Classified Past Papers Physics Edexcel Unit 1

 Past papers for unit 1 

Unit 1: 

Forces and motion 

The paper has question from paper 1 and 2 from higher tier.

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Weight and mass:

The strength of the Earth’s gravitational field, g, is 10 N/kg.The weight, W, of an object is the force that gravity exerts on it, which is equal to the object’s mass × the pull of gravity on each kilogram

.

The size of g also gives us the gravitational acceleration, because:

Weight is the pull of gravity on an object. Weight is a force measured in newtons, N. The weight of an object depends on the gravitational field strength of a planet.The mass of an object is always the same anywhere and is not affected by a planet’s gravitational pull; it depends on the amount of matter in the object. Mass is measured in kilograms, kg.

Falling and parachuting:

The size of the air resistance on an object depends on the area of the object and its speed:

·        The larger the area, the larger the air resistance.

·        The larger the speed, the larger the air resistance.

The size and shape of the two balls are same but the mass of red ball is 0.1 kg and mass of blue ball is 1 kg. The pull of gravity on the red ball is balanced by air resistance, so it now moves at a constant speed. It will not go any faster and we say it has reached terminal velocity. Nevertheless, the blue ball has more gravitational pull than red ball.

Another example for parachute falling:

Stopping distance:

The stopping distance is the sum of your thinking distance and braking distance. The thinking distance is the distance the car travels while you react to a hazard ahead – it takes time for you to take your foot off the accelerator and apply the brake.

What affects your reaction time?

·        Tiredness from long driving.

·        Medications which make a person less concentrative.

·        Alcohol and drug alter the conscious levels and affects thought process.

·        Distractions while driving.

Braking distance:

·        Road condition: Icy or wet roads reduces the grip on the tyres.

·        Area of car being driven: sand, muddy and gravels reduce the grip of the tyres.

·        Condition of the tyre: Damaged tyres can reduce their grip on the road.

·        Condition of brakes: old brake pads reduce the braking force and increases the braking time.

·        Loaded vehicle: A heavily loaded vehicle will have more mass, so there is less deceleration. Thus, braking time is increased.

·        Hills/slope: braking distance increases as the gravity pulls the car down.

·        Visibility: affected in certain weather conditions such as heavy rain, fog and so the braking distance is increased.

Stretching:

When the forces remain balanced, the objects remain at rest (stationary).However when two balance forces are applied, they can:

• stretch a spring.

            Or

• compress a beam.

If three or more balanced forces are applied to a beam it can bend.

When an object has been stretched there are two possibilities after the removal of load:

a)   It might return to its original length this is known as elastic deformation.

b)   It might not return to its original length, this results in permanent stretching, and this is known as inelastic deformation.

Elastic deformation:

An equation linking force exerted on spring and extension of the spring is shown below:

F=Force in Newtons (N)

K= Spring constant in Newton per metre (N/m)

e= extension in metres (m)

K i.e spring constant gives idea about the stiffness of a spring how easy it is to deform.If the K value is low it is easy to stretch the spring, if it is high it needs more amount of force to deform the spring.

The same equation can  be used when spring is getting compressed in that case e is the distance by which the spring has been compressed.

Investigating the relationship between force and the extension of a spring

Method:

a)   Set up apparatus as shown in picture. Make sure that the ruler is in line and close to the spring.

b)   Take the reading from the bottom of spring on the metre rule. This is initial length (L1).

c)   The spring is extended by placing a measured (100g) of weight . This exerts a force of 1.0 N. Take the reading from new position of the bottom of the spring. This is Final length.

d)   Calculate the extension of the spring (final length-initial length)

e)   Now construct a table to record the applied force, original length,extended length and the extension.

f)    Add on particular weight so that the spring is stretched by a force of 2 N. Record the new length Length-2 (L2) , and new extension L2 – L1 .

g)   Repeat the procedure up to a weight of 10 N. and record the data.

Analysing the results:

a.   Plot a graph of extension ( y-axis ) versus force applied (x-axis).

b.   Draw a smooth and straight line of best fit. ( Line o best fit is a line which touches maximum number of plotted points)

c.   Draw a suitable conclusion from the graph.

Advisory notes:

·      Before beginning the experiment, investigate that 10 N is an appropriate force for the experiment.

·        New springs are compressed and so it is better to begin with one weight on the spring and use that length as L0.

·        Recheck that the retort stand is placed properly to avoid any chances of toppling over.

 

Limit of proportionality:

The increase in extension of the spring is proportional to increase in load applied as shown in the graph. Beyond a certain point, the extension is not linear and that point is known as limit of proportionality, After this point the extension of the spring much greater. Spring constant (K) can be found by dividing force by extension as shown below:

Initially the spring obeyed Hooke’s law; i.e. there was linear increase in growth as force applied was increased.

Momentum: Is the product of velocity and mass.

Measured in kilogram metres per second (kg m/s).

Velocity is vector and momentum comprise velocity so momentum is also a vector quantity.

Forces and change of momentum:

Force is a product of mass and acceleration.

Acceleration formula sustituted in the above equation becomes:

As mass times change in velocity is equal to change in momentumthe equation that can be used :

 

Momentum (same as energy) is conserved. When two bodies collide, the total momentum before = total momentum after collision.

Newton’s third law:

For every force, there is an equal and opposite force.

Turning moments:

A turning moment is result of force exerted about a point. Thus, turning moment is increased by applying a large force and by using a long lever.

Perpendicular means at right angles from the pivot.

Centre of gravity

The point that the weight (W) acts through is called the centre of gravity. As an object is stationary, the pivot must exert an

upwards force R on it, which is equal to W.

In equilibrium of moments:

the sum of the anti-clockwise moments = the sum of the clockwise moments. The sum of the forces are same on both sides.

Newton’s 3rd law tells us that every force on an object , will have an equal and opposite force.