ACCELERATION, FORCE, MOMENTUM, ENERGY : solutions to higher level questions

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1 ACCELERATION, FORCE, MOMENTUM, ENERGY : solutions to higher level questions 015 Question 1 (a) (i) State Newton s secon law of motion. Force is proportional to rate of change of momentum (ii) What is the principal energy conversion that is taking place as the skier travels along the course? (Gravitational) potential (energy) to Kinetic (energy) (iii)ignoring friction, calculate her maximum velocity when she has travelle 400 m. mgh = ½mv // v = u + as v = 4 m s 1 (iv) She then ploughe into a snow rift an came to a stop in a time of 0.8 secons. What is the force that she exerts on the snow rift? Ft = change in momentum = (mv mu) F = N (v) What force oes the snow rift exert on her? F = (N) in opposite irection 014 Question 6 (i) Compare vector an scalar quantities. Give one example of each. Vectors have irection Scalars have no irection e.g. velocity is a vector an spee is a scalar (ii) Describe an experiment to fin the resultant of two vectors. Apparatus an arrangement e.g. 3 weights an pulleys Proceure an measurements e.g. ajust an rea each force Observation an result e.g. statement (iii)calculate the net force acting on the trolley an bag. Net force = Fforwar -Fbackwar Forwar force = horizontal force applie by golfer = 77 Cos N Backwar force = force of friction = 5 N Net force 0 N (iv) What oes the net force tell you about the golfer s motion? The golfer is travelling at constant spee (v) Use Newton s secon law of motion to erive an equation relating force, mass an acceleration. F proportional to (mv mu)/t F proportional to ma F = kma k = 1 (by efinition of the newton) F = ma (vi) Calculate the spee of the ball as it leaves the club. There are a number of ways to o this. The following isn t necessarily the shortest, but it might be the most familiar: F = ma 5300 =.045 a a = m s -1 Now use vuast: v = u +at v = 0 + ( ) ( ) v = 63.6 m s 1 (vii) Calculate the maximum height reache by the ball. uy = u sinθ = 63.3 sin 15 0 = m s 1 v = u +as 0 = (16.46) +(-9.8)s height = 13.8 m OR you coul have use: ½mv = mgh 1

2 013 Question 1 (a) (i) State the law of conservation of energy. The Principle of Conservation of Energy states that energy cannot be create or estroye but can only be converte from one form to another. (ii) Calculate the height through which the bob has been raise an the potential energy that it has gaine. h = (1 l cos θ) h = 8 8 cos 30 = 1.07 m E = mgh = = 63 J (iii)what is the maximum velocity it attains? Kinetic energy at the bottom = Potential energy at the top ½mv = 63 J v = 4.58 m s 1 (iv) Calculate the force applie. W = F F = = N OR F = ma F = = N 01 Question 6 (i) Calculate the acceleration ue to gravity at a height of 31 km above the surface of the earth. g = GM = ( ) + ( ) = g = ( )( ) ( ) g = 9.76 m s - (ii) What was the ownwar force exerte on Kittinger an his equipment at 31 km, taking their total mass to be 180 kg? W = F = mg F = 180(9.715) = N (iii)estimate how far he fell uring the first 13 secons. What assumptions i you take in this calculation? s = ut + ½ at s = ½ (9.715)(13) s = 80.9 m u taken as zero / g is constant / no atmospheric resistance / no buoyancy ue to atmosphere (iv) What was his average spee uring the next 4 minutes an 36 secons? Average spee = istance time Distance = = 5180 m Average spee = = 91.3 m s 1 (v) How much was the force on a hemispherical parachute of iameter 8.5 m greater than that on a similar parachute of iameter 1.8 m? F = (PA) = (4.5) Or F = (πr ) = 8.5 F =.3 N F (PA) (0.9) F (πr ) 1.8 (vi) Calculate the upthrust that acte on Kittinger when he reache constant velocity in the last stage of his escent (assume g = 9.81 m s uring this stage). Pressure = force / area Upthrust (U) = mg = (180)(9.81) = 1766 N

