θ with respect to the horizontal. The horizontal and vertical components of the initial velocity can be expressed as:

• Horizontal component: u_x = u * cos(θ)
• Vertical component: u_y = u * sin(θ)

Analyzing the Trajectory

The shell reaches its maximum height when its vertical velocity becomes zero. At this point, we can calculate the time taken to reach the peak height using the formula:

t_{peak} = u_y / g

where g is the acceleration due to gravity (approximately 9.81 m/s²). The maximum height H can be calculated using:

H = u_y * t_{peak} - (1/2) * g * t_{peak}^2

Explosion and Fragment Motion

At the peak of its trajectory, the shell explodes into two fragments, P and Q, of equal mass. Given that fragment P comes to a complete stop immediately after the explosion, it will fall straight down from the height H. Fragment Q, however, will continue moving horizontally with the same horizontal velocity as the shell had at the peak, which is u_x.

Center of Mass Calculation

To find out where the center of mass of the two fragments lands, we need to consider their positions after the explosion. The center of mass (CM) of two objects can be calculated using the formula:

CM = (m_P * x_P + m_Q * x_Q) / (m_P + m_Q)

Since both fragments have equal mass, we can simplify this to:

CM = (x_P + x_Q) / 2

Now, let's denote:

• x_P = 0 (since fragment P falls straight down)
• x_Q = u_x * t_Q (where t_Q is the time taken for fragment Q to hit the ground)

Time of Flight for Fragment Q

Fragment Q will take the same time to fall to the ground as it takes for the shell to reach the ground from height H. The total time of flight T for the shell can be calculated as:

T = 2 * t_{peak}

Thus, the time taken for fragment Q to hit the ground is:

t_Q = T = 2 * (u_y / g)

Final Position of the Center of Mass

Substituting the values into the center of mass equation:

CM = (0 + u_x * t_Q) / 2

Replacing t_Q with 2 * (u_y / g) gives:

CM = (u_x * 2 * (u_y / g)) / 2 = (u_x * u_y) / g

Now substituting back the expressions for u_x and u_y:

CM = (u * cos(θ) * u * sin(θ)) / g = (u² * sin(θ) * cos(θ)) / g

Using the identity sin(2θ) = 2 * sin(θ) * cos(θ), we can further simplify this to:

CM = (u² * sin(2θ)) / (2g)

Conclusion

Therefore, the center of mass of the fragments will land at a horizontal distance of (u² * sin(2θ)) / (2g) from the point directly below where the shell exploded. This result shows how the initial velocity and the angle of projection influence the landing point of the center of mass after the explosion.