Problem 14 - Applied Physics

Question

A cannon is mounted on a railroad car. The cannon shoots a \(1.75-\mathrm{kg}\) ball with a muzzle velocity of \(30 \overline{0} \mathrm{~m} / \mathrm{s}\). The cannon and the railroad car together have a mass of \(45 \overline{0} \mathrm{~kg}\). If the ball, cannon, and railroad car are initially at rest, what is the recoil velocity of the car and cannon?

Step-by-Step Solution

Answer

The recoil velocity of the car and cannon is \(-1.167 \text{ m/s}\).

Step 1: Understand the Conservation of Momentum

The problem involves the principle of conservation of momentum. When the cannon fires the ball, the total momentum before and after firing remains the same (since there are no external forces). We can express this as: \[ m_1v_1 + m_2v_2 = 0 \] where \( m_1 \) and \( v_1 \) are the mass and velocity of the ball, and \( m_2 \) and \( v_2 \) are the mass and velocity of the cannon and railroad car.

Step 2: Assign Values to Known Quantities

From the problem statement, we know:- Mass of the ball, \( m_1 = 1.75 \text{ kg} \)- Velocity of the ball, \( v_1 = 300 \text{ m/s} \)- Combined mass of the cannon and car, \( m_2 = 450 \text{ kg} \)The velocity of the cannon and car, \( v_2 \), is what we are trying to find.

Step 3: Set Up the Momentum Equation

Using the conservation of momentum, we set up the equation:\[ m_1v_1 + m_2v_2 = 0 \]This can be expanded to:\[ (1.75)(300) + (450)(v_2) = 0 \]

Step 4: Solve for the Recoil Velocity \( v_2 \)

Rearrange the equation from Step 3 to solve for \( v_2 \):\[ 450v_2 = -1.75 imes 300 \]This simplifies to:\[ v_2 = \frac{-1.75 imes 300}{450} \]Calculate the result:\[ v_2 = \frac{-525}{450} = -1.1667 \text{ m/s} \]The negative sign indicates that the cannon car moves in the opposite direction to the ball.

Key Concepts

Recoil Velocity

Imagine standing at the edge of a swimming pool and jumping off a floating raft. As you leap forward, the raft moves backward. This opposite motion is due to recoil velocity, a fascinating phenomenon observed in many physics problems. When a cannon fires a ball, like in our exercise, it experiences a "kick" backward—a recoil.
This recoil occurs because, to conserve momentum, the cannon and ball must move in opposite directions upon firing.
Let's clarify further:

• The cannon and the ball start at rest, meaning total initial momentum is zero.
• Upon firing, the ball moves forward, which requires an equal and opposite reaction for the cannon and car.
• This reaction results in the recoil velocity, pushing the cannon backward.

Understanding recoil velocity helps in figuring out how launching systems, like cannons or rocket boosters, affect their surroundings.

Physics Problem Solving

Physics problems may seem tricky, but breaking them down step by step can make them much more manageable. Let's take our cannon problem as an example. It's a straightforward application of physics principles, starting with understanding the core ideas.
Here's a simple approach:

• Identify known and unknown values. For the cannon problem, known values are the mass and speed of the ball and the total mass of the car and cannon.
• Determine the physics principle that applies, in this case, the conservation of momentum.
• Write down the relevant equations based on this principle.
• Solve the equations by substituting the known values and solving for the unknowns—recoil velocity in our case.

This structured approach to problem-solving not only solves the problem but builds confidence and understanding of the topic.

Momentum Equation

The concept of momentum is crucial in understanding many physical phenomena, especially in collision and explosion scenarios. Let's delve into the momentum equation used in our problem.
Momentum is the product of an object's mass and velocity, represented mathematically as:

• For the ball: \( p_1 = m_1 \cdot v_1 \)
• For the cannon and car: \( p_2 = m_2 \cdot v_2 \)

In a closed system, like our stationary cannon and railcar before firing, the total momentum is conserved. That means:

• The total momentum before firing is zero.
• After firing, the forward momentum of the ball is balanced by the backward momentum of the cannon and car, maintaining a total momentum of zero.

The equation to express this conservation is:\[ m_1v_1 + m_2v_2 = 0 \]Using this equation, we were able to solve for the recoil velocity, showcasing how fundamental physics principles allow us to predict and understand motion.

Projectile Motion

While projectile motion isn't directly the focus when solving for recoil velocity, it plays a significant role in understanding the ball’s path after it leaves the cannon. Projectile motion explains how objects move in a curved trajectory under the influence of gravity.
In any projectile motion:

• An object experiences a horizontal velocity component, which is constant, and a vertical velocity component, affected by gravity.
• The ball from the cannon follows a curved path as its initial horizontal velocity interacts with Earth's gravitational pull.
• Analyzing this motion can reveal how far and high the ball will travel, useful information for applications like predicting the land point of a cannonball or engineering purposes.

Therefore, understanding projectile motion complements our primary problem, broadening our grasp of how forces and motions interplay in real-world scenarios.