Chapter 9: Momentum, Work, Power Flashcards

Question

Define projectile and give example

Answer 1

Object projected outwards such as a cannonball

Answer 2

* You need to resolve the weight into vertical and horizontal components * Use the component acting in the direction of the motion i.e horizontal component mg cos theta = mg \* cos theta

Answer 3

Rate at which work is done Rate at which energy is transferred

Answer 4

P = work done / time P = ΔE / t P = Force \* velocity

Answer 5

1. Same type of force 2. Same magnitude 3. Act along same line of action 4. Act in opposite directions 5. Act on two different bodies

Answer 6

One where momentum is conserved and kinetic energy is conserved

Answer 7

One where some of the kinetic energy is converted into other forms during the collision. But momentum is always conserved

Answer 8

Contact friction between solid surfaces Fluid friction, drag or fluid resistance or air resistance

Answer 9

They always act in the opposite direction to the motion of the object They can never speed things up or start something moving They convert kinetic energy into heat

Answer 10

A driving force that stays the same all the time e.g. weight A fictional or drag force that increases with speed

Answer 11

When the frictional force equals the driving force e.g. when weight = drag due to air resistance

Answer 12

Curve, where the gradient decreases, eventually turning into a horizontal line. Graph starts at the origin and increases

Answer 13

Acceleration starts off high then decreases. The rate of decrease of acceleration starts slow then speeds up then slows again. The acceleration finally reaches zero. The graph looks like an unexaggerated S shape

Answer 14

First half of the graph looks like the normal velocity against time for an object reaching terminal velocity. Then the graph suddenly drops to a lower velocity and flattens out, where it reaches a new terminal velocity

Answer 15

Whenever energy is transferred

Answer 16

The rate of doing work – it’s the amount of energy transferred from one form to another per second

Answer 17

1. resolve initial velocity into horizontal and vertical components 2. use vertical component to find how high it goes and and to work out how long it's in the air for 3. use horizontal component to find distance travelled horizontally while in air

Answer 18

If the graph is symmetrical: at the halfway point, vertical velocity is 0 for an instantaneous time, so the value of v_v must be multiplied by 2 To get the second time v_v = 0 which is when it will hit the ground so v_v = 0, u_v = resolve into vertical and horizontal to find out a = g = -9.81 (upwards as neg) t = ? use v = u + at to get time This is time taken to get to highest point; multiply by 2 to get the time to reach ground again

Answer 19

There is no acceleration horizontally so a = g = 0 this means that u_H = v_H (speed) time for it to hit ground found from vertical component = t use this to find s, distance travelled you can use speed = distance / time distance = speed \* time

Answer 20

break it into vertical and horizontal vertical: thrown horizontally with speed Vh so for vertical u = o a = g = (-) 9.81 if done above ground level s = xm so use s = ut + 1/2 at^2 usually just goes to s = 1/2 at^2 use this to find time horizontal: Vh isnt affected by gravity so Uv = Vv and a = N/A use time t from vertical component usually can use speed = distance / time distance = speed \* time s = Vv (Uv) \* t

Answer 21

if falling at constant speed, forces are equal, drag = mg

Answer 22

Work done is not necessarily the total energy. If you move an object higher up you have increased its potential energy, but it already had some potential energy to start with Force, F, will usually be fixed. Equation assumes force is in same direction as direction of movement

Answer 23

Work done = force \* distance \* cos theta

Answer 24

``` Power = Force \* velocity P = Fv ``` ``` Power = Force cos theta \* velocity P = F cos theta \* velocity ```