3 010 Question 1 (a) (i) List the instructions... Stan 1 m from wall (an select START) Stationary for 5 s Move back to 3 m (from wall) in 6 s / accept specific velocity an time/istance Stationary for 7 s Approach to 1 m in 4 s (ii) Using the graph, calculate the istance travelle by the cyclist an the average spee for the journey. Distance travelle = area uner the graph 18 km h -1 = 18000/60 m s -1 = 300 m min -1 S1 = = 900 m S = = 400 m S3 = = 750 m Total istance = 4050 m Average spee = total istance ivie by total time = 4050/19 = 13. m min -1 = 3.55 m s Question 6 (i) State Newton s law of universal gravitation. Force between any two point masses is proportional to prouct of masses an inversely/inirectly proportional to square of the istance between them. (ii) Use this law to calculate the acceleration ue to gravity at a height above the surface of the earth, which is twice the raius of the earth. Note that above surface is 3 from earth s centre GM g GM g new ( 3) where = m gnew = 1.09 m s - (iii)explain why the spacecraft continues on its journey to the moon, even though the engines are turne off. There are no external forces acting on the spacecraft so from Newton s 1 st law of motion the object will maintain its velocity. (Strictly speaking the spacecraft will always experience the force of gravity from nearby planets, stars etc, but this is consiere negligible). (iv) Describe the variation in the weight of the astronauts as they travel to the moon. Weight ecreases as the astronaut moves away from the earth an gains (a lesser than normal) weight as she/he approaches the moon (v) At what height above the earth s surface will the astronauts experience weightlessness? Gravitational pull of earth = gravitational pull of moon Gm E m Gm m m M M 1 E M 9 = ( 81) 1 1 3

4 E = 9 m an E + m = m 10 m = m = E = Height above the earth = ( ) ( ) = m (vi) The moon orbits the earth every 7.3 ays. What is its velocity, expresse in metres per secon? v = πr T π( ) v = v = 10.9 m s -1 (vii) Why is there no atmosphere on the moon? The gravitational force is too weak to sustain an atmosphere. 009Question 6 (i) State Newton s laws of motion. Newton s First Law of Motion states that every object will remain in a state of rest or travelling with a constant velocity unless an external force acts on it. Newton s Secon Law of Motion states that the rate of change of an object s momentum is irectly proportional to the force which cause it, an takes place in the irection of the force. Newton s Thir Law of Motion states that when boy A exerts a force on boy B, B exerts a force equal in magnitue (but) opposite in irection (on A). (ii) Show that F = ma is a special case of Newton s secon law. From Newton II: Force is proportional to the rate of change of momentum Force rate of change of momentum F (mv mu)/t F m(v-u)/t F ma F = k (ma) [but k = 1] F = ma (iii)calculate the average acceleration of the skateboarer on the ramp. v = u + as (1.) = 0 +a(5) a =.98 m s (iv) Calculate the component of the skateboarer s weight that is parallel to the ramp. W = mgsin = mgsin0 = N (v) Calculate the force of friction acting on the skateboarer on the ramp. Fnet = ma Force own (ue to gravity) Resistive force (ue to friction) = ma Gravitational force ma = friction force Gravitational force = N (see part (ii) above) an ma = 70(.98) = N = friction force = 6.5 N What is the initial centripetal force acting on him? mv F c r = 70 (10.5) /10 = N (vi) What is the maximum height that the skateboarer can reach? v = u + as u = g(s) s = 5.63 m (vii) Sketch a velocity-time graph to illustrate his motion. Velocity on vertical axis, time on horizontal axis, with appropriate numbers on both axes. 4

5 008 Question 1 (a) (i) State the principle of conservation of energy. The Principle of Conservation of Energy states that energy cannot be create or estroye but can only be converte from one form to another. (ii) Draw a velocity-time graph to illustrate the athlete s horizontal motion. See iagram (iii)use your graph to calculate the istance travelle by the athlete before jumping. Distance (s) = area uner curve s = ½ (3)(9.) + (9.) / / 3. m (iv) What is the maximum height above the groun that the athlete can raise his centre of gravity? K.E. = P.E. ½ mv = mgh h = v /g = (9.) /(9.8) = 4.3 Max height above the groun = = 5.4 m. 007 Question 1 (a) (i) What is friction? Friction is a force which opposes the relative motion between two objects. (ii) Calculate the net force acting on the car. Fnet = ma = (750)(1.) = 900 N east. (iii)calculate the force of friction acting on the car. Fnet = Fcar - Ffriction 900 = Ffriction Ffriction = 1100 N west (iv) If the engine is then turne off, calculate how far the car will travel before coming to rest? Friction causes eceleration: a = F m a = (-1100) 750 = ms - v = u + as 0 = 5 +(-1.47) s or s = 13 m 005 Question 1 (a) (i) State the principle of conservation of energy. Energy cannot be create or estroye but it can only be change from one form to another. (ii) What is the maximum kinetic energy of the ball as it falls? KE = PE (at height of 3.05 m) v = u +as v = 0 + (9.8)(3.05) v = {you coul also have use P.E. = mgh} Ek = ½mv = Ek = 17.9 J (iii)on bouncing from the groun the ball loses 6 joules of energy. What happens to the energy lost by the ball? It changes into soun an heat. (iv) Calculate the height of the first bounce of the ball. [retaine energy = 17.9 E = 11.9 J E = mgh h = E / mg h = 11.9 /(0.600)(9.8) h =.0 m 5

