How well do you understand the concepts presented in this book? Answers are here.
Chapter 1: Position, velocity and acceleration
1.A scalar quantity is one that:
a) is described by a magnitude only
b) is described by a magnitude and direction
c) calculates movement along a rectilinear path
d) calculates movement along a curvilinear path
2.If a track cyclist sprints on a lap of the oval track, which quantity equals 0 m as they cross the line?
a) their average speed
b) their average acceleration
c) the distance travelled
d) their displacement
3.If a basketball player ran 6 m along the baseline before running 10 m down the court (see figure) in 5.2 s, what would his average velocity be?
a) 0.45 m·s-1
b) 2.24 m·s-1
c) 2.24 m·s-1 at 36.9°
d) 2.24 m·s-1 at 59°
4.If a runner starts from a stationary position and reaches a velocity of 4.8 m·s-1 in 2 s, what is her average acceleration?
a) 0.42 m·s-2
b) 2.4 m·s-2
c) 9.6 m·s-2
d) none of the above
5.A lacrosse player starts from a stationary position and runs 5 m straight ahead, then turns 180° and runs 10 m before coming to a stop. If the initial forward direction is designated as the positive direction, the acceleration in the last metre before coming to the stop at the end of the 10 m segment would be defined as:
a) positive acceleration
b) negative acceleration
c) zero acceleration
d) this cannot be determined
Chapter 2: Angular position, velocity and acceleration
1.During running, the legs predominately move in which plane?
a) frontal
b) coronal
c) transverse
d) sagittal
2.With respect to anatomical references, anything to the front of the body is referred to as:
a) caudal
b) cranial
c) anterior
d) posterior
3.If a baseball bat was swung such that it travelled through a 260° angle in 0.16 s, what would its angular velocity have been?
a) 0.73 rad·s-1
b) 28.4 rad·s-1
c) 41.6 rad·s-1
d) 1625 rad·s-1
4.If a diver was spinning in the tuck position at 800°·s-1 and then slowed to 40°·s-1 over a 0.3 s period as they opened out before hitting the water, what would be the diver’s angular acceleration?
a) 44.2 rad·s-2
b) -44.2 rad·s-2
c) 2533.3 rad·s-2
d) -2533.3 rad·s-2
5.What would be the linear velocity of the foot if a person’s leg was 100 cm long and was swung at an angular velocity of 3.2 rad·s-1?
a) 0.3125 m·s-1
b) 3.2 m·s-1
c) 320 m·s-1
d) this cannot be determined from the information given
Chapter 3: Projectile motion
1.Assuming wind resistance is negligible the projection range of an object fired from the ground is maximum when the projection angle is:
a) 35°
b) 45°
c) 55°
d) 90°
2.If the projection height of an object is positive (i.e. it is projected from a point higher than on which it lands) then the optimum projection angle:
a) is less than when the object is projected from the ground
b) is the same as when the object is projected from the ground
c) is greater than when the object is projected from the ground
d) cannot be determined mathematically
3.According to one of Galileo’s equations of constant acceleration (i.e. projectile motion equations), the final velocity of an object is equal to the product of acceleration and time (a × t) plus:
a) the initial velocity
b) displacement
c) acceleration due to gravity
d) the square of initial velocity multiplied by displacement
4.A child was impressed with how far in the air he could throw a tennis ball and wanted to know how fast he was throwing it. You timed that, on average, the ball was reaching the top of its trajectory in 1.4 s. How fast must the ball have left his hand?
a) 0.14 m·s-1
b) 7.0 m·s-1
c) 13.7 m·s-1
d) 14.0 m·s-1
5.With respect to conducting a video analysis of an athlete, the minimisation of perspective and parallax errors can be best achieved by:
a) keeping the camera as close as possible to the athlete
b) using the highest shutter speed possible
c) keeping the camera as far from the athlete as practical and zooming in on them
d) none of the above
Chapter 4: Newton’s laws
1.Newton’s First Law is often referred to as the law of:
a) acceleration
b) action–reaction
c) inertia
d) cosines
2.According to Newton’s Second Law, the greatest change in an object’s state of motion will occur when:
a) the object’s mass and the force applied to it are reduced
b) the object’s mass is increased and the force applied to it is reduced
c) the object’s mass and the force applied to it are increased
d) the object’s mass is reduced and the force applied to it is increased
3.We apply a force against the ground in running, but the force that propels us, the ground reaction force, is directed upwards. This principle is consistent with Newton’s:
a) law of inertia (first law of motion)
b) law of acceleration (second law of motion)
c) law of action–reaction (third law of motion)
d) law of gravitation
4.At the mid-point of a vertical jump an 80 kg athlete was producing a vertical force of 1869 N. What would the athlete’s acceleration have been at this point?
a) 0.04 m·s-2
b) 23.4 m·s-2
c) 149 520 m·s-2
d) none of the above
5.The diameter of a baseball is 0.075 m. How would this be written in scientific notation?
a) 7.5 × 10-2 m
b) 7.5 × 10-3 m
c) 7.5 × 102 m
d) 7.5 × 103 m
Chapter 5: The impulse–momentum relationship
1.Which of the following football players has the greatest momentum?
a) a 70 kg player running at 6 m·s-1
b) a 100 kg player running at 5 m·s-1
c) an 80 kg player running at 8 m·s-1
d) a 60 kg player running at 10 m·s-1
2.What was the vertical impulse provided by a 70 kg runner whose vertical ground reaction force averaged 1100 N over a 0.18 s contact period?
a) 12.6 Ns
b) 198 Ns
c) 6111 Ns
d) 77 000 Ns
3.Assuming a shot-putter was stationary before a throw, what would be the release velocity of a 7.26 kg shot projected by the thrower who produced, on average, 460 N of force over 0.22 s during the throw?
