AL Syllabus Notes CHAPTER Force. Work (as a measure of energy transform). Power. Newtonian mechanics. Conservation of linear momentum. Transformations between P.E. & K.E. Resolution of coplanar forces. Revision & consolidation of lower form work. Useful Mathematics in Physics (I) Vector / and Scalar / There are tow kinds of physical quantities :. Scalars have magnitude only. They can be added algebraically.. Vectors have both magnitude and direction. They can be added geometrically or analytically and vector symbol is usually specified with an arrow ( ). The length of the arrow represents its magnitude and the head shows its direction. Examples of scalars and vectors: Scalar quantities Vector quantities length l Displacement x r r time t force F speed v velocity v r r volume V momentum p mass m angular momentum L r current I torque / τ r pressure P acceleration a r r density ρ area A (A) Addition of Vectors () Geometric method (a) Parallelogram(triangular) method The resultant of two vectors a r, b r, which acted at angle θ can be represented by the diagonal of parallelogram. The vectors are linked up tip to tail and the direction is as shown below. o a + b abcos(( 80 θ ) = R a + b + abcosθ = R a sinθ tanφ = b + a cosθ φ R θ a b CSKMS AL PHYSICS M 004 by Leo Chung P.
(b) Resolution Multiple vectors can be resolved into x and y components. The x and y components of them can be added up algebraically. Then the resultant vector can be calculated from the x and y components based on Pythagoras Theorem. Example From the diagram, fond the resultant vector F of the following three vectors of forces acting on a point O. F F F F x x y y F = tanθ = 50 cos 45 = 3.N = 50 sin 45 = 53. N F x + F 53. 3. y o o 00 cos 30 + 00 sin 30 = 3. o o θ = 76. or03.8 (B) Subtraction of vectors o o + 53. 60 cos 40 60 sin 40 o = 54.8N o 60 N 00 N F 50 N 30 o 45 o 40 o O There are two methods to draw the resultant vector of vector a r - vector b r. r () tail to tail drawing of two vectors a and b r. Then, then resultant vector is formed by drawing a vector from r b to a r. () The resultant vector can be treated as vector a r add up with vector (- b r ). -b b s a s = a b = a + ( - b ) Class work:. A man is walking due west at a speed m/s on the deck and the ship is travelling due north at 4 m/s. Find the velocity of man with respect to (w.r.t. ) the ground. (4.47 m/s,6.5 o ). A man is travelling due north at m/s. He feels that the wind is blowing towards him from east at 4 m/s. What is the velocity of the wind w.r.t. to the ground? 4.47 m/s, N 63.4 o W) CSKMS AL PHYSICS M 004 by Leo Chung P.
(C) Multiplication of vectors I ) Dot Product (scalar product) the product become a scalar : e.g. work = force x displacement = F. S = F S cos θ a. b = a. b cos θ II) Cross Product (vector product) the product is still a vector : c = a x b = a b sin θ ( right hand rule ) II. Newtonian Mechanics I Definition () Distance : it is the length of actual travelling path. () Displacement : it is the distance moved in a constant direction. Suppose a particle travelled from A to B by the path l. Displacement is independent of the path of motion and just dependent on the final and initial position. Displacement = x r = x r x r f i There is no need to account the path l. (3) Average speed is defined as the ratio of total distance moved and the time required. Average speed = total distance moved / total time (4) Average velocity is defined as the rate of change of displacement over a period of time. average velocity = total displacement / time taken = x t (5) Instantaneous velocity is defined as the rate of change of displacement in a very short time interval. lim x dx v = = = slope in displacement vs time graph t 0 t dt (6) Instantaneous acceleration is defined as the rate of change of velocity in a very short interval. lim v dv a = = = slope in velocity vs time graph t 0 t dt CSKMS AL PHYSICS M 004 by Leo Chung P.3
Class work The motion of a particle is given by the following equation x=t - t + Find (a) The average of velocity from t= to t=s, (b) the average velocity from t= to t=. s, (c) the average velocity from t= to t=.0s, (d) the instantaneous velocity at t=s. II Graphical representation CSKMS AL PHYSICS M 004 by Leo Chung P.4
III Equation for uniform acceleration v u a = t v = u + at v u x = + t s = ut + at v = u = as IV Equations of non-uniform acceleration If the acceleration is a function of time, the velocity and displacement should be calculated by integration. dv a = dt v = x = adt vdt CSKMS AL PHYSICS M 004 by Leo Chung P.5
Class work ()The velocity of an object is v=4t + t +, find (a) the acceleration at t=0 and t=3s (ms -, 6ms - ) (b) the displacement from the time t= to 4 seconds. (0m) () A stone is thrown vertically upward from a height of.0 m with a speed of 5.0 ms -. Find the veloc ity of the stone as it reaches the ground and the time of flight.(-8.06 ms -,.3s) V Terminal velocity When an object is falling within a fluid (liquid or gas), its velocity will increase to a maximum value. This maximum velocity is called terminal velocity. It is because there exists viscosity (resistance of fluid). The magnitude of viscosity force is directly proportional to on its shape, size and surface area. the velocity of the falling object and depends At first, weight is greater than the sum of upthrust and viscous force, so the object accelerates. W > U + f Later, as the viscous force increases with velocity, weight will be equal to the sum of upthrust and viscous force. W = U + f Therefore, the net force become s zero and the object reaches a constant velocity. CSKMS AL PHYSICS M 004 by Leo Chung P.6
