To solve the problem of a shell fired from a cannon that explodes into two equal masses at the highest point of its trajectory, we need to apply the principles of conservation of momentum and the characteristics of projectile motion. Let's break this down step by step.
Understanding the Motion of the Shell
When the shell is fired, it follows a parabolic trajectory due to the influence of gravity. The initial velocity \( V \) can be broken down into two components: the horizontal component \( V_x = V \cos(\theta) \) and the vertical component \( V_y = V \sin(\theta) \). At the highest point of its path, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged.
Momentum Conservation at the Highest Point
At the peak of its trajectory, the shell has only horizontal motion. When it explodes into two pieces of equal mass, we can denote these pieces as \( m_1 \) and \( m_2 \), each with a mass of \( \frac{m}{2} \). According to the law of conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion.
- Before the explosion: The momentum is entirely horizontal, given by \( P_{initial} = m V_x = m \left( V \cos(\theta) \right) \).
- After the explosion: The momentum of the two pieces must also sum to the initial momentum. If \( v_1 \) is the speed of piece 1 (which retraces its path) and \( v_2 \) is the speed of piece 2, we have:
\( P_{final} = \frac{m}{2} (-v_1) + \frac{m}{2} v_2 \)
Analyzing the Motion of the Pieces
Since piece 1 retraces its path back to the cannon, its speed \( v_1 \) is equal to the horizontal component of the shell's velocity at the highest point, which is \( V \cos(\theta) \). Therefore, we can substitute \( v_1 \) into our momentum equation:
\( m V \cos(\theta) = \frac{m}{2} (-V \cos(\theta)) + \frac{m}{2} v_2 \)
Solving for the Speed of the Second Piece
Now, we can simplify the equation. First, we can cancel out the mass \( m \) from both sides:
\( V \cos(\theta) = -\frac{1}{2} V \cos(\theta) + \frac{1}{2} v_2 \)
Next, we can rearrange this to isolate \( v_2 \):
\( V \cos(\theta) + \frac{1}{2} V \cos(\theta) = \frac{1}{2} v_2 \)
\( \frac{3}{2} V \cos(\theta) = \frac{1}{2} v_2 \)
Multiplying both sides by 2 gives us:
\( 3 V \cos(\theta) = v_2 \)
Final Result
The speed of the second piece immediately after the explosion is:
\( v_2 = 3 V \cos(\theta) \)
This result shows that the second piece moves away from the explosion site with a speed that is three times the horizontal component of the initial velocity of the shell. This example illustrates the principles of momentum conservation and how they apply to projectile motion and explosions. Understanding these concepts is crucial in physics, especially in analyzing motion and forces in various scenarios.
