St. Joseph’s Anglo-Chinese School 0 St. Joseph’s Anglo-Chinese School HKDSE CS – Physics Notes W aves W aves W aves W aves M echanics M echanics M echanics M echanics E lectricity E lectricity E lectricity E lectricity H eat H eat H eat H eat St. Joseph’s Anglo-Chinese School 1 Chapter 1 Temperature and Thermometers 1.1 Temperature Lower fixed point (ice point) Temperature of pure melting ice at normal atmospheric pressure Upper fixed point (steam point) Temperature of steam over pure boiling water at normal atmospheric pressure Celsius temperature scale Divide 100 equal divisions between the lower and upper fixed point. Each division is 1 o C 1.2 Kinetic theory (a) All matter is made up of very tiny particles. (b) These particles are constantly in motion. (c) Forces between particles: (i) When particles are close together, they attract/repel each other strongly. (ii) When particles are far apart, they hardly attract/repel each other. Solid Liquid Gas Particle arrangement (1) close together (2) arranged in regular pattern (1) close together (2) not in fixed position (1) very far apart Particle motion Vibrate in fixed positions Can move freely from one place to another Move at random at very high speed (~500 ms -1 ) 1.3 Heat and Internal energy * Heat is the energy transferred from one body to another due to a temperature difference. internal energy = kinetic energy (K.E.) + potential energy (P.E.) of all particles K.E (depends on temperature) P.E. (depends on the state of matter) ↑ ⇔ ↑ E K T (particles vibrate more rapidly at higher temperature) * Temperature is a measure of average kinetic energy of the particles. heat warmer cooler solid liquid gas P.E. of particles increases St. Joseph’s Anglo-Chinese School 2 Chapter 2 Heat Capacity and Specific Heat Capacity Heat capacity (C) Specific heat capacity (c) Definition Energy required to raise the temperature of a substance by 1 o C Energy required to raise the temperature of a 1 kg substance by 1 o C Formula E = C ∆ T E = mc ∆ T Unit J o C -1 J kg -1 o C -1 C = mc Example 1 After absorbing 1000 J of energy, the temperature of a substance increases by 4 o C. If the mass of the substance is 2 kg, find (a) the heat capacity, and (b) the specific heat capacity of the substance. Solution (a) For heat capacity, E = C ∆ T 1000 = C(4) C = 250 J o C -1 (b) For specific heat capacity, E = mc ∆ T 1000 = (2)(c)(4) c = 125 J kg -1 o C -1 Power = rate of energy transferred t E P = or E = Pt Example 2 2 kg of water is heated by a heater of power 1500 W. Find the time it takes for the temperature of water to increase from 20 o C to 98 o C. Given: the specific heat capacity of the water = 4200 J kg -1 o C -1 Solution ∵ E = Pt and E = mc ∆ T ∴ Pt = mc ∆ T 1500t = (2)(4200)(98 – 20) t = 436.8 s St. Joseph’s Anglo-Chinese School 3 Example 3 The figure below shows the variation of temperature of an object with time. If the power of the heater is 800 W, find the heat capacity of the object. Solution ∵ E = Pt and E = C ∆ T ∴ Pt = C ∆Τ (800)(5 × 60) = C(60 – 20) [5 minutes = 5 × 60 s] t = 6000 J o C -1 Example 4 A piece of 0.1 kg hot copper is put into a pond of water of 2 kg. If the initial temperatures of the copper and water are 500 o C and 20 o C respectively, find the final temperature of the copper. Given: the specific heat capacity of the water = 4200 J kg -1 o C -1 the specific heat capacity of the copper = 370 J kg -1 o C -1 Solution Copper: 500 o C → T Water: 20 o C → T Assume no heat loss to the surroundings Energy lost by hot object (copper) = Energy gained by cold object (water) (0.1)(370)(500 – T) = (2)(4200)(T – 20) 18500 – 37 T = 8400 T – 16800 T = 22.1 o C Remark Since the specific heat capacity of water is large, water can absorb a large amount of energy with only a small temperature rise. ( ∆ T = 22.1 – 20 = 2.1 o C) Uses of high specific heat capacity of water (1) coolant (2) body temperature regulation 20 0 60 5 time / minute Temperature / o C St. Joseph’s Anglo-Chinese School 4 Example 5 In an experiment to find the specific heat capacity of aluminium, the following results are obtained. Mass of aluminium block = 1 kg Initial joulemeter reading = 98 300 J Final joulemeter reading = 104 900J Initial temperature of aluminium block = 28.5 o C Final temperature of aluminium block = 35.0 o C (a) Find the specific heat capacity of aluminium. (b) The standard value of the specific heat capacity of aluminium is 900 J kg -1 o C -1 , find the percentage error of the experiment. (c) How to improve the accurate of the experiment. Solution (a) By E = mc ∆ T 104900 – 98300 = (1)(c)(35 – 28.5) c = 1020 J kg -1 o C -1 (b) Percentage error = % 8 12 % 100 900 900 4 1015 = × − (c) (1) Wrap the aluminium block with cotton to reduce heat loss to the surroundings. (2) Add a few drop of oil to the holes in the aluminium block to ensure a good thermal contact between the heater, thermometer and the block. St. Joseph’s Anglo-Chinese School 5 Chapter 3 Change of State 3.1 Latent heat * E = ml , where l : specific latent heat * Unit: J kg -1 Example 1 How much energy is required to melt 2 kg of ice at 0 o C and to raise the temperature to 30 o C? Given that: the latent heat of fusion of ice = 3.34 × 10 5 J kg -1 , and the specific heat capacity of water = 4200 J kg -1o C -1 Solution Energy required T mc ml ∆ + = ) 0 30 )( 4200 )( 2 ( ) 10 34 3 )( 2 ( 5 − + × = 5 10 2 9 × = J Example 2 A coffee machine injects 0.03 kg of steam at 100 o C into a cup of cold coffee of mass 0.17 kg at 20 o C. Find the final temperature of the coffee. Given that: the latent heat of vaporization of ice = 2.26 × 10 6 J kg -1 , and the specific heat capacity of coffee = 5800 J kg -1o C -1 Solution Steam (0.03 kg): Coffee (0.17 kg): Energy lost by steam = Energy gained by coffee (0.03)(2.26 × 10 6 ) + (0.03)(4200)(100 – T) = (0.17)(5800)(T – 20) T = 90.0 o C No change in state E = mc ∆ T Heating Change in state E = ml Gas Liquid Solid Absorb latent heat (P.E. ↑ ) ∆ T = 0 (K.E. remains unchanged) Absorb latent heat (P.E. ↑ ) ∆ T = 0 (K.E. remains unchanged) Release latent heat (P.E. ↓ ) ∆ T = 0 (K.E. remains unchanged) Release latent heat (P.E. ↓ ) ∆ T = 0 (K.E. remains unchanged) E = ml E = mc ∆ T Ice (0 o C) Water (0 o C) Water (30 o C) E = ml E = mc ∆ T 100 o C 100 o C T (final temperature) water water E = ml T (final temperature) coffee 20 o C coffee St. Joseph’s Anglo-Chinese School 6 3.2 Measure the specific latent heat of fusion of ice Procedures (1) Set up the apparatus as shown. (2) Fill both funnels with roughly equal amounts of crushed melting ice. (3) Record the initial joulemeter reading ( E 1 ). (4) Switch on the heater for a period of time. (5) Find the mass of ice ( m ) melted by the heater and record the final joulemeter reading ( E 2 ). (6) Calculate the specific latent heat of fusion of ice ( l f ) by E 2 – E 1 = ml f Precautions - Ice should be crushed to increase the contact area with the heater. - Melting ice is used so that it is at 0 °C. - Before switching on the heater, pack the crushed ice in the two funnels so that the drip rates are steady and about the same. - After switching off the heater, do not remove the beakers; wait until the drip rates have become steady and about the same. - A small piece of wire gauze or steel wool at the neck of the funnels can prevent the crushed ice from dropping into the beakers directly. Example 3 The following results are obtained from the above experiment: Mass of water in experimental cup = 0.050 kg Mass of water in control cup = 0.014 kg Initial joulemeter reading = 15 000 J Final joulemeter reading = 29 200 J (a) Find the specific latent heat of fusion of ice. (b) Calculate the experiment percentage error. Account for any difference of the value obtained from the standard value, 3.34 × 10 5 J kg -1 Solution (a) By ml E = f l ) 014 0 050 0 ( 15000 29200 − = − 5 10 94 3 × = f l J kg -1 (b) Percentage