VI Newton`s Law of Motion (A) Newton`s First Law Issac Newton (64-77) summarized Galileo s ideas as Newton s First Law: Everybody remains at rest or in a state of straight line uniform motion unless a resultant force acts on it. This seems to be contradictory to our daily experience. Everything will stop its motion if it is not act by an external force. For example, a cyclist has to exert a force in order to keep the bicycle in motion. This explained by the existence of frictional force. If a body moves in a frictionless surface, it will continue with a uniform velocity. What is force? Force is an invisible thing which changes the state of motion of a body. First law introduces the concept of inertia. Every has a tendency to keep its state of motion. When a car stops suddenly, the driver has a forward tendency. Inertia of a body is the reluctance of it to change its state of motion. Greater inertia,means a body needs a greater force to change its state of motion. In our daily life, we treat mass as the amount of matter. While in Newtonian mechanics, mass is the measure of inertia. Greater mass means a large force to produce certain acceleration. Here, we can use the Newton`s second law to measure the mass of a body. The mass(inertia) of a body is defined by the ratio of resultant force acting on it and its acceleration. (B) Newton`S Second Law The rate of change of momentum is directly proportional to the resultant force and it takes place in the same direction of the resultant force. F F F d ( mv) dt k d ( mv ) d ( mv) = = dt dt = m dv + v dm = m dv = ma dt dt dt if k= Actually, first law is a typical case of second law. If the resultant force acting on a body is zero, the body will keep its state of motion as its acceleration is zero. It is important to know that the force will not only change the momentum of a body but also the direction of momentum. The direction of acceleration is the same as the direction of force. The direction of velocity has no connection with the direction of force. CSKMS AL PHYSICS M 004 by Leo Chung P.7
(C) Newton`s Third Law If a body A exerts a force on body B, then B exerts an equal magnitude but opposite direction force on A. F A F B F A = -F B Consider a system containing two bodies A and B, if there is no external force acting on this system, the net force is zero. When the body A exerts a force F B on the body B. A must receive a force F A in order to maintain the net force on the system is zero, such that F A + F B = 0 this pair of forces is called action and reaction force. They exist mutually and have a cause and effect relationship. It is important to note that they are acting on different bodies. CSKMS AL PHYSICS M 004 by Leo Chung P.8
Class work: A worker of mass 60 kg standing on a lift pulls down a rope with a force of T. If the mass of lift is 0 kg and the normal reaction on the man is 300 N. (a) Draw free body diagrams of the man and lift. (b) Find acceleration of the lift and the tension of the rope. VI Friction When the surface of a body moves or tends to move over another surface, each body experiences a frictional force. These forces are action and reaction pair forces. The frictional force has the following properties. (a) Frictional force always opposes motion and does not have a fixed value. It increases as the other forces tending to produce the motion increase, always trying to make the net force zero and to keep the object at rest. (b) The frictional force cannot increase indefinitely, but has a maximum value L called Limiting static friction. If the resultant of other forces exceeds L, the object will move. (c) In general f < L, but when the object starts to move, f = L. (d) The value of L is independent of the area of contact, but is roughly proportional to the norma l force N pressing the two surface together. L= µ s N, where µ s, is called coefficient of static friction. The value µ s, depends on the surfaces in contact and 0 < µ s <. (e) Once the object to move, the frictional force f is slight less then L but is still approximately proportional to N. f = µ k N where µ k, is called coefficient of kinetic friction. (f) The frictional force between in the equation of part (e) is essentially independent of velocity v. This is great contrast to air resistance, which is roughly proportional to v. CSKMS AL PHYSICS M 004 by Leo Chung P.9
Class work: N F () An object of mass m = kg is pulled up an incline of angle θ=30 o by a force F= 5 N. If the frictional force is 3 N. (a) Find the normal reaction N and the acceleration of the object (b) Find the coefficient of kinetic friction µk. (c) If µ s = µ k, what is the steeper inclined on which the object can rest without slipping. (8.5 N, 7. ms -,0.353,9.4 o ) CSKMS AL PHYSICS M 004 by Leo Chung P.0
() A large trolley of mass M with a flat top can roll frictionless on a horizontal bench. A mass m is projected horizontally on the top of the trolley with speed v 0. If the coefficient of kinetic friction is µ k. Find (a) the acceleration a of m, (-µ k g) (b) the acceleration a` of the trolley, (c) The distance s covered by the m on the trolley top before stopping, and (d) the final common velocity of the system. (v=m v 0 /m +M) VII Moments & Couples A. The moment or torque of a force about a point is measured by the product of the force and the perpendicular distance from the line of action of the force to the point. Moment τ = F r τ = F r sin θ In equilibrium, upward force = downward force, However, effect is to turn the rod.(i.e. turning effect) CSKMS AL PHYSICS M 004 by Leo Chung P.
