Mechanics: Newton's Laws of Motion

Question 1

Newtons first law of motion.

Answer 1

Objects either stay at rest or move with constant velocity unless acted on by a resultant force.

Question 2

Newton’s second law of motion.

Condition required?

Answer 2

By defining force (N) as resultant force that will give an object of mass 1kg an acceleration of 1 m/s^2 , F = ma.

Only true for constant mass.

Question 3

Acceleration direction?

Answer 3

Acceleration is always in the same direction as the resultant force, no matter the direction of motion.

negative acceleration = positive deceleration

Question 4

When object in eqm… S and W?

Answer 4

Support force on it is equal and opposite to its weight.

Question 5

What is inertia?

Answer 5

The mass of an object is a measure of its inertia, which is its resistance to change of motion.

Question 6

F = ma problems coming up!

Answer 6

Rocket problems, lift problems, pulley problems, sliding down slope problems.

Question 7

Rocket:
If T is the thrust of the rocket engine when its mass is m, and the rocket is moving upwards, its acceleration a is ?
T is ?
Condition for takeoff?

Answer 7

a = T - mg = ma
Rocket thrust, T = mg + ma

Therefore rocket thrust must overcome the weight of the rocket for the rocket to take off.

Question 8

Lift:
What is resultant force in lift, if T is tension in lift cable and m is total mass?

If lift moving up and accelerating vs down and dec vs .. vs.. then tension eq where ma </> mg.

Answer 8

T - mg = ma
T = mg + ma

Eg up + acce T = mg + ma > mg

Tension in cable less than weight if:
- up + dece (v > 0 and a < 0)
- down + acce (v < 0 and a < 0)
Tension in cable greater than weight if:
- up + acce (v > 0 and a > 0)
-down and dece (v< 0 and a > 0)

Question 9

Pulley:
2 masses M and m attached to either side of string on pulley.
When released, M accelerated down and m accelerates up

Answer 9

M: Mg - T = Ma
m: T - mg = ma

Therefore Mg - mg = (M+m) a

Question 10

Sliding down slope:
Block sliding down slope resultant F?

Answer 10

Resultant force on block as it accelerates down against resistance = mg sinθ - F0

Question 11

Drag force (when moving through fluid) depends on…

Answer 11

-Shape of object (streamline vs not)
-Its speed,
-The viscosity of the fluid which is a measure of how easily the fluid flows past a surface.

Question 12

Explain terminal velocity. (6)

Answer 12

-Speed of object as it falls through fluid increases as it falls;
-So drag force increases;
-Resultant force is difference between weight and drag force.
-As drag force increases, resultant force decreases, therefore acceleration decreases.
-If continues to fall, attains terminal velocity;
-which is when D = - W. a = 0 and speed is constant.

Question 13

Acceleration eq at any instant during this process?

Answer 13

At any instance F = mg - D
so a = mg - D / m = g - D/m

Question 14

Initial acceleration in this process?

Answer 14

Initial acceleration = g because speed = 0 so D = 0.

Question 15

At terminal speed, what’s the energy transfer that occurred?

Answer 15

At terminal speed, potential energy of object is transferred, as it falls, into internal energy of fluid by the drag force.

Question 16

Cars:
What is motive force?
What is resistive force?
What is resultant force, and a?

Answer 16

Motive force, Fe = driving force provided by the engine.
Resistive force, Fr = sum of drag forces acting on the vehicle.
Resultant force, F = Fe - Fr
So acceleration = Fe - Fr / m

Question 17

If you reach terminal velocity freefalling, the open parachute, what happens?

Answer 17

Drag force increases therefore speed drops until drag force = weight again. New lower terminal velocity reached.

Question 18

What is thinking distance, S1?

(assume constant speed)

Answer 18

The distance travelled by a vehicle in the time it takes the driver to react. If speed is constant, S1 = speed x reaction time = u x t0

Question 19

What is braking distance, S2?

(assume constant deceleration)

Answer 19

The distance travelled by a car in the time it takes to stop safely from when the brakes are first applied. Assuming constant deceleration, a, to zero speed from speed u;
S2 = u^2 / 2a

Question 20

What does braking distance depend on?

