A shell is fired from a cannon with a velocity v (m/s) at an angle theta with the horizontal directi

Q: Wait, why is the velocity at the highest point $v \cos\theta$?

In projectile motion, gravity only pulls downwards, affecting the vertical velocity. The horizontal velocity is unaffected by gravity (ignoring air resistance), so it remains constant at $v \cos\theta$ throughout the flight. At the highest point, the vertical velocity becomes zero, leaving only the horizontal component.

Q: Why does the piece that retraces its path have a velocity of $-v \cos\theta$?

To exactly retrace a parabolic path back to the launch point, a particle must have the exact same speed but in the opposite direction at any given point. Since the original shell was moving forward with $v \cos\theta$, the returning piece must move backward with $-v \cos\theta$.

Q: Why don't we use conservation of kinetic energy here?

An explosion is an inelastic process in reverse. Chemical potential energy is converted into kinetic energy during the blast. Therefore, the total kinetic energy of the system actually *increases* after the explosion. Only momentum is conserved.

Q: What happens to the center of mass of the two pieces after the explosion?

Because the explosion only involves internal forces, the center of mass of the system continues to follow the exact same parabolic trajectory it would have followed if the shell had never exploded!

Q: Why do we ignore the vertical momentum during the explosion?

The explosion happens instantaneously. While gravity is an external force acting downwards, the time interval of the explosion is so infinitesimally small ($\Delta t \to 0$) that the vertical impulse ($mg \Delta t$) is negligible. Thus, momentum is conserved in all directions just before and just after the blast.

Question

A shell is fired from a cannon with a velocity $v$ (m/s) at an angle $	heta$ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (m/s) of the other piece immediately after the explosion is

Options:

(A) $3 v \cos 	heta$

(B) $2 v \cos 	heta$

(C) $\frac{3}{2} v \cos 	heta$

(D) $\sqrt{\frac{3}{2}} v \cos 	heta$

Sigma Percentile · Accepted Answer

### The Setup: A Parabolic Journey

Imagine standing on a battlefield, watching a heavy shell being fired from a cannon. The shell, which we will assign a total mass of $2m$, is launched with an initial velocity $v$ at an angle $	heta$ to the horizontal.

As it soars through the air, it traces a perfect parabolic trajectory. Gravity constantly pulls it downwards, steadily decreasing its vertical velocity. However, in the horizontal direction, there are no forces acting on it (assuming we ignore air resistance). This means its horizontal velocity remains perfectly constant at $v \cos	heta$ throughout its entire flight.

### The Climax: Explosion at the Peak

The most critical moment of this problem occurs when the shell reaches the absolute highest point of its trajectory. At this exact peak, the shell has momentarily stopped moving upwards. Its vertical velocity is exactly zero.

Therefore, the only velocity it possesses is its constant horizontal velocity, $v \cos	heta$. Just before the dramatic event, the total initial momentum of the shell in the horizontal direction is simply its mass times its velocity:

$$p_{	ext{initial}} = 2m(v \cos	heta)$$

Suddenly, the shell explodes into two equal fragments, each with a mass of $m$. An explosion is a violent event, but crucially, it is driven entirely by **internal forces**. Because there is no external force pushing the system in the horizontal direction, the fundamental law of physics applies: **linear momentum must be conserved**.

### The Master Equation: Conservation of Momentum

The problem gives us a fascinating piece of information about one of the fragments: it perfectly retraces its path back to the cannon.

Think about what this means physically. For a projectile to travel backward along the exact same parabolic path it just took, its velocity at the highest point must be exactly reversed. Since the original shell was moving forward with a velocity of $v \cos	heta$, this returning fragment must be moving backward with a velocity of $-v \cos	heta$.

Let's call the unknown velocity of the second fragment $v'$. We can now set up our master equation by equating the total momentum just before the explosion to the total momentum just after:

$$p_{	ext{initial}} = p_{	ext{final}}$$

$$2m(v \cos	heta) = m(-v \cos	heta) + m(v')$$

### The Resolution: Finding the Missing Velocity

Now, we simply let the algebra guide us to the solution. Notice how the mass $m$ appears in every single term of our equation. We can divide the entire equation by $m$, effectively canceling it out:

$$2v \cos	heta = -v \cos	heta + v'$$

To isolate our target variable, $v'$, we move the negative term to the other side of the equation. It becomes positive and adds to the existing velocity:

$$v' = 2v \cos	heta + v \cos	heta$$

$$v' = 3v \cos	heta$$

And there we have it! The second fragment shoots forward with a velocity of **$3v \cos	heta$**, which is three times the horizontal speed of the original shell. The elegance of momentum conservation allows us to solve a seemingly complex explosion with just a few lines of algebra.

Select Answer:

Visualized Solution

Projectile Motion Setup

Velocity at the Highest Point

Conservation of Linear Momentum

Analyzing the Fragments

Applying the Momentum Equation

Simplifying the Equation

Solving for $v^{'}$

The Sigma Insight: Conservation of Linear Momentum

The Setup

The Climax

The Master Equation

The Resolution

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