F F F sin θ Pivot A θ O A θ Moment of F about O = F x O A (clockwise) Moment of F about O = F x O C = F x O B sin θ If we resolve F into F sin θ & F cos θ, i.e. the nd figure, then we have, Moment of F about O = moment of F sin θ about O + moment of F cos θ about O = F sin θ O B + F cos θ O = F sin θ O B Usually, clockwise moment is +ve, Anti-clockwise moment is ve (Unit : Nm ) B. Couple : Two equals but opposite parallel forces whose lines of action do not coincide. F F Moment of a couple about any point = F l = one force dist. between the forces CSKMS AL PHYSICS M 004 by Leo Chung P.
C. Equilibrium of Coplanar forces: The necessary & sufficient conditions for a rigid body to be in equilibrium when subjected to a system of coplanar forces:. Translational, the algebraic sum of all forces in any given direction is zero F = 0 ( F x = 0, & F y = 0). Rotational, the algebraic sum of moments of all the forces about any point must be zero (Principle of Moment) M = 0 AL Syllabus Applications of the Principle of conservation Of linear momentum in one & two dimensions. Notes Distinction between elastic & inelastic collisions Principle of measuring Inertial mass e.g. using m x /m v =?V v /?V x for explosive separation of two masses initially at rest. Equivalence of inertial & gravitational mass. E.g. of linear conservation include recoil of rifles, collision of aparticles with He atoms.(analysis of cloud chamber photographs.) VIII. Momentum A ) Conservation of Linear Momentum The concept of linear momentum, or momentum for short, can be introduced using Newton s Laws. r r r r r r r F ma m v v mv mv ( mv ) = = = = t t t We then defined the momentum p = mv The Newton s second law stated as : The net force acting on an object is equal to the rate of change of momentum. in symbols r r r p r dp F = orf = t dt If F ext = 0, dp/dt = 0, p = constant. Another Proof of The principle of Conservation of Momentum It state that the total linear momentum of colliding bodies remains constant if there is no external force acting on them. CSKMS AL PHYSICS M 004 by Leo Chung P.3
Proof : Consider two bodies collide with each other. m u + m u = m v + m v t= contact time By nd Law: rate of change momentum is equal to net force acting on the body mv mu F = t m v mu F = t By 3 rd Law, the forces acting on mass and mass are equal in magnitude but opposite in direction. F = F then, m u + m u = m v + m v m v m u t m v m u = ( ) t Class Work Find V, the impulse on m, and the average force exerted by m on m during collision. Given that m = kg, m = kg; u =- m/s, u = - m/s; V = 0,?t = 0.05 s. B ) Types of Collisions - Perfectly elastic collision : the K.E. of the whole system is conserved. - Perfectly inelastic collision : the colliding bodies move with a common velocity after collision. - all collisions are between perfectly elastic and perfectly inelastic. For Elastic: by conservation of momentum p & conservation of K.E. m u m + m u u = + m m u v = + m m v v + m v CSKMS AL PHYSICS M 004 by Leo Chung P.4
For perfectly inelastic: m u + m u = (m + m )v Collision in dimension: y v m u m x v gas molecules under elastic collision. By cons. of p: x direction : m u = m v cos + m v cos y direction : m v sin m v sin m u = m v + m v K ( 3) 3 equations, 4 unknowns (v, v,, ). So the equations can be solved if one of the unknowns is specified. C). Applications of the principle of conservation of momentum: a) To measure the inertial mass of an object: (explosion occurs by using a spring in between ticker tape timer is used to measure the V and V u. ) Q 0 m = u mv = mv V u + ( ) m u V u V u m u unknown mass m V b) To calculate the recoil velocity of a rifle. c) Collision of particles with He atoms: In a cloud chamber, particle is made colliding with He atoms and the subsequent path of collision is photographed. CSKMS AL PHYSICS M 004 by Leo Chung P.5
