7.2. Assessment in Diploma Program Physics 281

Transcription

1 7.2. Assessment in Diploma Program Physics 281 figures in 100 kmh 1, we round the answer to two significant figures. Note that we have to use at least one significant figure more than the final result in 27.8 ms 1 to avoid the rounding error. Solution to Exercise 3-45 A. The velocity is v = u + at = 5.2ms ms 2 3.0s = 12.4ms 1 12ms 1. B. The distance travelled is (v + u)t s = 2 = (12.4ms m/s) 3.0s 2 = 26m. Solution to Exercise 3-46 If the direction of motion is positive, the acceleration is negative a = 1.9ms 2 and the initial velocity positive u = 7.8ms 1. A. The velocity after 5.0 s of rolling is v = u + at = 7.8ms 1 1.9ms 2 2.0s 4.0ms 1. Because the velocity is positive, the ball is still rolling up the ramp. B. The velocity after 5.0 s of rolling is v = u + at = 7.8ms 1 1.9ms 2 5.0s 1.7ms 1. Because the velocity is negative, the ball is going down the ramp. C. The ball has stopped, when the final velocity is v = 0ms 1. Substituting the value into equation v = u + at and solving the time gives the stopping time t = u a = 7.8ms 1 = s 4.1s ms The stopping distance is (v + u)t s = 2 = (0ms m/s) 4.105s 2 16m.

2 282 Constants, Equations and Assessment Solution to Exercise 3-47 Solution to Exercise 3-48 The final speed is v = 160kmh 1 = 160kmh ms 1. A. The average acceleration is a av = v t = 44.44ms 1 0ms 1 52ms 2. (7.21) 0.86s B. The final speed is v = a av t = 51.68ms 2 1.5s 78ms 1 280kmh 1. (7.22) Solution to Exercise 3-49 Solution to Exercise 3-50 Solution to Exercise 4-1 By Newton s second law of motion, the unit of force is The correct answer is D. [F ] = [m][a] = kgms 2. (7.23) Solution to Exercise 4-2 By Newton s third law of motion, the correct answer is B. hen velocity is constant, the direction of movement remains the same, and the object moves along a straight line. Solution to Exercise 4-3 The correct answer is C. According to Newton s second law F = ma. If force F is constant, the acceleration is constant as well. Solution to Exercise 4-4 The correct answer is D. The object is in accelerated motion. Solution to Exercise 4-5 The correct answer is C. On a rough surface, the friction is unbalanced, and the object is in decelerated motion. Solution to Exercise 4-6 The correct answer is B. There are no unbalanced forces and the net force is zero.

3 7.2. Assessment in Diploma Program Physics 283 Solution to Exercise 4-7 The correct answer is C. Force and reaction force act on different objects and, as a result, cannot cancel each other out. Solution to Exercise 4-8 The examples include the normal forces from the objects I touch, weight that pulls me towards the Earth, electric force by the comb on dry hair, and magnetic forces from power lines. Solution to Exercise 4-9 The examples of a deforming force include hen a ball bounces from the floor, the fall deforms due to the normal reaction force exerted by the floor on the ball. The high tide and low tide are caused by the variations of the gravitational force by the Moon on the oceans. A bridge bends down, when the weight of a car pulls the car down, and the normal reaction force from the car on the bridge pushes the bridge down. In some cases, it is possible to sense the bridge vibrations, as you walk on the bridge. Solution to Exercise 4-10 The examples of a rotating force include hen an athlete throws a discus, the contact force from the athlete s hand makes the discus spin. The static friction from the ground makes a bicycle front tyre to spin. The contact force from the chain makes a bicycle back tyre to spin. Solution to Exercise 4-11 In theory, the answer is no: the range of gravitational interaction is infinite, so every object in the universe interacts with all other objects in the universe. In practise, however, in outer space far away from the galaxies even gravitational forces are very weak. So, from a practical point of view, we could also answer yes.

