# Momentum and movement basic GCSE Quest

• nonthesecond
In summary, when a gun is fired, a very large force acts on the bullet for a very short time according to the formula force x time = change in momentum (kg m/s). An average force of 4000 acts for 0.01 seconds on a bullet of 50g mass. When the bullet is fired horizontally, in the short time it takes to reach its target, its horizontal speed decreases by 80%. This is due to air friction, which is a function of the bullet's speed. To calculate the percentage of its original kinetic energy the bullet still has when it reaches its target, we can use the fact that the kinetic energy decreases by 96% when the speed decreases by 80%.
nonthesecond
when a gun is fired a very large force acts on the bullet for a very short time.

formula that's given: force x time = change in momentum (kg m/s)

an average force of 4000 acts for 0.01 seconds on a bullet of 50g mass.

DONE THIS PART

b. the bullet is fired horizontally. in the short time that it takes for it to reach it's target, its horizontally has fallen by 80% of its initial speed.

explain why the speed of the bullet has decreased so quickly.

c. calculate the percentage of its original Kinetic energy the bullet still has when it reaches its target.

hi nonthesecond!
nonthesecond said:
b. the bullet is fired horizontally. in the short time that it takes for it to reach it's target, its horizontally has fallen by 80% of its initial speed.

explain why the speed of the bullet has decreased so quickly.

c. calculate the percentage of its original Kinetic energy the bullet still has when it reaches its target.

show us what you've tried, and where you're stuck, and then we'll know how to help!

I'm a little confused. What part have you done already? Also, please show us your attempt at a solution.

tiny-tim said:
hi nonthesecond!

show us what you've tried, and where you're stuck, and then we'll know how to help!

for b i think because it already has reached it's fastest speed it must slow down from then on. what's the best way to word this?

c. for this question i think it's o.5(0.005) x velocity^2 but i don't know how to work out velocity i know it's v=d/t.

tiny-tim said:
hi nonthesecond!

show us what you've tried, and where you're stuck, and then we'll know how to help!

tal444 said:
I'm a little confused. What part have you done already? Also, please show us your attempt at a solution.

i've finished the first part (a)

i can't do b and c

for b i think because it already has reached it's fastest speed it must slow down from then on. what's the best way to word this?

c. for this question i think it's o.5(0.005) x velocity^2 but i don't know how to work out velocity i know it's v=d/t.

hi nonthesecond!
nonthesecond said:
for b i think because it already has reached it's fastest speed it must slow down from then on.

but why?
c. for this question i think it's o.5(0.005) x velocity^2 but i don't know how to work out velocity i know it's v=d/t.

you know that vf/vi is 0.8 …

so how much KE is lost?

tiny-tim said:
hi nonthesecond!

but why?

you know that vf/vi is 0.8 …

so how much KE is lost?

sorry but you have to dumb this down I'm doing GCSEs and i don't understand what you mean by the Vs, i also have no idea why it slows down

vf = final velocity, vi = initial velocity.

nonthesecond said:
sorry but you have to dumb this down I'm doing GCSEs and i don't understand what you mean by the Vs, i also have no idea why it slows down

ah, the question says that the horizontal speed has fallen to 20% of its initial value

ie vf/vi = 0.2, where vi means initial speed and vf means final speed

(i put vf/vi = 0.8 before, i misread the question)

is the answer 40% ?

(0.5x 0.005) x 40^2

You have to model air friction right? If not I don't see how the bullet is supposed to loose speed (momentum)

Once the bullet is flying, the only horizontal force acting on it is air friction. This force is a function of the bullet's speed, and since the speed decreases constantly, the force also decreases constantly. The model you're supposed to use for air friction I don't know, but it's most probably a linear model, something like F= k · v

Use Newton's law of conservation of momentum to obtain an ODE (an easy one), solving that ODE will have your problem solved.

For the kinetic energy, I might be confusing the question because it's too easy, if they say the speed has decreased by 80%, than the energy has decreased by:

v1= v -> v2= 0,2 · v

E1 proportional to v1^2= v^2 -> E2 prop. to v2^2 = (0,2·v)^2 = 0.04·v^2

Thus the kinetic energy has decreased by 96% if I understood the question properly.

## 1. What is momentum?

Momentum is a measure of an object's motion, specifically the quantity of motion that an object has. It is the product of an object's mass and its velocity.

## 2. What is the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a closed system remains constant, meaning that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

## 3. How is momentum different from velocity?

Momentum is different from velocity because velocity only describes an object's speed and direction of motion, while momentum also takes into account the object's mass. Two objects can have the same velocity but different momentum if they have different masses.

## 4. What is the relationship between force and momentum?

Force is directly proportional to the rate of change of momentum, meaning that the greater the force applied to an object, the greater its change in momentum will be. This can be represented by the equation F=ma, where F is force, m is mass, and a is acceleration.

## 5. How is momentum used in real-life applications?

Momentum is used in many real-life applications, such as in sports, car safety, and rocket propulsion. In sports, athletes use their momentum to gain speed and power in their movements. In car safety, features like airbags and crumple zones work to reduce the change in momentum during a collision. In rocket propulsion, the thrust generated by the rocket's engines creates a change in momentum that propels the rocket forward.

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