Class Work Prove that in an elastic collision between particles of equal mass, one of which is initially at rest, the recoiling particles always move off at 90 0 to one another. particle & He atom have equal masses and their collision is elastic. But in a cloud chamber, the angle between the tracks may not be 90 IX. Work, Energy and Power (I) Work A dot product of Force and displacement (unit : J, Nm) Work is the energy transferred by the application of force. Working is a process of transferring energy. F r r W. D. = F x = Fxcosθ (F is constant) θ x () θ = 90 o Work by the force is zero. () θ < 90 o Work done by the force on the object is positive. It means that the energy of the object is increased (usually the K.E. of the object gained) or the force supports motion. (3) 80 o > θ > 90 o Work done on the force on the object is negative. It means that the energy of object is decreased (Lost of PE or K.E.) or the force is against the motion of the object. CSKMS AL PHYSICS M 004 by Leo Chung P.6
(4) Suppose a hand exerts an upward force F on a mass and raises it up. Positive work is done by the hand on the mass. Also the work done by the force F, or the work done by the force F against gravity. (5) Suppose a hand exerts an upward force F on a mass and lowers it down. The work done by the hand on the mass is negative but that done by the mass on the hand is positive. Example A man uses his hand to stop the motion of a moving block. He exerts an average force of 50 N for a distance of 0.3 m and the initial kinetic energy of the block is 5 J. Work done by the man = 50 x 0.3 x cos 80 o = -5 J Kinetic energy gained by the block = work done by the man = -5J Total energy of the block = initial K.E. + W.D. on it = 5-5 = 0 So the block comes to rest. If a varying force acts on a body, work done by the force will be W = F x = Fdx x Work done by a varying force = area under the force vs displacement graph. (II) Energy Energy is just a concept in our brain. We cannot see or touch energy. The energy of a body is its capacity to do work. The amount of energy possessed by a body is equal to the amount of work that it can do. ) Power P (unit : J/s, W) rate of doing work = Energy / time = E / t. ) Energy (unit : J) that enable a body to do work. Kinetic energy Kinetic energy of a body is the energy due to its motion. The kinetic energy of a moving body is the amount of work that it can do in coming to rest or the amount of work that we must do in increasing its velocity from rest to the velocity it possesses. Consider a constant force F acts on a mass m which is initial at rest for a distance, s. K.E. = ½ mv. v = u 0 = u + as + as u a = s Kinetic energy of body consider a body of mass m moving with vel. u & final vel. V = 0 = W = Fs = mas u = ms s = mu In general, if the velocity of body of mass m increases from u to v when work is done on it by a force F acting over a distance s, then Fs = mv mu x CSKMS AL PHYSICS M 004 by Leo Chung P.7
This is called the work energy equation and may be stated as: Work done by the forces = Change in kinetic energy acting on the body of the body (b) Potential energy Potential energy of a body is the energy due to its position. () Gravitational potential energy If a mass m moves upwards in gravitational field, an external force is required to acts against gravitational force and hence work has to be done in moving a body from lower position to higher position. We say that the potential difference between these two positions is the work done in moving a mass m from one position to another. Consider a mass m being lifted by a constant force F from A to B to a height h and the force F is equal to the weight of the mass. F A Work done by the force in moving it from B to A mg = Fs = mgh h P.E. of the mass at A with respect to B = work done by the force from B to A = mgh B One point should be note that h is measured from an arbitrary level. So potential is referred to a reference point. () Strain potential energy Hook`s Law : The extension of a wire is proportional to the