error = % 1 18 % 100 10 34 3 10 34 3 10 9444 3 5 5 5 = × × × − × Possible sources of error include: (1) Water dripping down the two funnels at different rates. (2) Energy is lost to the surroundings. [Since energy is lost to the surroundings, less amount of ice is melted by the heater. By m E l f = , the measured l f is greater than the standard value.] St. Joseph’s Anglo-Chinese School 7 heater electronic balance kilowatt-hour meter 3.3 Measure the specific latent heat of vaporization of water Procedures (1) Set up the apparatus as shown. (2) Take the reading from the electronic balance ( m 1 ) and the kilowatt-hour meter ( E 1 ) after the water boils (3) Boil the water for a few minutes and turn off the heater. (4) Wait until the water becomes steady and take the final reading of the balance ( m 2 ) and the kilowatt-hour meter ( E 2 ). (5) Calculate the specific latent heat of vaporization of water by v l m m E E ) ( 2 1 1 2 − = − from the results. Precaution Do not switch on the heater unless the heating part is totally immersed in water. Possible sources of error (1) Steam condensing on the heater and drips back into the cup ⇒ Larger experimental l v (2) Energy is lost to the surroundings ⇒ Larger experimental l v (3) Some water ‘bubbles’ out of the cup ⇒ Smaller experimental l v Example 4 The following results are obtained from the above experiment: Mass of water boiled away = 0.10 kg Energy supplied to the heater = 246 000 J (a) Find the specific latent heat of fusion of ice. (b) Account for any difference of the value obtained from the standard value, 2.26 × 10 6 J kg -1 Solution (a) By ml E = v l 1 0 246000 = 6 10 46 2 × = f l J kg -1 (b) Possible sources of error include: (1) Steam condensing on the heater and drips back into the cup. (2) Energy is lost to the surroundings. [These will cause a smaller amount of water boiled away. By m E l v = , the measured l v is greater than the standard value.] St. Joseph’s Anglo-Chinese School 8 3.4 Evaporation and Boiling (a) Common: absorb latent heat ( E = ml ) to change from liquid state to gas state (b) Difference: Evaporation Boiling Occurs at any temperature Occurs at a definite temperature – the boiling point Occurs at surface Occurs with liquid No bubbles formed Bubbles appear (c) Evaporation and particle motion (i) Some of the particles in a liquid have greater K.E. while the other have smaller K.E. (ii) Some of the particles at the liquid surface may gain enough KE to escape into the space above the liquid and become particles of vapour. (iii) As fast-moving particles fly away, the average KE of the remaining particles is lowered; so the liquid becomes colder. (d) Ways to increase the rate of evaporation (i) increase temperature of the liquid (ii) increase the surface area of the liquid (iii) decrease the humidity of air (iv) increase the movement of air Example 5 A person is wearing a wet shirt. There is 0.1 kg of water on the shirt in total. (a) How much energy is required to evaporate the water? (b) Where is the energy required taken from? Specific latent heat of vaporization of water = 2.26 × 10 6 J kg –1 Solution (a) Energy required evaporating 0.1 kg of water E = ml = (0.1)(2.26 × 10 6 ) = 2.26 × 10 5 J (b) The energy required is taken from the surroundings (or the skin of the person). St. Joseph’s Anglo-Chinese School 9 Chapter 4 Heat Transfer Conduction Convection Radiation Description – Particles at the hot end vibrate faster. – The fast vibrating particles bump into the slower neighbouring particles and make them vibrate more rapidly. – Fluid (gas or liquid) expands, rises and is replaced by the surrounding cooler fluid. – Such movement of fluid is called convection. The solar energy is transferred to the earth by radiation Medium required Solid, liquid or gas (Vacuum ) Fluid (liquid or gas) No medium is required Rate of heat transfer Conductor – faster Insulator – slower Dark colour – good absorber and radiator Daily