HKAL Test Bank (Essay) : Part 1 Mechanics
HKAL Test bank (Essay) : Part 1 Mechanics
Chapter 1 Kinetmatics
1.1Kinematics Definitions
1.1.1Falling in a Viscous Liquid
A raindrop which falls at terminal speed.
(a)Constant velocity →no net force.
(b)The weight of the raindrop, pointing downwards is balanced by the air resistance (drag force), pointing upwards. (Should mention either direction.)
1.1.2Bouncing Ball with / without Energy Loss
A small ball is released stationary from a height of 1 m above a smooth expanse of ground. The ball falls under gravity, hits the ground and bounces up.
(A)Assuming that no energy is lost in the process
(i)By v2 = u2 + 2 as, vy = = 4.5 m s-1. By v = u + at, t = = 0.45 s.
(ii)During the flight of the ball when it is not in contact with the ground, only gravitational force (weight) is acting on the ball. When the ball is in contact with the ground, the reaction force from the ground also acts on it.
(iii)Before the rebound, the acceleration is constant as the ball is falling freely and so its velocity increases uniformly from zero to 4.5 m s-1 before hitting the ground.
(iv)When the ball is in contact with the ground, the acceleration of the ball is very large and directed upward as the upward reaction force is greater than the weight of the ball. The velocity of the ball changes abruptly from downward (+ ve) to upward (-ve).
(v)During the upward flight, the acceleration of the ball is constant and itsvelocity increases uniformly from –4.5m s-1(upwards)to zero at the highest point.
(vi)The momentum of the ball is not conserved when it hits the ground since there is a resultant external force acting the ball – reaction force minus gravitational force. (Remark : However, if the ball and the earth are considered as a single system, the total momentum of the system is conserved since the gravitational forces and reaction forces acting on them are now internal forces.)
(B)If a ball of greater mass is used.
As the acceleration due to gravity is independent of the mass of the ball, the acceleration-time graph and hence the velocity-time graph remain the same.
(C)If some kinetic energy is lost when the ball hits the ground
The magnitude of the velocity decreases after the rebound and hence the time taken for the upward flight decreases.(Accept graphical explanation) The shape of the acceleration-time graph remains unchanged.
1.2Linear Motion (Constant Acceleration)
1.2.1Derivation of the Expression for Kinetic Energy
(a)Let acceleration = a and time 1 2 be t. Force = ma.
(b)Then, a = v/t, s = (0+v)t/2 = vt/2.
(c)Work done = ma s = ½mv2 = K.E., which is the stored energy of body.
1.2.2Stopping Distance
(a)Apparatus : 1 trolley, 1 runway and 1 light gate connected to a timer.
(b)Set up the tilting runway as shown. Arrange a light gate for measuring the speed of the trolley near the lower end of the tilting runway.
(c)The speed of the trolley is calculated from the time taken for the card to pass the light gate.
(d)Measure the stopping distance of the trolley, which is from the light gate up to the place where it stops. Repeat the experiment by releasing the trolley at different heights.
(e)Plot a graph of stopping distance against the square of the speed recorded (representing the kinetic energy of the trolley). A linear graph should be obtained showing the stopping distance is directly proportional to the kinetic energy.
(f)Source of error: the friction at the wheels of the trolley is not constant.
1.3Parabolic Motion (Graphs)
(a)A small ball is projected horizontally with a certain speed from a height of 1 m above a smooth expanse of ground. The ball falls under gravity, hits the ground and bounces up.
(b)The acceleration due to gravity is not affected by the horizontal speed of the ball, both acceleration-time and vertical velocity-time graph remain the same.
1.3.1Daily Example
(a)Shot put (parabolic)
(b)Neglect air resistance, the shot is only acted upon by its weight (mg) whose magnitude is constant and its direction is always towards the ground. As a result, the vertical component of the motion is under constant acceleration towards the ground while the horizontal component is unaffected (i.e. uniform motion).
HKAL Test bank (Essay) : Part 1 Mechanics
Chapter 2 Newton’s Laws of Motion
2.1Newton’s Laws and Momentum
2.1.1Inertia
2.2.1.1Definition
(a)The inertia of a body is its reluctance to change the velocity.
(b)A force is required to change the velocity of the body.
2.1.1.2Example of a Moving Object but with Zero Net Force
A trolley moves (with uniform velocity) on a friction-compensated runway. No net force is acting on the trolley, however it is moving but is not at rest.
2.1.2Newton’s Second Law of Motion
2.1.2.1Meaning
2nd law - The rate of change of momentum of a bodyis proportionalto the resultant force acting on the body and this occurs in the direction of this force.
Alternative Description
The rate of change of momentum produced in a body is proportional to the resultant force acting on it and occurs in the direction of the force.