6 004 Question 6 (i) Define force. Force is that which causes an acceleration. (ii) Define momentum. Momentum is the efine as the prouct of mass multiplie by velocity. (iii)state Newton s secon law of motion. Newton s Secon Law of Motion states that the rate of change of an object s momentum is irectly proportional to the force which cause it, an takes place in the irection of the force. (iv) Hence, establish the relationship: force = mass acceleration. From Newton II: Force is proportional to the rate of change of momentum F (mv mu)/t m(v-u)/t ma (v) Calculate the velocity of the bob just before the collision. Loss in P.E = gain in K.E mgh = ½ mv v = gh = (9.8)(0.) v = 1.98 m s -1 (vi) Calculate the velocity of the block immeiately after the collision. m1u1 + mu = m1v1 + mv (0.01)() = (0.008) v v=.48 m s -1 (vii) What was the average horizontal force exerte on the block while travelling this istance? v = u + as / 0 = (.5) + a() a = 1.56 m s - F = ma = (0.008)(1.6) = N 003 Question 1 (a) (i) State Newton s secon law of motion. Newton s Secon Law of Motion states that the rate of change of an object s momentum is irectly proportional to the force which cause it, an takes place in the irection of the force. (ii) Calculate the average vertical acceleration of the skyiver. v = u +as (50) = 0 + (a)(1500) a = 0.83 m s - (iii)what is the magnitue an irection of the average resultant force acting on him? F = ma = = 75 N Down (iv) Use a iagram to show the forces acting on the skyiver an explain why he reaches a constant spee. Weight acting own on iagram Air resistance / friction / buoyancy acting up on iagram Air resistance = weight, therefore resultant force = 0 Therefore acceleration = 0 6

7 003 Question 6 Give the ifference between vector quantities an scalar quantities an give one example of each. A vector has both magnitue an irection whereas a scalar has magnitue only. (i) Describe an experiment to fin the resultant of two vectors. 1. Attach three Newton Balances to a knot in a piece of threa.. Ajust the size an irection of the three forces until the knot in the threa remains at rest. 3. Rea the forces an note the angles. 4. The resultant of any two of the forces can now be shown to be equal to the magnitue an irection of the thir force. (ii) Calculate the istance travelle by the cyclist. The isplacement is equivalent to one quarter of the circumference of a circle = πr/4 = 5π/ = 1.5π = 39.3 m. (iii)calculate the isplacement unergone by the cyclist. Using Pythagoras: x = (iv) Calculate the force require to keep the wheelchair moving at a constant spee up the ramp. (You may ignore the effects of friction.) If the wheelchair is moving at constant spee then the force up must equal the force own. So to calculate the size of the force up, we just nee to calculate the force own: F = mgsinϑ = 900 Sin 10 o = N (v) The ramp is 5 m long. Calculate the power exerte by the person in the wheelchair if it takes her 10 s to travel up the ramp. Power = work/time Work = Force isplacement = = 780 J Power = 780/10 = 78 W 00 Question 1 (a) (i) State the principle of conservation of momentum. The principle of conservation of momentum states that in any collision between two objects, the total momentum before impact equals total momentum after impact, provie no external forces act on the system. (ii) Calculate the mass of gas that the spacecraft must expel at a spee 50 m s -1 for the spacecraft to lock onto the space station. (The change in mass of the spacecraft may be ignore.) m1u1 + mu = m1v1 + mv (50000 ) = ( ) + (50m) m =1500 kg (iii)in what irection shoul the gas be expelle? Forwar (towar the space station). (iv) Explain how the principle of conservation of momentum is applie to changing the irection in which a spacecraft is travelling. As the gas is expelle in one irection the rocket moves in the other irection. 7