Problem 3 - Master Resource Book in JEE Main Physics

Question

A shell is fired from a cannon with a velocity \(v\) at angle \(\theta\) with horizontal. At the highest point, it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed of the other piece just after explosion is (a) \(3 v \cos \theta\) (b) \(2 v \cos \theta\) (c) \(\frac{3}{2} v \cos \theta\) (d) \(\frac{\sqrt{3}}{2} v \cos \theta\)

Step-by-Step Solution

Answer

(a) \(3 v \cos \theta\)

Step 1: Analyze Initial Conditions

The initial velocity of the shell is given as \(v\) and it's fired at an angle \(\theta\) to the horizontal. The velocity components are \(v_x = v \cos \theta\) and \(v_y = v \sin \theta\). At the highest point, the vertical component of velocity is zero; only the horizontal component \(v_x = v \cos \theta\) remains.

Step 2: Consider Explosion at Highest Point

Since the shell is at the highest point, it's moving horizontally with speed \(v \cos \theta\). The shell explodes into two pieces of equal mass. Assuming mass before explosion is \(m\), each piece has a mass of \(\frac{m}{2}\) after explosion.

Step 3: Apply Conservation of Momentum

Before explosion, the horizontal momentum is \(m \cdot v \cos \theta\). After explosion, let the velocity of the piece retracing its path be \(-v \cos \theta\). Apply conservation of momentum: \[m \cdot v \cos \theta = \frac{m}{2} \cdot (-v \cos \theta) + \frac{m}{2} \cdot v_2\]Solving for \(v_2\), we have \[m \cdot v \cos \theta = -\frac{m}{2} v \cos \theta + \frac{m}{2} v_2\]\[\frac{3}{2} m v \cos \theta = \frac{m}{2} v_2\]\[v_2 = 3v \cos \theta\]

Step 4: Conclusion

The speed of the other piece just after explosion is thus \(3v \cos \theta\). Hence, the correct choice is (a) \(3 v \cos \theta\).

Key Concepts

Conservation of Momentum

In the scenario of a shell being fired from a cannon, the principle of conservation of momentum plays a crucial role. This principle states that within a closed system, the total momentum before an event must equal the total momentum after the event, provided no external forces act on it.

For our shell, it begins with a certain momentum determined by its mass and velocity. Before the explosion at the highest point of its trajectory, the shell has a horizontal velocity component. It's essential to remember that momentum is a vector quantity, meaning it has both magnitude and direction.

• The shell's initial horizontal momentum is given by the product of its total mass (\(m\)) and its horizontal velocity (\(v \cos \theta\)).
• During the explosion, the total momentum along the horizontal axis remains consistent.
• One piece retraces its path with a velocity of \(-v \cos \theta\), affecting how the second piece moves.

Conservation of momentum ensures that even after the explosion, the combined momentum of both pieces still equals the initial momentum just before the blast.

Explosion

An explosion is a sudden and rapid release of energy, causing the system involved to break apart. In our problem, the shell explodes into two pieces at the highest point of its trajectory.

This occurrence transforms the movement of the shell, requiring us to look at how each piece behaves independently post-explosion. Here are some key aspects:

• The mass of the shell is equally divided into two pieces post-explosion.
• The explosion doesn't add external momentum; it just redistributes the existing momentum.
• One piece retraces its path, which is crucial for balancing the momentum in the system.

It is through breaking this balanced system that we are able to resolve for the velocity of the second piece, which turns out to be more significant due to the retracing piece moving backward.

Horizontal Velocity

In projectile motion problems like this one, breaking down the velocity into horizontal and vertical components is essential, especially when the projectile reaches its peak.

At the highest point, the vertical component of the velocity is zero because gravity acts downwards, momentarily stalling vertical movement. However, the horizontal component \(v_x = v \cos \theta\) remains constant without air resistance.

• The horizontal velocity component does not change, as no external horizontal forces are present.
• This steady velocity becomes the base for calculating any post-explosion motion of the shell's fragments.
• It determines the momentum of the fragments after the shell explodes.

Understanding that the horizontal motion persists despite gravity's forces is vital for resolving the problem at hand, as it allows us to predict where and how the pieces travel post-explosion.

Angle of Projection

The angle of projection refers to the initial angle at which a projectile is launched relative to the horizontal plane. This angle is significant as it dictates the projectile's trajectory and its ability to reach certain points in its path, like the peak where the explosion occurs.

In the context of our exercise, the angle \(\theta\) affects both the horizontal and vertical components of the velocity. Here's why it's important:

• It determines the ratio between the horizontal and vertical velocities at launch time.
• The angle itself does not change during the flight, but it sets how much of the velocity contributes to horizontal motion.
• The cosines of this angle, \(\cos \theta\), is used to compute the horizontal component, which is crucial after the explosion.

This angle impacts how far and high the projectile travels and ultimately influences the explosion dynamics, particularly the momentum transferred to each fragment.