Further Mechanics

Question 1

Is momentum vector or scalar?

Why?

Answer 1

Vector.

Because p=mv,
m is scalar but v is vector therfore momentum is vector in the direction of the velocity.

Question 2

What is an impulse?

Answer 2

The effect of a force on the rate of change of momentum.

Question 3

What is the formula for impulse?

Answer 3

FΔt = Δp

Question 4

Derive the formula for impulse from newtons second law.

Answer 4

The rate of change of momentum can be expressed as:
Δp = dp/dt

The equation for momentum:
p=mv

The rate of change of momentum can now be written as:
dp/dt = d(mv)/dt

Assuming that mass is constant, equation that can be written as:
dp/dt = m dv/dt

The rate of change of velocity is equal to acceleration, so it can be rewritten as:

dp/dt = m a.

QED

Question 5

How do you find impulse on a Force-Time graph?

Answer 5

Find the area under the graph.

Question 6

To improve the safethy of a car, the manufacturer sets out to design a more effective crumple zone. Why would the manufacturer try to achieve this?

Answer 6

To incease the impact time.

As impulse = mΔv = FΔt, for the same car (same m) travelling at the same speed (v) the force of impact will be reduced if the time of impact Δt can be increased.

Question 7

What is the law of conservation of momentum?

Answer 7

The momentum before an interaction is equal to the momentum after the interaction.

Question 8

What are the two types of interactions to discuss momentum in?

Answer 8

Collision: where two objects collide.

Explosions: where one object separates into two.

Question 9

A 0.2kg ball is fired from a gun at 10ms⁻¹.

The ball lands in a 1.2kg wood block.

What is the velocity of the ball and wood block as they move off together?

Write the method to work this out.

Answer 9

Σp before = Σp after

Σp before = ( m[ball] x v[ball] ) + ( m[wood] x v[wood] )
Simplify.

Σp after = ( m[ball] + m[wood] ) x v
Make equal to Σp before, solve for v.

Question 10

A firework travels vertically and reaches maximum hight. It explodes into two pieces. One piece has mass 250g and moves with speed 8ms⁻¹. The other piece has a mass of 120g. Calculate its velocity.

Answer 10

Σp before = Σp after

Σp before = 0 as the firework is stationary

Σp after = (m₁ x v₁) + (m₂)v₂
Make equal to Σp before, make v₂ the subject, solve.

Question 11

How does the law of conservation of momentum applied in interactions within 2 planes?

Answer 11

Momentum is conserved in each direction seperately.

Question 12

What is the method for modeling collisions in two planes?

Answer 12

Draw out before and after.

Resolve vectors (the velocity).

Using the method from 1D interactions find the horizontal Σp before and after to find the unknown.

Using the method from 1D interactions find the verticle Σp before and after to find the unknown.

Combine the vectors for the component.

Question 13

What are elastic and inelastic collisions?

Answer 13

Elastic: collsions where there is no loss in kenetic energy.

Inelastic: collisions where there is a change in kinetic energy, usually loss in KE and gain in internal energy.

MOMENTUM MUST ALWAYS BE CONSERVED.

Question 14

Derive Eₖ = p²/2m from the equations form kenetic energy and linear momentum.

Answer 14

Eₖ = 0.5mv²
p = mv

v² = Eₖ/0.5m = 2Eₖ/m
p² = m²v²

p² = m² 2Eₖ/m
p² = m2Eₖ
Eₖ = p²/2m

QED

Question 15

What does a radian describe?

Answer 15

An angle using the ratio of the arc length (s) to the radius (r)

θ=s/r

Question 16

How many radians are there in a circle?

What equation shows this?

Answer 16

2π radians,

shown by circumference = 2πr

Question 17

What is the angular displacement?

Answer 17

The angle through which a point is rotated.

Question 18

What is the angular velocity and how is it calculated?

Answer 18

The rate of change of angle

SYMBOL: ω

ω = Δθ/Δt

UNITS: rad s⁻¹

Question 19

What is the time period and how is it calculated?

Answer 19

The time taken for one full revolution.

T = 2π/ω

( this links to ω = Δθ/Δt because a full if θ is one full revolution t will be the time taken for it, and so it relates ω)

Question 20

Link velocity (linear) to angular velocity.

Answer 20

ω = Δθ/Δt
Δθ = Δs/r

therefor…

ω = Δs/Δt r
ω = (1/r)Δs/Δt
ω = v/r

Question 21

What is the difference between linear velocity and angular velocity?

Answer 21

• Linear velocity is the rate of change of POSITION.

- Angular velocity is the rate of change of ANGLE.

Question 22

What forces act on a spinning object?

Answer 22

• A velocity acting in the direction of spin. tangential to the radius.
• This component of velocity is CHANGING,
• Therefore (due to newtons second law) it must be accelerating.
• This acceleration is always centripetal.
• And so there must be a centripetal force.