examples A cotton jacket traps air (air is a poor conductor) – Air conditioners are installed high on the wall – Heating element is fixed near the bottom of an electric kettle – Most transformers are black in colour – Car engines are painted black. – Fuel storage tanks are painted silvery. Example 1 Solar heater (a) What is the function of the glass? (b) The temperature of the 0.3 kg water increases to 70 o C after 30 minutes. If the initial temperature of water is 30 o C, find the power of the solar heater. (c) Give a suggestion to improve the design of the solar heater in order to obtain a greater temperature rise of water. Explain your answer briefly. Solution (a) The glass traps warm air to reduce heat loss by convection. (b) ∵ E = Pt and E = mc ∆ T ∴ Pt = mc ∆ T P(30 × 60) = (0.3)(4200)(70 – 30) P = 28 W (c) (1) Paint the bowl black in colour since dark colour object is a good radiation absorber. (2) Stick slivery paper (aluminium paper) onto the inner wall of the container so that more radiation is reflected to the bowl. Hot water rises Cold water falls bowl glass container water St. Joseph’s Anglo-Chinese School 10 Vacuum flask Plastic or cork stopper reduces heat loss by conduction and convection outer case Silvery glass or stainless steel walls reduce heat loss by radiation A vacuum between the double walls reduces heat loss by conduction and convection insulated support cap Vacuum cooker Inner and outer lids reduce heat loss by conduction and convection Silvery steel walls reduce heat loss by radiation A thin film of air between inner pot and outer pot. Reduce heat loss by conduction and convection Outer vacuum insulated pot (the vacuum between the double steel walls of the outer pot reduces heat loss by conduction and convection ) Food is kept at a high temperature for a long time and is cooked without a fire. Inner pot Outer pot Inner pot St. Joseph’s Anglo-Chinese School 11 Chapter 1 – 2 Motion (a) Distance and Displacement Speed and Velocity Example 1 A person takes 4 s to walk from A to B along the following path. Find the distance, displacement, speed and velocity Solution Distance = length of the path = 2 10 π = 15.7m Displacement = the length of straight line AB = 10 m (N50 o E) Speed time Distance = 93 3 4 15.7 = = ms -1 Velocity time nt Displaceme = 5 2 4 10 = = ms -1 (N50 o E) Scalar: Distance, Speed Vector: Displacement, velocity (b) Acceleration time velocity in change on accelerati = t u v a − = Unit: ms -2 Example 2 It takes 6 s for a racing car to attain a speed of 100 km h -1 from rest. Find the average acceleration of the car. Solution Initial velocity u = 0 ms -1 (start from rest) Final velocity v = 100 km h -1 = 6 3 100 ms -1 = 27.78 ms -1 Time t = 6 s Acceleration t u v a − = 63 4 6 0 78 27 = − = ms -2 5 m B A 40 o St. Joseph’s Anglo-Chinese School 12 (c) Motion graphs (s – t, v – t, a – t graph) (i) Uniform velocity (ii) Uniform acceleration (iii) Slope and area under graph slope = velocity slope = acceleration Area under v-t graph = displacement (d) Equations of uniformly accelerated motion (1) at u v + = u = initial velocity v = final velocity (2) as u v 2 2 2 = − a = acceleration (3) 2 2 1 at ut s + = s = displacement (4) t v u s 2 + = t = time t s 0 t v 0 t a 0 Displacement changes at a constant rate Velocity remains constant. No acceleration. a = 0 ms -2 t s 0 t v 0 t a 0 Displacement: Beginning: increases a little. End: increases significantly. Velocity increase at a constant rate. Acceleration remains constant. t s 0 t s 0 t v 0 St. Joseph’s Anglo-Chinese School 13 (e) Motion of free fall In the absence of air resistance, all objects fall with an acceleration of g (9.81 ms -2 ) Example 3 A stone is thrown vertically upwards from the ground with a speed of 20 m s -1 (a) How high does the stone rise? (b) How long does it take the stone to reach the ground? (c) Draw a v-t graph for the stone unit it returns to the ground. Solution (a) Consider the