2.1.2.2Demonstrating Experiment
(a)Set up a friction-compensated runway.
(b)To investigate the relation between force and acceleration, a trolley is pulled by one, two and three identical elastic strings which are stretched by the same amount.
(c)The corresponding accelerations are recorded and a graph of the force (number of elastic strings) is plotted against the acceleration, which shows a straight line passing through the origin (linear relationship).
(d)To investigate the relation between mass and acceleration, use one elastic string to pull one, two and three trolleys. The corresponding accelerations are recorded and a graph of is plotted against the acceleration, which shows a straight line passing through the origin (linear relationship).
(e)Thus, acceleration .
(f)For a body of mass 1 kg and moves with acceleration 1 m s-2, the force acting on it is 1 N.
(g)So demonstrating Newton’s second’s law : force mass ×acceleration.
2.1.2.3Example
(a)For a body, initially at rest, is subject to a constant force.
(b)The body will move with an acceleration a in the direction of the resultant force F and a= F/m, where m is the body mass.
2.1.3Newton’s Third Law of Motion
2.1.3.1Meaning
3rd law - if a body A exerts a force on a body B, then body B exerts an equal but opposite force on body A.
2.1.3.2Example
(A)Stationary rock
(i)A man pushes a heavy rock resting on the ground, but it does not move.
(ii)
The pushing force acting on the rock is balanced by the friction/resistance from the ground, but not by the reaction force which acts on the man.
(B)Jumping boy
(i)W : weight of the boy,W’ : gravitational force acting on the earth by the boy,
R : reaction force acting on the boy by the ground,
R’ : reaction force acting on the ground by the boy.
(ii)The reaction R gives a resultant upward force (R-W) that enables the boy to accelerate vertically.
(C)Launching rockets
(i)When the fuel in a rocket burns, a stream of gas is produced and then escapes at high velocity through the exhaust nozzle. The exhaust gases are pushed backwards by the rocket.
(ii)Therefore a reaction acts on the rocket in the opposite (upward) direction and it is this force that overcomes its weight and enables it to accelerate.
2.1.4Feeling of One’s Weight
2.1.4.1Mass and Weight
(i)The mass m of a body is constant and is a measureof its inertia to any enforced change of state,either stationary or moving. For a force F thebody experiences an acceleration 1/mass.
(ii)The weight W of a body can be variable and is theforce which would act if the body were allowed tofall freely under the influence of gravity.
W = mg, where g is the local 'free-fall'acceleration due to gravity.
(iii)In a lift moving downwards with an acceleration(or for a lift moving up, suddenly stops), therewill be less reaction between the weight (force)of the passenger and the floor of the lift, givingthe impression of weightlessness.
2.1.4.2Example
A car travelling over a hump which is an arc of a vertical circle.
(i)The car and hence the passenger is undergoing circular motion, therefore he is experiencing a centripetal acceleration a = v2 / r.
(ii)There are two forces acting on the passenger:
(I)the force of gravity, W = mg, acting downwards,
(II) the force of the seat, R, acting upwards.
(iii)As mg – R = mv2 / r (part of the weight is used as centripetal force)
i.e. R = m(g – v2/r) < mg (force of the seat is smaller than weight/normal) (iv) Thus the passenger would feel lighter than normal.
2.2Momentum and Impulse
2.2.1Proof of the Principle of Conservation of Momentum
For two objects having a head-on collision.
A B
Before collision m1, u1 and m2, u2
After collision m1, v1 and m2, v2
(a)For time of impact t and u1u2.
(b)Body A will exert a force F on body B for a time t and by Newton's3rd law body B will exert a force -F on body A (opposite direction).
(c)By Newton's 2nd Law, force = rate of change of momentum (constantof proportionality = 1)
(d)Hence since forces on bodies are equal and opposite.
(e)So m1u1 + m2u2 = m1v1 + m2v2,and linear momentum is conserved.
Alternative Description
Consider the head-on collision between two billiard balls A and B moving with velocity u1 and u2 respectively (u1u2) in the same direction.
(a)Let FA and FB be the average forces acting on A and B respectively during collision and Δt be the time during which the two balls are in contact.
(b)By Newton’s second law, the impulse FAΔt will change the momentum of ball A, i.e.
FAΔt = m1v1 – m1u1
(c)Similarly, the momentum of ball B will change by FBΔt,
FBΔt = m1v2 – m2u2
(d)By Newton’s third law, FA = -FB (equal in magnitude but are opposite in direction), therefore
m1v1 – m1u1= -(m2v2 – m2u2),m1u1 + m2u2 = m1v1 + m2v2.