applied force, provided that the elastic limit is not exceeded. F x x= extension of the wire F= kx k = force constant, unit : Nm -. Consider a spring attached with a mass is being extended by a force F. Work done by the force in moving a distance x is x x W = Fdx = kxdx = kx 0 0 applied force/n extension x/m Work done by the force is stored in the spring due to extension. We call the hidden energy or stored energy as strain/elastic potential energy. If we release the mass., the hidden energy will converted to kinetic energy of the mass. (3) Transformation of potential energy and kinetic energy In a mechanical system, if there is no frictional force and external force acting on it, the total mechanical energy is conserved CSKMS AL PHYSICS M 004 by Leo Chung P.8
Potential energy + kinetic energy = constant U + K = constant U + K = 0 By differentiation, du + dk = 0 du = - dk Hence change in kinetic energy will converted to potential energy. Moreover, the work done on a system W will increases the energy of the system. i.e. the work done will converted to the kinetic energy and potential energy W = E = U + K Principle of Conservation of Mechanical Energy The total amount of mechanical energy (k.e. + p.e.) which the bodies in an isolated system possess is constant. Loss of K.E. = gain in P.E. + gain of internal energy Energy van be transferred from one form to another, but it cannot be created or destroyed, i.e. the total energy of a system is constant. CSKMS AL PHYSICS M 004 by Leo Chung P.9
Class Work. Find h and the rise in internal energy of the whole system. Given that m = 0 g; M = 990 g; u = 00 ms - u M h. Trolley A of mass m with velocity u collides elastically with trolley B of mass M at rest. Find their velocities v and V after collision. Discuss the motion of the trolleys when (a) m > M, and m < M (b) m >> M (c) m << M (d) m = M 3. A rife fires a bullet with velocity of 900ms - and the mass of the bullet is 0.0kg. The mass of the rife is 4 kg and the momentum of gas ejected is about 4 kgms -. Find the velocity of the recoil of rife. (-3.7 ms - ) CSKMS AL PHYSICS M 004 by Leo Chung P.0
Past Papers Practices MC Questions. 983-I- A ball bounces up and down from the floor. Which of the following graphs shows the variation of its velocity v with time t? A. B. C. D. E.. 983-I-5 The velocity v of a particle varies with time t as shown. Which of the following graphs best represents the variation of the displacement s of the particle with time t? A. B. C. D. E. CSKMS AL PHYSICS M 004 by Leo Chung P.
MC Questions 3. 989 I 3 A ball rolls down an inclined plane. The ball is first released from rest from P and then later from Q. Which of the following statements is/are correct? () The ball takes twice as much time to roll from Q to O as it does to roll from P to Q. () The acceleration of the ball at Q is twice as large as the acceleration at P. (3) The ball has twice as much K.E. at O when rolling from Q as it does when rolling from P. A. (), () and (3) B. () and () only C. () and (3) only D. () only E. (3) only 4. 995 IIA (AL) When given a slight push, a toy car moves freely with constant velocity down a plane inclined at 0 o to the horizontal. If the mass of the car is 0.5 kg, find the force parallel to the inclined plane for pulling the car up the plane with constant velocity. A..7 N B. 3.4 N C. 4.7 N D. 6.7 N E. 9.4 N 5. 995 IIA (AS)/IIA-(AL) Two small spheres A and B of masses kg and kg respectively are released from rest at heights 4h and h above the ground as shown. Which of the following statements is/are correct? (Assume air resistance is negligible) () The acceleration of sphere A doubles that of sphere B. () The time taken for sphere A to reach the ground is double that of sphere B. (3) The kinetic energy of sphere A when reaching the ground is double that of sphere B. A. () only B. (3) only C. () and () only D. () and (3) only E. (), () and (3) CSKMS AL PHYSICS M 004 by Leo Chung P.