motion of the stone when it is thrown upward until it reaches the highest point. u = 20 ms -1 a = -9.81 ms -2 (uniform deceleration) v = 0 ms -1 (momentarily at rest at the highest point) s = ? By as u v 2 2 2 = − s ) 81 9 ( 2 20 0 2 2 − = − s = 20.4 m ∴ greatest height is 20.4 m (b) Method 1 Method 2 Consider the motion of the stone when it is thrown upward until it reaches the highest point. Consider the motion of the stone when it is thrown upward until it returns to the ground. u = 20 ms -1 a = -9.81 ms -2 (uniform deceleration) v = 0 ms -1 (momentarily at rest) t = ? By at u v + = t ) 81 9 ( 20 0 − + = 04 2 = t s Time required for stone to reach the ground = 2 × 2.04 = 4.08 s u = 20 ms -1 a = -9.81 ms -2 (uniform deceleration) s = 0 (return to ground) t = ? By 2 2 1 at ut s + = 2 ) 81 9 ( 2 1 20 0 t t − + = 0 ) 20 905 4 ( = − t t 0 = t (rejected) or 0 20 905 4 = − t ∴ 08 4 = t s (c) 20 ms -1 v = 0 ms -1 (highest point) 20 ms -1 v = 0 ms -1 20 ms -1 s = 0 m t / s v / ms -1 -20 20 0 2.04 4.08 St. Joseph’s Anglo-Chinese School 14 Remark: Take upward direction as positive Stage 1: upward motion Stage 2: at highest point Stage 3: downward motion decelerate uniformly Momentarily at rest accelerate downward uniformly a = -9.81 ms -2 a = -9.81 ms -2 a = -9.81 ms -2 Misconception (1) The acceleration of the object at the highest point a = 0 ms -2 (2) We are certain that at the highest point, v = 0 ms -1 (3) If at the highest point a = 0 ms -2 and v = 0 ms -1 , the object will remain at rest forever and will not fall down. Example 4 The v - t graph of a car is shown. (a) Describe the motion of the car from t = 0 s to t = 80 s. (b) Find the deceleration of the car from t = 40 s to t = 70 s. (c) Find the total displacement of the car. Solution (a) From t = 0 s to 40 s, the car moves with a uniform velocity. From t = 40 s to 70 s, the car decelerates uniformly and becomes momentarily at rest at t = 70 s. From t = 70 s to 80 s, the car accelerates uniformly in opposite direction. (b) Method 1 by t u v a − = Method 2 a = slope of v-t graph From t = 40 s to 70 s u = 50 ms -1 , v = 0 ms -1 , t = 30 s 67 1 30 50 0 − = − = a ms -2 ∴ Deceleration is 1.67 ms -2 From t = 40 s to 70 s a = slope of v-t graph = 67 1 40 70 0 50 − = − − − ms -2 ∴ Deceleration is 1.67 ms -2 (c) Total displacement of the car = area under v – t graph = ) 25 )( 10 ( 2 1 2 ) 50 )( 70 40 ( − + = 2625 m t / s v / ms -1 -25 50 0 70 40 80 St. Joseph’s Anglo-Chinese School 15 Chapter 3 Force and Motion 3.1 Newton’s law of motion Newton’s 1 st law Every object remains in a state of rest or uniform motion (i.e. constant velocity) unless acted on by a net force, or an unbalanced force. Newton’s 2 nd law The acceleration of an object is directly proportional to, and in the same direction as, the net force acting on it, and inversely proportional to the mass of the object. [ F a ∝ , m a 1 ∝ ⇒ F = ma ] Newton’s 3 rd law Action and reaction pair (1) Equal in magnitude (2) Opposite in direction (3) Acting on different bodies. 3.2 Forces in daily life (a) Weight ( W ) (b) Normal reaction ( R ) (c) Friction ( f ) (d) Tension ( T ) 3.3 Mass m and Weight W W = mg Mass m Weight W Scalar Vector A measure of inertia Gravitational force acting on the object Remain unchanged Vary in different planets (g is different) 3.4 Addition and resolution of forces Addition: parallelogram of forces method Resolution: F x and F y are components of F θ cos F F x = θ sin F F y = F R (resultant force) F 1 F 2 x F y F y F x θ St. Joseph’s Anglo-Chinese School 16 30 ο 3.5 Typical exam questions (a) Inclined plane Resolve W into two components: Example 1 A block is placed on the inclined plane and remains at rest. (a) (i) Draw a free diagram of the object. (ii) If the mass of the block is 5 kg, find the friction acting on the block (iii) Find the normal reaction acting on the block (b) Now, a force of 120 N is applied on the block and pulls the block upwards along