4 284 Constants, Equations and Assessment Solution to Exercise 4-12 Air resistance and rolling resistance slow down the motion. The forces are, however, so small that the cart is almost in uniform motion. Solution to Exercise 4-13 No, there are not. According to Newton s firs law motion there are no unbalanced forces in uniform motion. Solution to Exercise 4-14 A. The luggage tends to stay at rest unless acted upon by an unbalanced force. As the train gets going, the shelf moves under the luggage. If there is not enough friction, or the luggage is not supported well, it may fall. B. As the train decelerates, the luggage tends to keep its state of motion. As the shelf decelerates, the luggage carries on, if there is not enough friction, or the luggage is not well supported. Solution to Exercise 4-15 Because the stone moves along a straight line, it is in uniform motion. Then, by Newton s first law of motion, there are no unbalanced forces acting on the stone. Solution to Exercise 4-16 A. It is easier the jump, because the force of gravity is only one sixth of the gravity on the Earth. You can jump longer and higher in the Moon. The flight time is longer in the Moon. There is no air resistance on the Moon. The acceleration vector points towards the centre of the Moon, just like it points towards the Earth s centre here on Earth. The shape of the trajectory is a parabola, although of different form. B. In the absence of gravity, the astronaut would continue moving along a straight line at constant speed. Solution to Exercise 4-17 A. The forces acting on a book lying on a table are the weight and the normal reaction force R.

5 7.2. Assessment in Diploma Program Physics 285 v R B. The forces acting on a book sliding on a table are the weight, the normal reaction force R, and the sliding friction F µ. a v R F µ C. The forces acting on a freely apple are the weight and the air resistance F air. a v F air

6 286 Constants, Equations and Assessment Solution to Exercise 4-18 A. At the time of impact the forces are the weight, the normal reaction force R and the contact force F foot from the foot. F foot a v B. In going up the forces acting on the ball are the weight and the air resistance F air. F air a v C. At the highest point the ball is momentarily at rest, so there is no air resistance. The only force acting on the ball is the weight.

7 7.2. Assessment in Diploma Program Physics 287 a D. In going down the forces acting on the ball are the weight and the air resistance F air. a v F air Solution to Exercise 4-19 A. The forces acting on the ice puck are the weight and the normal reaction force R. There are no unbalanced forces, and the puck is in uniform motion. v R B. The forces acting on the ice puck are the weight, the normal reaction force R, and the sliding friction F µ. The sliding friction is an unbalanced force, and the puck is in decelerated motion.

8 288 Constants, Equations and Assessment a v R F µ Solution to Exercise 4-20 A. The forces acting on the piece of wood are the weight G and the buoyant force B. B B. Because the block floats at rest, there are no unbalanced forces acting on it. The weight and the buoyant force are equal in magnitude. Solution to Exercise 4-21 v 1 = 0ms 1, v 2 = 100kmh 1, t = 5.7s,m = 278kg For the net force we need first solve the average acceleration of the motorcycle. Since the speed is expressed in kmh 1, we convert to ms 1 by dividing the value by 3.6. The average acceleration is a av = v t = v v = ms 1 0ms ms 2. t 5.7s

9 7.2. Assessment in Diploma Program Physics 289 The magnitude of the (average) net force on the motorcycle system is ΣF = ma = 278kg 4.873ms N. The direction of the net force vector is in the direction of the motion of the motorcycle (i.e the direction of the acceleration vector). Solution to Exercise 4-22 F = 123N,m = 0.450kg By Newton s second law of motion, the magnitude of the acceleration of the ball is a = F m = 123N 0.450kg 270ms 2 which is approximately 27 times the acceleration due to the gravity! The direction of the acceleration is the direction of the net force acting on the ball. Solution to Exercise 4-23 A. The weight is the gravitational force the Earth exerts on the apple. The reaction force to the weight is thus the gravitational force the apple exerts on the Earth. B. The normal reaction force the chair exerts on you is the contact force the chair exerts on you. The reaction force to that is the normal force by which you push the chair down. C. The reaction force to the static friction force the ground exerts on a bicycle tyre is the friction force the tyre exerts on the ground. Solution to Exercise 4-24 The forces are equal and opposite, but the mass of the Earth is so large that the Earth does not move. Solution to Exercise 4-25 By Newton s third law of motion, the teams exert equal and opposite forces on one another. The which team wins, depends, for example, on the total mass of the teams, their grip on the rope, and the friction between the team members and to ground.

