10. An object of mass 5 kg is projected with a velocity of 20 m/s at an angle of 60° to the horizont

Q: Why is the vertical velocity zero at the highest point?

At the peak of a projectile's trajectory, it stops moving upwards and is about to start falling downwards. For that brief instant, its vertical velocity component is exactly zero.

Q: Why do we only conserve momentum in the horizontal direction?

During the explosion, the internal forces are much larger than gravity, so momentum is conserved in all directions *during* the brief explosion. However, after the explosion, gravity acts vertically, changing the vertical momentum. Since there are no horizontal forces, horizontal momentum is conserved throughout.

Q: Does it matter which fragment moves left or right?

No, the math takes care of the directions. If you assume a fragment moves left and the math gives a negative velocity, it simply means it actually moved right. The relative velocity of separation will be the same.

Q: Why is the time of fall the same as half the time of flight?

The explosion only affects the horizontal velocities. The vertical motion of both fragments is identical to an object dropped from rest from that maximum height. The time to fall from the maximum height is exactly half the total time of flight of the original unbroken projectile.

Question

An object of mass $5	ext{ kg}$ is projected with a velocity of $20	ext{ m/s}$ at an angle of $60^\circ$ to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses $1	ext{ kg}$ and $4	ext{ kg}$. The fragments separate horizontally after the explosion. The explosion releases internal energy such that the kinetic energy of the system at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground.

Sigma Percentile · Accepted Answer

### The Explosive Projectile: A Dance of Momentum and Energy

Imagine a $5	ext{ kg}$ object launched into the air with an initial velocity of $20	ext{ m/s}$ at an angle of $60^\circ$ to the horizontal. It traces a beautiful parabolic path. When it reaches the very peak of its trajectory, something dramatic happens: it explodes into two fragments of $1	ext{ kg}$ and $4	ext{ kg}$.

At this highest point, the vertical velocity of the projectile is momentarily zero, meaning it is only moving horizontally. The explosion is an internal event, meaning no external horizontal forces are acting on the system. This is our cue to use one of the most powerful tools in physics: the conservation of linear momentum.

### Conservation of Momentum

Just before the explosion, the entire $5	ext{ kg}$ mass is moving horizontally with a velocity of $u \cos 60^\circ$.

$$p_i = M u \cos 60^\circ = 5 	imes 20 	imes \frac{1}{2} = 50	ext{ kg m/s}$$

After the explosion, let's assume the $4	ext{ kg}$ fragment moves to the right with velocity $v_1$, and the $1	ext{ kg}$ fragment moves to the left with velocity $v_2$. The total horizontal momentum must remain the same:

$$4v_1 - 1v_2 = 50$$

This gives us our first crucial equation, linking the velocities of the two fragments.

### The Energy Boost

The problem states that the explosion releases internal energy, causing the kinetic energy of the system to double. Let's calculate the initial kinetic energy at the peak:

$$K_i = \frac{1}{2} M (u \cos 60^\circ)^2 = \frac{1}{2} 	imes 5 	imes 10^2 = 250	ext{ J}$$

Since the kinetic energy doubles, the final kinetic energy of the two fragments must be $500	ext{ J}$. We can express this as:

$$\frac{1}{2} (4) v_1^2 + \frac{1}{2} (1) v_2^2 = 500$$

Simplifying this, we get our second equation:

$$4v_1^2 + v_2^2 = 1000$$

### The Quadratic Crossroads

We now have a system of two equations. By substituting $v_2 = 4v_1 - 50$ into the energy equation, we get a quadratic equation in terms of $v_1$:

$$4v_1^2 + (4v_1 - 50)^2 = 1000$$

Expanding and simplifying this yields:

$$20v_1^2 - 400v_1 + 1500 = 0$$

Dividing by 20 gives a much friendlier equation: $v_1^2 - 20v_1 + 75 = 0$. Factoring this reveals two possible roots: $v_1 = 15	ext{ m/s}$ or $v_1 = 5	ext{ m/s}$.

### The Invariance of Separation

Here is where the physics gets truly elegant. Let's calculate the relative velocity of separation for both cases.

If $v_1 = 15	ext{ m/s}$, then $v_2 = 4(15) - 50 = 10	ext{ m/s}$. The fragments are moving in opposite directions, so their relative velocity is $15 - (-10) = 25	ext{ m/s}$.

If $v_1 = 5	ext{ m/s}$, then $v_2 = 4(5) - 50 = -30	ext{ m/s}$. The negative sign means the $1	ext{ kg}$ fragment is actually moving to the right, faster than the $4	ext{ kg}$ fragment. The relative velocity is $|5 - 30| = 25	ext{ m/s}$.

In both mathematically valid scenarios, the fragments separate from each other at exactly **$25	ext{ m/s}$**!

### The Final Descent

To find the final separation on the ground, we need to know how long the fragments are in the air. Since the explosion only affected horizontal velocities, their vertical motion is identical to an object dropped from the peak. The time to fall is exactly half the total time of flight of the original projectile:

$$t = \frac{u \sin 60^\circ}{g} = \frac{20 	imes \frac{\sqrt{3}}{2}}{9.8} \approx 1.77	ext{ s}$$

Finally, the horizontal separation is simply the relative velocity multiplied by the time of fall:

$$x = v_{rel} 	imes t = 25 	imes 1.77 = 44.25	ext{ m}$$

And there we have it, a beautiful synthesis of momentum, energy, and kinematics!

Enter Numerical Value:

Visualized Solution

Projectile Motion and Explosion

Conservation of Linear Momentum

Momentum Equation

Simplifying Momentum Equation

Kinetic Energy Condition

Energy Equation

Simplifying Energy Equation

Solving for Velocities

Roots of the Quadratic

Relative Velocity of Separation

Time of Fall

Calculating Time

Final Separation

The Sigma Insight: Conservation of Linear Momentum

The Explosive Projectile

Conservation of Momentum

The Energy Boost

The Quadratic Crossroads

The Invariance of Separation

The Final Descent

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