Question 23

How can you work out centrapetal acceleration?

Answer 23

a = v²/r (linear)

a = ω²r (angular)

Question 24

Given two velocities on a spinning object, how can you calculate the change in velocity?

Answer 24

• Combine v₁ and v₂ in a vector diagram.
• Work out the distance between v₁ and v₂, this is Δv.
  (hint: finding right angles and using small angle aprox)

Question 25

Derive a = v²/r from Δv = vΔθ

Answer 25

Δv = vΔθ
(accleeration is rate of change of velocity, so you can introduce time to get…)

a = Δv / Δt = vΔθ / Δt
(angular velocity is the rate of change of angle)

vΔθ / Δt = a = v/ω
(ω = v/r)

a = v²/r

Question 26

Derive a = ω²r from Δv = vΔθ

Answer 26

Δv = vΔθ
(accleeration is rate of change of velocity, so you can introduce time to get…)

a = Δv / Δt = vΔθ / Δt
(angular velocity is the rate of change of angle)

vΔθ / Δt = a = vω
(ωr = v)

a = ω²r

Question 27

What forces are casuing these exmples to undergo circular motion?

• Cyclist on a banked veladrom.
• Single sock in a spin drying mechine.
• Fairground car at the bottom of a roller coaster dip.
• The Moon cicling the Earth.
• Gymanst on high bar.

Answer 27

Cyclist on a banked veladrom:
- Inward component of the force of the track in the cyle’s wheel.

Single sock in a spin drying machine:
- Inward force of the drum of the spin dryer on the sock.

Fairground car at the bottom of a roller coaster dip:
- Upwards force if the track on the car.

The Moon cicling the Earth:
- Pull of Earth on the Moon.

Gymanst on high bar:
- Inwards force of the bar on the gymnast’s hands.

Question 28

What is important to consider when drawing a free body diagram for an object moving in a circle?

Answer 28

• The free body diagram must not be balanced (in the radius plane).
• This is because there must be a resultant force cetrapeltally for circular motion to be maintained.

Question 29

What are all the quantities that should be remembered when answering a question of circular motion?

Answer 29

θ 	= Anglular displacement
t 	= time
T	= Timer period
r 	= radius
ω 	= circular velocity
v 	= linear velocity
a 	= acceleration
F 	= force

Question 30

What are conatants in uniform circular motion?

Answer 30

• Constant angular displacement and angular velocity.
• Constant time period and frequency.
• Constant magnitude of centripetal acceleration and force.

Question 31

Descrive the free body diagrams of an object moving in a vertical circle at the following points (attached by a string)…

• At the bottom.
• At the top.
• Horizontally level with the centre of the cirle.

Answer 31

At the bottom:

• Wieght (mg) acting downwards.
• A larger tension (T) acting upwards.

At the top.

• Both weight (mg) and tension (T) acting downwards.
• Its magnitudes are dependent on the total centrepetal force and the mass of the object.

Horizontally level with the centre of the circle.

• Wieght (mg) acting downwards.
• Tension (T) acting inwards - either left or right.

Question 32

What effects the centrapetal force of an object moving in a verticle circle (attached by a string)?

Answer 32

Centrapetal force (mv²/r) = Weight (mg) + Tension (T)

Weight = constant (always downwards)

Tension = variable (to meet conditions of top equation)

Question 33

Describe and explain how the tension in string of an object being spun in a vertical circle varies at different points in the circle.

Answer 33

• Tension is at a maximum at the lowest point as it must provide the centrapetal force and act against mg.
• Tension is at a minimum at the highest point as a portion of the centrapetal force is provided by mg.
• Between these point tnesion varies sinusoidally as the centrapetal force of mv²/r must always be provided to keep the object moving in a circle.
• At other points the tension varies depending on thee component of mg acting towards the centre (at bottom half tension > mv²/r, at top half tension < mv²/r)

Question 34

What forces act on objects in virtical circular motion…

• on a string?
• on the inside surface of a loop?
• on the outside surface of a loop?

Answer 34

On a string - Tension and weight.

On the inside surface of a loop - Contact force and weight.

On the outside surface of a loop - Reaction force from the ground and weight.

Question 35

Explain why the water does not fall out of the bucket when it is upside down, if it is being swung in a circle at a high enough speed v.

Answer 35

• The bucket experiences a force Fᵦ towards the centre of the circle from the tension in the hand.
• This force Fᵦ provides a contact force F𝓌 on the water accelerating the water towards the centre (acting against the linear velocity acting along the line of motion).
• The water in the bucket has a weight W acting towards the ground.
• The water must experience a centripetal resultant force equal to mv²/r to travel in a circle.
• Provided that the speed v is high enough, the forces F𝓌 and W acting together on the water only provide a resultant force equal to mv²/r.
• The forces acting on the water are enough to maintain circular motion, but not large enough to accelerate it out of the bucket.