Horizontal Distance of Second Fragment of Split Projectile from Origin

To find the horizontal distance $x$ from the origin $O$ where the second fragment lands, we analyze the motion in two phases: prior to the explosion at the peak, and after the explosion.

Phase 1: Motion before the explosion

The projectile is launched from the origin $O(0,0)$ with:

• Initial speed: $u = 5\sqrt{2} \text{ m/s}$
• Angle with the vertical: $\theta_v = 45^\circ \implies$ Angle with the horizontal: $\theta = 45^\circ$
• Acceleration due to gravity: $g = 10 \text{ m/s}^2$

The initial velocity components are: $u_x = u \cos 45^\circ = 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5 \text{ m/s}$ $u_y = u \sin 45^\circ = 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5 \text{ m/s}$

The time taken $t_H$ to reach the highest point (peak of the trajectory) is: $t_H = \frac{u_y}{g} = \frac{5}{10} = 0.5 \text{ s}$

At the peak, the position coordinates $(x_H, H)$ are: $x_H = u_x t_H = 5 \times 0.5 = 2.5 \text{ m}$ $H = u_y t_H - \frac{1}{2} g t_H^2 = 5(0.5) - \frac{1}{2}(10)(0.5)^2 = 2.5 - 1.25 = 1.25 \text{ m}$

Just before the split, the velocity vector of the projectile of mass $M$ is purely horizontal: $\vec{v}_0 = 5\hat{i} \text{ m/s}$

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Phase 2: The explosion and motion after the split

At the peak, the projectile breaks into two equal masses $m_1 = m_2 = \frac{M}{2}$.

Finding the velocity of the first fragment:

The question states that the first part falls vertically downwards, which means its horizontal component of velocity is zero ($v_{1x} = 0$).

Let its vertical velocity immediately after the split be $v_{1y}$ (taking downward as positive). It strikes the ground from height $H = 1.25 \text{ m}$ in time $t_1 = 0.5 \text{ s}$: $H = v_{1y} t_1 + \frac{1}{2} g t_1^2$ $1.25 = v_{1y}(0.5) + \frac{1}{2}(10)(0.5)^2$ $1.25 = 0.5 v_{1y} + 1.25 \implies v_{1y} = 0 \text{ m/s}$

Thus, the velocity of the first fragment immediately after splitting is: $\vec{v}_1 = \vec{0} \text{ m/s}$

Finding the velocity of the second fragment:

By conservation of linear momentum at the instant of splitting: $M \vec{v}_0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$ $M (5\hat{i}) = \frac{M}{2}(\vec{0}) + \frac{M}{2}\vec{v}_2$ $\vec{v}_2 = 10\hat{i} \text{ m/s}$

So, the second fragment has:

• Initial horizontal velocity: $v_{2x} = 10 \text{ m/s}$
• Initial vertical velocity: $v_{2y} = 0 \text{ m/s}$

Calculating the final horizontal distance $x$:

The second fragment drops from height $H = 1.25 \text{ m}$ under gravity. The time $t$ to reach the ground is: $H = \frac{1}{2}g t^2 \implies 1.25 = 5t^2 \implies t = 0.5 \text{ s}$

During this time $t$, the horizontal distance covered by the second fragment from the peak position $x_H$ is: $\Delta x = v_{2x} \cdot t = 10 \times 0.5 = 5 \text{ m}$

Therefore, the total horizontal distance $x$ from the origin $O$ is: $x = x_H + \Delta x = 2.5 + 5 = 7.5 \text{ m}$