Answer 20

-Speed of vehicle
-Condition of vehicle tyres
-Road conditions eg if road is icy, skidding is more likely because limiting friction is reduced from dry value.

Question 21

What is stopping distance?

Answer 21

Thinking distance + braking distance.
ut0 + u^2/2a

Question 22

True or false: a car can only move forward by pushing backwards on the road?

Answer 22

True!

Question 23

Skidding - friction between tyres and road prevent wheel spin (slipping) so the driving wheels therefore roll along the road.

Answer 23

If driver tries to accelerate too fast, the wheels skid. This is because there’s an upper limit to the amount of friction between the tyres and the road.

Question 24

Skidding - when the brakes are applied, wheels slowed so vehicle slows, provided wheels don’t skid.

Answer 24

If braking force increased, the friction force between the tyres and the rod increase. However, if upper limit of friction/limiting frictional force reached, wheels skid. When this happens, the brakes lock and the vehicle slides uncontrollably forward.

Question 25

Skidding - what do most vehicles now have to stop brake locking?

Answer 25

Anti-lock brake system (ABS). It reduces the brake pressure on the wheel to stop it locking.

Question 26

How to measure limiting friction between underside of block and surface ?

Answer 26

Limiting friction is equal and opposite to pull force on block just before sliding occurs. Use newton-metre.

Question 27

What does braking distance depend on?

Answer 27

-Speed of vehicle
-Condition of vehicle tyres
-Road conditions

Eg if road is icy, skidding is more likely because limiting friction is reduced from dry value. To stop vehicle, brake must be applied with less force than on dry otherwise skidding. THEREFORE braking distance longer than on dry road.

Question 28

How to easily relate impact force of wall on vehicle with weight:

Answer 28

By expressing acceleration or deceleration in terms of g. Eg a = -3g means impact force = 3mg (3 times weight)

Question 29

When two objects collide and bounce off each other, they’re in contact for a certain time. The shorter the contact time…

Answer 29

the greater the impact force for the same initial velocities of the two objects.

Question 30

If the objects remain tangled..

Answer 30

they exert forces on each other until they’re moving at the same velocity.

(here duration of impact force does not equal contact time)

Question 31

Impact time, t (the duration of the impact force) can be worked out by…

Answer 31

applying t = 2s / u+v to one of the vehicles, where s is distance moved by vehicle during impact.

Question 32

If vehicle mass known, can work out impact force by..

Answer 32

t = 2s / u+v
a = v-u / t
F = ma

Question 33

The work done by the impact force over an impact distance (=Fs) is equal to what?

Answer 33

The change in Ek of the vehicle.

The impact force can also be worked out using
F = change of Ek / impact distance

Question 34

Reduce impact force by increasing impact time by what 5 interventions?

Answer 34

1) Vehicle bumpers
2) Crumple zone
3) Seat belts
4) Collapsible steering wheel
5) Airbags

Question 35

Vehicle bumpers

Answer 35

Vehicle bumpers - give way a little in low-speed impact so increase impact time

Question 36

Crumple zone

Answer 36

Crumple zone - gives way in front-end impact so increase impact time.

Question 37

Seat belts

Answer 37

Seat belts - restrains wearer from crashing into vehicle frame after sudden stop, without holding them rigidly. Restraining force &laquo_space;impact force on wearer if hit frame. Seat belt stops wearer more gradually than without.

Question 38

Collapsible steering wheel

Answer 38

Collapsible steering wheel - if driver contacts steering wheel, impact force lessened as a result of wheel collapsing in impact.

Question 39

Airbags

Answer 39

Airbags - acts as cushion and increases impact time, and also force on impact spread over contact area which is greater than contact area with seatbelt so pressure on body is less!

Question 40

Explain how these actually reduce injury?

(real 5 marker)

Answer 40

Moving driver has momentum.
In sudden impact momentum must be lost in very short time.
F = Δ (mv) / Δ t
Above interventions increase stopping contact time/impact time.
Hence reducing impact force.