the plane. If the friction between the block and the plane becomes 60 N, find the acceleration of the block. Solution (a) (i) Do not accept W , mg , f , R (ii) f θ θ sin sin mg W = = ° = 30 sin ) 81 9 )( 5 ( 5 24 = N (iii) R θ θ cos cos mg W = = ° = 30 cos ) 81 9 )( 5 ( 5 42 = N (b) Consider the direction along the plane By F = ma 120 – W sin θ – f = (5)(a) 120 – (5)(9.81)sin θ – 60 = 5 a a = 7.10 ms -2 along the plane θ sin W or θ sin mg perpendicular to the plane θ cos W or θ cos mg Weight ( W = mg ) Friction ( f ) Normal reaction ( R ) W θ θ W cos θ W sin θ f = 60 N θ 120 N W sin θ W θ W cos θ W sin θ f R St. Joseph’s Anglo-Chinese School 17 * W = mg * a = -9.81 ms -2 * R = 0 (b) Apparent weight in a lift Example 2 (a) A boy of mass 55 kg stands on a weighing machine in a lift. Find the reading on the weighing machine when (i) the lift moves upwards with an acceleration of 1 ms -2 , and (ii) the lift moves downward with a deceleration of 1.5 ms -2 (b) When the lift moves upward and the reading on the weighing machine is 480 N, find the acceleration of the lift. Solution (a) (i) The lift moves upward and accelerate , a = + 1 ms -2 ma W R = − ) 1 )( 55 ( ) 81 9 )( 55 ( = − R 595 = R N [The boy feels heavier] (ii) The lift moves downward and decelerate , a = + 1.5 ms -2 ma W R = − ) 5 1 )( 55 ( ) 81 9 )( 55 ( = − R 622 = R N [The boy feels heavier] (b) ma W R = − a ) 55 ( ) 81 9 )( 55 ( 480 = − 08 1 − = a ms -2 ∴ The deceleration of the lift is 1.08 ms -2 Remark (1) When applying ma W R = − (2) When the cable of the lift break, Reading of weighing machine = normal reaction = R When the lift at rest or moves with uniform velocity R = W Feel normal weight accelerate upward R > W Feel heavier accelerate downward R < W Feel lighter Weight is always constant ( W = mg ) R – W = ma (Always take upward direction as +ve) upward with acceleration a = +ve upward with deceleration a = – ve downward with acceleration a = – ve The lift moves downward with deceleration a = +ve Weighing machine W R a W R St. Joseph’s Anglo-Chinese School 18 3.6 The turning effect of a force (a) Moment = Fs Unit: Nm Moment = Fs = (5)(2) = 10 Ns Moment = Fs = (5 sin 30 o )(2) = 5 Ns (b) Equilibrium of a rigid body No net force: all forces acting on the body are balanced No net torque: clockwise moment = anticlockwise moment Example 3 The mass of the meter rule is 0.5 kg. (a) Find the distance between the school bag and the pivot to maintain balance. (b) Find the reaction from the pivot. Solution (a) Clockwise moment = anticlockwise moment (W 2 )(0.3) + (W 3 )(d) = (W 1 )(0.2) (0.5 × 9.81)(0.3) + (4 × 9.81)(d) = (10 × 9.81)(0.2) d = 0.463 m (b) 3 2 1 W W W R + + = = (10)(9.81) + (0.5)(9.81) + (4)(9.81) = 142 N Remark (1) Moment = Force × Distance (distance is usually measured from the pivot) (2) Must label all forces clearly on the figure. 2 m 5 N pivot 2 m 5 N pivot 5 cos 30 o 5 sin 30 o 30 o d W 2 W 3 W 1 R 0.2 m 0.3 m School bag 4 kg Block 10 kg St. Joseph’s Anglo-Chinese School 19 Chapter 4 Work, Energy and Power 4.1 Work (W) Definition: s F W // = Unit: J Example 1 Calculate the work done by F in each of the following case. (a) (b) Solution (a) W = (2)(5) = 10 J (b) W = (2 cos 30 o )(5) = 8.66 J Consider only the component of F along the direction of displacement. i.e. 2 cos 30 o 4.2 Energy Mechanic energy: (1) K.E. 2 2 1 mv = (2) P.E. mgh = (3) Elastic potential energy 4.3 Energy change Conservation of energy: Energy can be changed from one form to another, but it cannot be created or destroyed. Example 2 Find the speed of the particle at B. Assume that the track is smooth. Solution Total mechanical energy at A = Total mechanical energy at B P.E. A + K.E. A = P.E. B + K.E. B Common mistake m (9.81)(10) + ½ m (2) 2 = m (9.81)(3) + ½ m ( v ) 2 cannot use the formula as u v 2 2 2 = − 98.1 + 2 = 29.43 + ½ v 2 reason: not a uniformly accelerated motion v = 11.9 ms -1 F = 2 N 5 m 5 m F = 2 N 30 o 2 ms -1 B A 10 m 3 m