# Cannon and coefficient of friction between ground

• fishfish
In summary, this conversation discusses the calculation of the coefficient of friction between a 3000kg cannon and the ground it rests upon. The cannon fires a 14kg shell horizontally with a muzzle velocity of 450m/s and recoils for a distance of 2.5m before coming to a stop. Using momentum conservation, the initial velocity of the cannon can be found to be 0. After finding v2, the acceleration exerted by friction on the cannon can be calculated using the formula v12 = v22 - 1/2ad. The conversation also clarifies that initial velocity for the cannon and shell system is zero, and momentum conservation states that momentum before is equal to momentum after.
fishfish

## Homework Statement

A 3000kg cannon fires a 14 kg shell horizontally with a muzzle velocity of 450m/s. The cannon recoils for a distance of 2.5m before coming to a stop. What is the coefficient of friction between the cannon and the ground that it rests upon?

mcannon= 3000kg
mball=14kg
v1ball=0
v2ball=450m/s
v1cannon = 0
v2cannon= ?

I don't really know where to begin with this question.. Is it a momentum/impulse question or something else? Please help..

Use momentum conservation to find the initial velocity of the cannon.

kuruman said:
Use momentum conservation to find the initial velocity of the cannon.

Wouldn't the initial velocity of the cannon be zero though since both the ball and the cannon start off at rest? If I were to find the v2 of the cannon though, how would I proceed with the rest of the solution?

Yes, initial velocity is 0. What kuruman meant was the thing you call v2, that is, the velocity of the cannon at the instant it fired the cannonball.

After finding v2, can you find the acceleration that friction exerts on the object?

Yes, the initial velocity (and momentum) of the cannon and shell system is zero. The final momentum of the cannon and shell system will also be zero. However, before the shell is fired nothing is moving, but after the shell is fired, both cannon and shell are moving.

ideasrule said:
Yes, initial velocity is 0. What kuruman meant was the thing you call v2, that is, the velocity of the cannon at the instant it fired the cannonball.

After finding v2, can you find the acceleration that friction exerts on the object?

Would I use the formula v12 = v22 - 1/2ad?

Am I using the wrong formula..? Because I seem to get this massive number as the acceleration..

What is the definition of momentum? Momentum conservation says

Momentum before = Momentum after.

## What is a cannon?

A cannon is a large, heavy piece of artillery that is used to fire heavy projectiles at high speeds. It was commonly used in warfare before modern weaponry was developed.

## How does a cannon work?

A cannon works by using gunpowder to create a controlled explosion that propels a projectile, such as a cannonball, out of the barrel at a high velocity. The angle and force of the explosion determine the trajectory and distance the projectile will travel.

## What is coefficient of friction between ground?

The coefficient of friction between ground is a measure of the amount of friction that exists between two surfaces in contact with each other. It is a dimensionless number that represents the ratio of the force required to move one surface over the other to the force holding them together.

## Why is the coefficient of friction between ground important for cannons?

The coefficient of friction between ground is important for cannons because it affects the trajectory and distance of the fired projectile. A higher coefficient of friction means more resistance on the projectile's movement, resulting in a shorter distance traveled.

## How is the coefficient of friction between ground calculated?

The coefficient of friction between ground can be calculated by dividing the force required to move one surface over the other by the weight of the object. This calculation can be affected by factors such as the type of surfaces in contact and the force of gravity on the object.

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