2.2.2Three Types of Collision
2.2.2.1Definition
(a)In a perfectly elastic (or elastic) collision, the kinetic energy of the system is conserved.
(b)In a perfectly inelastic collision, the two objects move together after collision.
(c)In a partially elastic collision, the kinetic energy of the system decrease but the two objects move separately after collision.
Alternative description
The tracks of an α-source are observed in a diffusion cloud chamber in which a trace amount of helium is introduced.
or
(a)Elastic collision
(i)Total kinetic energy conserved,.
(ii)Total momentum (vector) is conserved, or.
(b)Non-elastic collision
(i)Total kinetic energy is not conservedsome of energy appearing in another form dueto work done against an internal force, increasing the internal energy such ase.g. heat.
(ii)Total momentum (vector) is conserved.
2.2.2.2Conservation of Kinetic Energy
For two objects having a head-on collision.
(a)K.E. may not be conserved since some energy may be converted in another form such as heat and sound.
(b)Total energy is, however, conserved.
2.2.2.3Perfectly Elastic Collision
Ball of mass m1 moving with velocity u1 undergoes head-on collision with another ball of mass m2 which is initially at rest on a smooth horizontal surface. The collision is perfectly elastic.
(a)Let the velocity of m1 after collision be v1.
By the conservation of linear momentum m1u1 = m1v1 + m2v2 orm1 (u1 – v1) = m2v2---(1).
As the collision is perfectly elastic, = + ,m1 (–) = m2 ---(2).
Combining (1) and (2) m1 (–)= ,
m2 (u1 + v1)= m1 (u1 – v1),
v1= .
(b)(i)For m1m2, therefore v1u1.
For example the velocity of a bowling ball is hardly affected by collision with an inflated beach ball of the same size.
(ii)For m1 = m2, therefore v1 = 0.
For example a billiard ball stops when it collides head-on with another stationary billiard ball.
(iii)For m1m2, therefore v1 -u1.
For example a ball drops vertically onto the ground (collision with the earth) and rebounds with a reversed velocity and will reach the same height.
(c)To slow down the fast neutrons effectively, the stationary targets for collision should be of nearly the same mass of neutrons. Therefore materials with numerous hydrogen centres (such as water) are preferred.
2.2.3Collisions in 2-dimensions
or
(a)Let = velocity of α-particle before collision, = velocity of α-particle after collision,
= velocity of the atom after collision.
(b)By conservation of momentum, mα+ 0 = mα+ mHe.
(c)As the track is right-angled, (mαu)2 = (mαv1)2 + (mHev2)2, u2 = v12 + (v2)2 --- (1)
(d)As the collision is elastic, K.E. is conserved
mαu2 = mαv12 + mHev22, u2 = v12 + v22 --- (2)
(e)Comparing (1) and (2), mα= mHe.
2.2.4Examples
(A)A billiard ball strikes the smooth cushion of a billiard table at an angle and rebounds with the same speed.
(i)Momentum of the billiard ball is not conserved.
(ii)Its momentum perpendicular to the cushion is altered due to the normal reaction.
(B)A rocket rises vertically upward during launching in the atmosphere near the earth’s surface.
Momentum of the rocket is not conserved as it is subjected to external forces such as thrust due to ejecting gases, gravity, air resistance.
(C)A radioactive nucleus emits an α-particle.
When an α-particle is emitted, the daughter nucleus recoils in opposite direction and therefore its momentum is not conserved.
(D)A ping pong ball which collides obliquely with a smooth wall.
(i)When a ping pong ball collides with the wall, the velocity/direction of the ball changes.
(ii)A net force must have been present while during collision, whose direction is perpendicular/normal to and away from the wall.
2.3Work and Energy
2.3.1Conservation of Mechanical Energy
(a)(i)A ball thrown vertically between two heights h1(h1 + h) will suffer a decrease of
velocityu1u2.
(ii)Since resistance of air can beneglected,decrease in K.E. = increase in P.E.
(iii)½mu12 - ½mu22 = mgh, where m is mass of bodyand g is free-fall acceleration.
(b)If body were thrown in a viscous medium -such as water, mechanical energy would not beconserved. In addition work done against thisopposing force would produce heating of the medium.
In fact loss of K.E. = gain of P.E. +gain of internal (heat) energy.
(c)Bouncing ball
(i)No equilibrium position between the two extreme ends and net force is acting.
(ii)The force on the ball is directed towards the centre of the earth except at the moment of rebound.
(iii)The force on the ball is constant except at the moment of rebound.
2.3.2Together with the Principle of Conservation of Momentum
The diagram shows a set-up used to measure the speed of a bullet in the laboratory.
The bullet of mass m is ‘fired’ horizontally towards a block of wood (of mass M, in which a hole has been drilled) suspended from two vertical inextensible strings (each of length L). On striking the block, the bullet is embedded and the block rises by swinging through an angle θas shown.
