Question

A \(750 .-\mathrm{kg}\) cannon fires a 15.0 -kg projectile with a speed of \(250 . \mathrm{m} / \mathrm{s}\) with respect to the muzzle. The cannon is on wheels and can recoil with negligible friction. Just after the cannon fires the projectile, what is the speed of the projectile with respect to the ground?

Short answer

Answer: The speed of the projectile with respect to the ground is 245.1 m/s.

Step-by-step solution

Understand the concept of conservation of momentum

The conservation of momentum states that the total momentum of a system of particles remains constant, as long as no external forces act on the system. In this problem, we have two objects: the cannon and the projectile. Before firing, both the cannon and projectile are at rest, so their total momentum is zero. After firing, the cannon recoils backward, and the projectile moves forward. The total momentum in the system should still be zero.

Set up the momentum equation

According to the conservation of momentum, the total momentum before firing is equal to the total momentum after firing. Let's denote the mass of the cannon as \(m_1\), the mass of the projectile as \(m_2\), the speed of the cannon after firing as \(v_1\), and the speed of the projectile with respect to the ground (i.e., what we want to find) as \(v_2\). The momentum equation can be set up as follows: \(m_1v_1+m_2v_2=0\)

Plug in the known values and solve the equation

We are given that \(m_1 = 750 kg\), \(m_2 = 15 kg\), and the speed of the projectile with respect to the muzzle (i.e., the cannon) is \(250 m/s\). We also know that the speed of the projectile with respect to the ground is the sum of the speeds of both the projectile and the cannon, so we can write \(v_2 = v_1 + 250 m/s\). Plug in these values into the momentum equation and we get: \(750 kg * v_1 + 15 kg * (v_1 + 250 m/s) = 0\)

Solve the equation for the speed of the projectile

First, distribute the mass of the projectile into the equation, then simplify by combining like terms: \(750 kg * v_1 + 15 kg * v_1 + 15 kg * 250 m/s = 0\) Now combine the terms with v_1 and isolate it on one side of the equation: \((750 kg + 15 kg) * v_1 = -15 kg * 250 m/s\) Now, divide both sides by the combined mass of the two objects to solve for \(v_1\): \(v_1 = \frac{-15 kg * 250 m/s}{765 kg} = -4.9 m/s\) Since \(v_1 = -4.9 m/s\), this means the cannon recoils with a speed of \(4.9 m/s\).

Find the speed of the projectile with respect to the ground

Now that we have the speed of the cannon, we can find the speed of the projectile with respect to the ground using the equation from step 3: \(v_2 = v_1 + 250 m/s = -4.9 m/s + 250 m/s = 245.1 m/s\) The speed of the projectile with respect to the ground is \(245.1 m/s\).

Key concepts

Momentum Equations

Understanding momentum equations is crucial for solving many physics problems. Momentum, a conserved quantity in physics, is defined as the product of an object's mass and its velocity. It is represented by the equation:
\( p = mv \),
where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In the absence of external forces, the total momentum of a system remains constant. This fundamental principle is known as the conservation of momentum. It is mathematically expressed when two or more objects interact, such as in a collision or explosion:
\( m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2 \).
Here \( m_1 \) and \( m_2 \) represent the masses of the objects, while \( v_1 \) and \( v_2 \) are their initial velocities, and \( v'_1 \) and \( v'_2 \) are their velocities after interaction. In a closed system, where the net external force is zero, these equations are indispensable for calculating the unknown variables involved in the motions of the objects.

Projectile Motion

Projectile motion is another fascinating concept which comes into play when an object is thrown into the air and moves under the influence of gravity alone. Any such object is called a projectile, and its motion is called projectile motion. The path followed by a projectile is known as a parabola. A key point to remember is that projectile motion is two-dimensional motion, so it has both horizontal and vertical components. These components are independent of each other; the horizontal motion proceeds at a constant velocity, while the vertical motion is affected by gravity, causing a constant acceleration downwards at \( 9.8 m/s^2 \).
Understanding these components is crucial for breaking down and analyzing projectile motion problems, where we often separate the motion into perpendicular directions and apply kinematic equations to each independently. With the right approach, we tackle complex motions by simply studying their separate, simpler motions.

Recoil Velocity

Recoil velocity is a term most commonly used in discussions about firearms or ballistic devices, but it's a broader physics concept that refers to the velocity with which one object moves backward after ejecting or throwing another object forward. This is a direct application of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. Moreover, recoil velocity is intimately tied to the principle of conservation of momentum. If an object of a certain mass is propelled forward, the object from which it is propelled must move backward with a velocity so that the total momentum of the system remains unchanged.
For instance, if you were standing on a frictionless surface and threw a heavy object forward, you would move backward. The speed at which you move is the recoil velocity, and its magnitude depends on the masses of the objects and the velocity of the ejection. It is calculated using the same momentum equations that we discussed earlier, providing an intricate example of these fundamental physics principles at work.

Physics Problem Solving

Physics problem solving is an art that requires understanding core concepts, developing strategies, and translating those strategies into mathematical equations. The general approach begins with identifying the relevant principles, such as the conservation of momentum or Newton's laws, and then carefully analyzing the given information. Another key is to work with vectors and their components, especially in two-dimensional problems like projectile motion.
An effective strategy involves creating a visual representation, such as a free-body diagram, to visualize forces, motion, and other aspects of the problem. Assumptions may also be required to simplify complex systems, allowing us to use ideal models for calculation purposes. Mathematics comes into play when translating the physical scenario into equations that can then be manipulated to find the solution. This often involves algebraic manipulation, solving systems of equations, or calculus, depending on the problem's complexity. An insightful tip is to always keep track of the units, as they provide a vital check on the consistency and correctness of your calculations.