MC Questions 6. 98-I-36 A particle of mass m strikes a barrier with speed v and rebounds with the same speed v. Which of the following statements is correct? () The angle φ must be equal to the angle θ. () The change in the component of momentum perpendicular to the barrier is mv sin θ (3) The change in the component of momentum parallel to the barrier is zero A. if (), () and (3) are all correct B. if () and () only are correct C. if () and (3) only are correct D. if () only is correct E. if (3) only is correct 7. 98-I-3 Two objects of masses m and 4 m move towards each other along a straight line with kinetic energies E and 4 E respectively. The total linear momentum of both masses taken together is A. 3 me B. 4 me C. 5 me D. 5 me E. 7 me 8. 986 I An object of mass 3 kg is placed on a smooth plane inclined at 30 to the horizontal. It is connected by a light string passing over a frictionless pulley to another object of mass kg, as shown above. Given that g = 0 m s -, when the system is released, the tension in the string will be A. 8 N. B. 0 N. C. 4 N. D. 5 N. E. 30 N. 9. 986 I A small block M of mass kg is transported across a small hill along the road ABC by an applied force F which is always parallel to the road. The speed of M is kept constant throughout the journey and the kinetic friction between the block and the road is.60 N. The total work done by F in transporting M from A to C is A. zero. B. 04 J. C. 5 J. D. 00 J. E. 304 J. CSKMS AL PHYSICS M 004 by Leo Chung P.3
MC Questions 0. 989-I-4 Two objects of weights N and 3N are suspended from a fixed point by two identical light springs A and B as shown in the diagram. The force constants of the springs are both N cm -. What are the extensions of spring A and B? Extension of spring A Extension of spring B A. 5 cm 3 cm B. 5 cm cm C. 3 cm cm D. 3 cm 5 cm E. cm 3 cm. 997-IIA-(AL)/(AS) Two wooden blocks A and B are connected by a string which passes over a smooth, fixed pulley as shown. The maximum friction between any two surfaces is N. If a horizontal force F is applied to block B, find its maximum value for moving B. A. N B. 4 N C. 6 N D. 8 N E. 0 N. 999-IIA-(AL)/(AS) Two objects A and B of equal mass m are connected by two identical light springs and are placed on a horizontal smooth surface. A horizontal force F is applied to B so that the system is in equilibrium. If the applied force F is suddenly removed, what are the magnitudes of the acceleration of each object at the instant when force F is removed? Acceleration of A Acceleration of B A. zero F/m B. zero zero C. F/m F/m D. F/m zero E. F/m F/m CSKMS AL PHYSICS M 004 by Leo Chung P.4
MC Questions 3. 999-IIA-(AL)/3(AS) A block of mass 5 kg is placed on the inclined surface of the wedge shown above. All contact forces are assumed to be smooth. What is the magnitude of the horizontal force F exerted on the wedge so that it remains stationary while the block is sliding down the inclined surface with acceleration? A. 0 N B. 8 N C. 4 N D. 30 N E. 3 N 4. 999-IIA-5(AL)/7(AS) The above graph shows the variation of the kinetic energy E with the square of velocity v of a moving mass m. What is the momentum of the mass when it is moving at a speed of m/s? A. Ns B. Ns C. 4 Ns D. 8 Ns E. 6 Ns. 5. 995-IIA-8(AL)/6(AS) The figure shows a uniform rigid beam AB, pivoted at A, held in horizontal position by a wire attached to a wall at point C, vertically above A. The beam carries a load W. If W is shifted gradually from A towards B, which of the following quantities will increase? () The tension in the wire. () The horizontal compression force in the beam. (3) The vertical component of the reaction A. A. () only B. (3) only C. () and () only D. () and (3) only E. (), () and (3) CSKMS AL PHYSICS M 004 by Leo Chung P.5
Essay Type Questions 6. 999 IIB (AL)/(AS) (a) A small ball is projected horizontally with a certain speed from a height of m above a smooth expanse of ground. The ball falls under gravity, hit the ground and bounces up. (i) Assuming that no energy is lost in the process, sketch graphs to show how the vertical component of the velocity and the acceleration of the ball vary with to the point when the ball bounces up to the original level. Label the axes wherever possible. Describe the force(s) acting on the ball and briefly explain the shape of the graphs. (ii) State the change(s), if any, to the graphs in (a) (i) for the following cases. Briefly explain your answer. (I) A ball of greater mass is used. (II) The projection speed is increased. (III) Some kinetic energy is lost when the ball hits the ground. (iii) Discuss whether the momentum of the ball is conserved when it hits the ground. (b) With the apparatus available in a school laboratory, describe a simple experiment to investigate the dependence of the stopping distance of a vehicle on its initial kinetic energy under the action of a constant resistive force. State and describe how to verify the expected result. State the source(s) of error. CSKMS AL PHYSICS M 004 by Leo Chung P.6
Essay Type Questions 7. 997 IIB (AS) (a) A man pushes a heavy rock resting on the ground, but it does not move. A student says that this is because the pushing force is balanced by the reaction of this force. Comment, with the aid of a diagram, on whether the student's argument is correct. (b) Draw a diagram to show the forces acting on a wooden block which slide down a rough incline with acceleration. Explain whether the work done by each of the forces is positive, negative or zero. State, in terms of energy, the effect of each force on the block. (c) Consider the cases in which (i) a man is inside a lift falling freely and (ii) he is inside a space-craft moving in a circular orbit round the earth. Identify THREE similarities between these physical environments. CSKMS AL PHYSICS M 004 by Leo Chung P.7