Question

A 750 -kg cannon fires a 15 -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 final velocity of the projectile with respect to the ground is 255 m/s.

Step-by-step solution

Understanding the conservation of momentum

The total momentum of a closed system is conserved when no external forces act upon it. In this case, we have a projectile and a cannon on wheels with negligible friction, so we can consider this a closed system. The conservation of momentum states that the total momentum before the interaction equals the total momentum after the interaction.

Define known quantities

We are given the following information: Mass of cannon, \(m_c = 750 \mathrm{~kg}\) Mass of projectile, \(m_p = 15 \mathrm{~kg}\) Speed of projectile with respect to the muzzle, \(v_p = 250 \mathrm{~m/s}\)

Calculate the total momentum before firing

Initially, the projectile and cannon are both at rest, so their total initial momentum is zero. Mathematically, this can be written as: \(P_{initial} = m_c \times 0 + m_p \times 0 = 0\)

Calculate the total momentum after firing

After firing the projectile, the cannon recoils, and the projectile moves forward. Let's assume the projectile moves in the positive direction and the cannon recoils in the negative direction. The total momentum after firing can be given by: \(P_{final} = m_c \times v_c + m_p \times v_p\) Here, \(v_c\) is the velocity of the cannon, which we need to find.

Apply conservation of momentum

Using the conservation of momentum, we know that the initial momentum must equal the final momentum: \(P_{initial} = P_{final}\) Substituting the values from steps 3 and 4, we get: \(0 = m_c \times v_c + m_p \times v_p\) Now, we need to isolate \(v_c\): \(v_c = -\frac{m_p \times v_p}{m_c}\) Substitute the given values: \(v_c = -\frac{15 \mathrm{~kg} \times 250 \mathrm{~m/s}}{750 \mathrm{~kg}} = -5 \mathrm{~m/s}\) Since the result is negative, the cannon moves in the opposite direction of the projectile, as expected.

Calculate the velocity of the projectile with respect to the ground

We've found the velocity of the projectile (\(v_p\)) with respect to the muzzle and the velocity of the cannon (\(v_c\)). To find the velocity of the projectile relative to the ground, we need to add the velocities of the projectile and cannon - remember that the cannon is recoiling, so the projectile velocity must be added to the absolute value of the cannon's velocity: \(v_{pg} = 250 \mathrm{~m/s} + |-5\mathrm{~m/s}| = 255 \mathrm{~m/s}\) The speed of the projectile with respect to the ground is \(255 \mathrm{~m/s}\).

Key concepts

Closed System

A closed system is one where no external forces influence the total momentum within the system. In the context of a cannon firing a projectile, the cannon and projectile together are considered a closed system. This means that since no outside forces like friction significantly affect them, the total momentum before and after firing remains constant.

• This assumption is important because it allows us to apply the principle of conservation of momentum to solve problems.
• The cannon is on wheels and the friction is negligible, ensuring minimal external interference.

In a closed system, if the system starts at rest, it must also have zero total momentum after any interaction within the system. This makes it possible to calculate the recoil velocity of the cannon and the velocity of the projectile relative to a static point like the ground.

Recoil Velocity

Recoil velocity refers to the backward movement of an object when it propels another object forward. In our scenario, the cannon fires a projectile, and as a result, the cannon itself moves in the opposite direction
going backwards.

• The recoil velocity of the cannon can be calculated using conservation of momentum, which helps us understand how different masses influence the velocities within a closed system.
• Since momentum is conserved, the mass and velocity of the projectile determine the recoil velocity of the cannon.

Here, the formula \( v_c = -\frac{m_p \times v_p}{m_c} \) was used to find that the cannon recoils at a speed of \(-5 \ \mathrm{m/s}\). The negative sign indicates the direction opposite to the projectile's motion.

Relative Velocity

Relative velocity is the velocity of one object as observed from the frame of reference of another object. In the exercise, we looked at the velocity of the projectile relative to both the cannon (muzzle) and the ground
, which is stationary.

• The projectile's speed with respect to the muzzle was given as \(250 \, \mathrm{m/s} \).
• To find the velocity with respect to the ground, we adjusted by taking into account the recoil velocity of the cannon.

The calculation shows that we add the gun's backward recoil velocity to the projectile's velocity to get the total velocity of the projectile relative to the ground:\( v_{pg} = 250 \, \mathrm{m/s} + |-5 \, \mathrm{m/s}| = 255 \, \mathrm{m/s} \).This shows how adjusting for the motion of the base (cannon) in translating the velocity of the projectile is essential for accurate measurements.

Projectile Motion

Projectile motion examines the trajectory that an object follows after being launched. It’s a form of motion where an object (the projectile) is thrown obliquely near the earth's surface, following a curved path, and is affected by gravity.

• Understanding projectile motion often involves analyzing velocities in both horizontal and vertical directions.
• In our exercise case, we interpreted the horizontal initial velocity component since the projectile is fired horizontally.

The example demonstrated here considers only the initial horizontal velocity component, which is crucial for solving motion problems where vertical motion isn’t affected initially. Thus, this simple model highlights how initial speeds are vital for predicting the initial pathways of such moving objects.