10 290 Constants, Equations and Assessment Solution to Exercise 4-26 a) By Newton s third law of motion, you exert a force on the raft that moves the raft in the opposite direction to your walking. b) The force you exert on the ship is so small, that there is no measurable motion of the ship which has large mass. Solution to Exercise 4-27 In all cases the forces exerted are equal and opposite by Newton s third law of motion. Solution to Exercise 4-28 No, I would not. alking is based on the reaction force from the ground due to friction: as I push the road backwards, the road pushes me onwards. In the absence of friction, there would be no force pushing me onwards. By Newton s first law of motion I would stay at rest. Solution to Exercise 4-29 I would not be able to walk on the surface, because there is no friction (see the answer to the previous question). But if I throw a stone, I exert a force on the stone. By Newton s third law of motion, the stone exerts on equal and opposite force on me. I would start moving in the opposite direction to the throw. The more stones I throw, the faster I move. Solution to Exercise 4-30 The correct answer C. The ball is in free fall. Solution to Exercise 4-31 The correct answer A. Acceleration is inversely proportional to mass. Solution to Exercise 4-32 The correct answer is C. In free fall the acceleration is the acceleration due to gravity which is independent of the mass. e see this from Newton s second law of motion. The gravitational force that accelerates an object is = mg. By Newton s second law of motion the acceleration is a = F m = mg m = g = 9.81ms 2 which is independent of the mass. In free fall, the objects fall with the same acceleration independently of their mass.

11 7.2. Assessment in Diploma Program Physics 291 Solution to Exercise 4-33 The correct answer is A by Newton s second law of motion. Solution to Exercise 4-34 According to Newton s First Law of Motion an object is in uniform motion when there are no unbalanced forces acting on an object. As a result, the acceleration is zero, and the object cannot be in accelerated motion. Solution to Exercise 4-35 A. hen an unbalanced force acts on an object, the object is in accelerated motion. In accelerated motion the velocity changes. The answer is no. B. The answer is yes. If a bike rounds a curve at constant speed, the speed remains constant, but direction of the velocity vector changes. The bike is in accelerated motion, but the speed is constant. Solution to Exercise 4-36 A. Graphs A. and B. could represent uniform motion, C. and D. uniformly accelerated motion. B. In uniform motion the net force zero. In C. and D. the velocity is increasing. The net force acts in the direction of motion. Solution to Exercise 4-37 A. The motorcyclist is in uniformly accelerated motion. B. The acceleration of the motorcycle is the slope of the line. From the graph we see that the motorcycle accelerates from u = 0ms 1 to v = 20ms 1 in 4.6s. The acceleration is a = v u t = 20ms 1 0ms 1 4.6s = ms 2 4.4ms 2. C. The net force is F = ma = 295kg 4.348ms N. D. hen the acceleration is one half of the acceleration in A, the equation of the line is v = 2.17ms 2 t. The graph is

12 292 Constants, Equations and Assessment ms 1 v t s Solution to Exercise 4-38 A. The forces acting on the system are the weight, normal reaction force R front on the front tyre, normal reaction force R back on the back tyre, air resistance F air, rolling resistance F roll on both tyres, and the static friction F µ0 between the back tyre and the ground. a v F air R back R front F roll F µ F roll B. The static friction between the back tyre and the ground. C. The average net force is Fave = maave = m v t = 92kg 18.2ms 1 0ms 1 230N. 7.4s

13 7.2. Assessment in Diploma Program Physics 293 Solution to Exercise 4-39 A. The weight on Earth is B. The weight on the Moon is = mg = 98kg 9.81ms 2 960N. = 1 6 mg = kg 9.81ms 2 160N. C. The mass is m = g = 120N 9.81ms kg. Solution to Exercise 4-40 A. The force diagram is R A box lies on a table. Normal reaction force R is equal in magnitude to the weight, but points in opposite direction. The point of application for the weight is the centre of mass. The normal reaction force starts from the middle of the bottom. B. No, they are not. Because the both forces act on the same object, they cannot be a force and the reaction force. See answers to Question 23 a) and b) for the reaction forces. C. The weight is G = mg = 0.370kg 9.81ms N. Note that for the calculation of the weight, the mass has to be in kilograms.