(A)Methods to ‘fire’ the bullet in laboratory
(i)Use a spring-loaded gun (i.e. a compressed spring).
(ii)Use cotton wool or plasticine inside the hole to ensure the collision to be inelastic (i.e. sticky).
(B)Experimental and Theoretical values of v
(i)The horizontal momentum of the system is conserved, therefore mv = (m + M)V where V is the common velocity just after impact.
(ii)After the collision, the only forces acting on the system (block + bullet) are the weight and the string tensions (which do no work), therefore the mechanical energy is conserved.
(iii)From energy conservation, (m + M)V2 = (m + M)gh.
As h = L(1 – cosθ), by eliminating V, we have .
(iv)During the impact the supporting strings may deviate a bit from vertical, therefore some horizontal external force would act on the system resulting in the experimental value lower than the theoretical value.
(v)L should be measured to the centre of gravity of he block instead to the top of it.
HKAL Test bank (Essay) : Part 1 Mechanics
Chapter 3 Circular Motion
3.1Definition
3.1.1Centripetal Acceleration
(a)For a satellite around the earth (uniform circular motion).
(b)The satellite is pulled by the earth (i.e. its own weight mg’) and this force is constant and its direction is always towards the centre of the earth/motion.
(c)The whole of this force is used for centripetal acceleration or merely changing its direction of motion, the (tangential) speed of the satellite remains unchanged.
3.1.2Derivation of the Centripetal Force Equation
(a)Draw vectors , and on completing , givesvector representation of velocity change AB.
(b)Since 0, direction at B is along , magnitude of velocity change = vΔ.
(c)Since AB = vΔt = rΔ, acceleration, a = vΔ/Δt = v2/r = r2, (v = r).
(d)Force = ma = mr2, directed towards O.
3.1.3In Circular Motion, the Speed can be a Constant
(a)A constant (centripetal) force F acts towards the centerO of the circle.
(b)This provides an acceleration in thisdirection for the rotating body.
(c)Only the speed v ofbody is constant, its velocity (and momentum) are continuously changing - due to the acceleratingforce F.
Alternative description
(a)A body (e.g. a satellite revolving round the earth) moving with uniform circular motion. Although the speed is unchanged, the body continuously changes its direction (velocity).
(b)The momentum changes. This is brought about by the (centripetal) force.
3.1.4Dependence of Period
(A)For period be a constant, when r increases, v also increases linearly with r as period T = is constant, therefore a = actually increases.
(B)For angular speed be a constant, when r increases, a = ω2r increases provided that ω= is constant, which is ensured by constant period T.
3.2Uniform Motion in a Horizontal Circle
3.2.1Conical Pendulum
(a)T is tension in string, F centripetal force.
(b)Resolving (1) vertically, mg = T cos, forces (2) horizontally F = T sin
(c)As F = ml sin 2, hence, cos = .
(d)Thus it follows that as increases, increases.
(e)As , /2, i.e. string becomes horizontal.
3.2.2Experiment to Verify the Equation for Centripetal Acceleration
(a)The glass tube is held vertically, the bung is whirledaround above his head by one student and the speed ofbung is increased until the marker is just below tube.
(b)Another student times, say, 50 revolutions of the bung.By moving marker the length l of the string can be variedand the relation between l and the angular velocity ()obtained. Since T (tension provided by screwnuts*) andmg are constant from T cos = mg it is clear that isconstant. Also T sin = mr², 1/.
(c)However there is friction between string and glass rodwhich may vary throughout experiment.
Alternative Description
(a)The string dips so that the vertical component of the tension balances the weight of the rubber bung.
(b) W cosθ= mω2r(as T = W)
W cosθ= mω2 (L cosθ)
W= mω2L
(c)Friction exists at the opening of the glass tube.
(i)The rubber bung is not swirled with constant speed.
(ii)The string is not inextensible.
(iii)The rubber bung is not swirled in a horizontal circle.
(d)Improvement
(i)Increase the weight W.
(ii)Reduce the mass m of the rubber bung.
(iii)Reduce the length L of the string.
(e)The rubber bung goes along tangential direction to A due to inertia and falls as a projectile to the ground due to gravity.
3.2.3Centrifuge
A closed tube containing a mixture of two liquids of densities ρand ρ’ (ρρ’) is attached at the end by a hinge (allowing vertical motion) to a rigid rod. If this rod, and also the tube, is rapidly rotated in a horizontal plane with an angular velocity ω.
(A)Excess force
(i)Due to rapid rotation, , tube ~ horizontal. Pressures P, P’ = P + P.
Net inward force on liquid density in elemental volume,