a) 0.07 m·s-1
b) 13.9 m·s-1
c) 280 m·s-1
d) 734 m·s-1
4.Large braking impulses provided by runners would likely:
a) reduce their running speed
b) have no effect on running speed because the benefits and costs are always equal
c) increase their running speed
d) influence running speed only when running on a declined surface
5.Relatively long strokes are used in pursuits such as rowing and swimming because this will result in:
a) a reduced momentum
b) a reduced impulse
c) a greater braking impulse
d) a greater propulsive impulse
Chapter 6: Torque and centre of mass
1.The point about which the mass of a body is evenly distributed in all directions is referred to as the:
a) torque
b) moment of force
c) centre of mass
d) gravitational axis
2.A person is holding a 10 kg weight in their hand, which is 35 cm from the elbow joint. What torque (moment of force) is generated about the elbow by the weight alone?
a) 0.29 Nm
b) 3.5 Nm
c) 28.6 Nm
d) 34.3 Nm
3.If the weight in question 2 had resulted in a torque of 10 Nm at the elbow and the biceps muscle had acted at a distance of 5 cm to maintain a constant elbow angle, what force would the biceps muscle have been producing?
a) 0.5 N
b) 2 N
c) 50 N
d) 200 N
4.When clearing the bar in a high jump or pole vault competition:
a) the centre of mass must pass over the bar
b) the centre of mass much reach a height equal to the height of the bar
c) a successful bar clearance may not require the centre of mass to be higher than the bar
d) the centre of mass is only required to be higher than the bar at the clearance mid-point
5.Using a picture of a person performing a skill, it is possible to calculate the centre of mass using the:
a) segmentation method
b) centre of mass calculation method
c) conservation of momentum method
d) least-squares equation
Chapter 7: Angular kinetics
1.The moment of inertia is influenced by what variables?
a) mass, angular velocity and angular acceleration
b) mass, the distance of the mass from the centre of rotation and angular velocity
c) mass and the distance of the mass from the centre of rotation
d) the distance of the mass from the centre of rotation and angular velocity
2.Children often hold a large bat closer to the centre of the bat (further from the end of the handle). This makes it easier to swing because:
a) the centre of mass is reduced
b) the radius of gyration is reduced
c) the effective mass is increased
d) the radius of gyration is increased
3.What torque must be provided to accelerate a bat weighing 2 kg at 18 rad·s-2 if the bat is held 0.8 m from its effective centre of mass (i.e. the radius of gyration is 0.8 m)?
a) 0.07 Nm
b) 23 Nm
c) 28.8 Nm
d) 3136 Nm
4.One way to reduce the moment of inertia of the leg during the recovery phase in running is to:
a) flex the knee to reduce the leg’s length
b) extend the knee to increase the leg’s length
c) increase the torque applied at the hip
d) decrease the torque applied at the hip
5.Objects that are swinging about an external axis have a ‘remote’ moment of inertia, but they also have a ‘local’ moment of inertia as they spin about their own axis. This complexity is considered in:
a) Newton’s Second Law
b) the dual mass theorem
c) the dual axis theorem
d) the parallel axes theorem
Chapter 8: Conservation of angular momentum
1.The fact that the total angular momentum of a body must remain constant unless an external force (i.e. not an internal one) acts on it is best described by which law?
a) law of conservation of momentum
b) law of conservation of energy
c) law of action–reaction (Newton’s Second Law)
d) law of rotations
2.A diver has left a diving tower and performs a somersault action. Given that no external forces can act on the diver until they hit the water, how can they increase their rate of spin while falling?
a) extend their arms or legs to increase their inertia
b) tuck their body tighter to increase their inertia
c) extend their arms or legs to decrease their inertia
d) tuck their body tighter to decrease their inertia
3.If a figure skater doing a pirouette is able to hold their arms closer to their body such that their moment of inertia is reduced by 10%, what will happen to their angular velocity?
a) it will increase by 10%
b) it will decrease by 10%
c) it will increase by 20%
d) it will decrease by 100% (i.e. 102)
4.At the end of a discus throw, throwers often rapidly bring the non-throwing arm as close as possible to their body. This increases their rate of spin, and thus the discus speed, because it:
a) increases the arm’s radius of gyration
b) decreases the arm’s radius of gyration
c) increases the arm’s moment of inertia
d) increases the arm’s centre of mass
5.A hurdler is about to clear a hurdle in a race. They therefore lift their lead leg (i.e. lift their front foot) over the hurdle. According to the law of conservation of momentum, in which direction is their upper body likely to rotate?
a) backwards, in the direction of the leg being lifted
b) forwards, in the opposite direction to the leg being lifted
c) laterally, to cancel the leg’s rotation
d) the upper body cannot be influenced by the movement of the legs
Chapter 9: Work, power and energy
1.A strength trainer lifts a 40 kg load upwards with constant velocity over 0.6 m. What work was done on the load?
a) 24 J
b) 66.7 J
c) 235.4 J
d) 654 J
2.A weightlifter applies an average force of 1400 N to lift a bar and his own centre of mass a distance of 0.6 m. If it took 0.25 s to perform the lift, what was the average power generated by the lifter?
a) 210 W
b) 466.7 W
c) 3360 W
d) 9333 W
3.By what amount does the kinetic energy of an object change if its velocity triples?
a) it increases 3 times
b) it decreases 3 times
c) it increases 9 times
d) it increases 30 times
4.What is the total energy of a non-rotating ball weighing 100 g that has a velocity of 20 m·s-1 and has 5 m left to fall?
a) 4.9 J
b) 20 J
c) 20.5 J
d) 24.9 J
5.According to the work–energy relationship:
a) the energy of an object decreases as the work done on it increases
b) the energy of an object increases proportionally with the work done on it
c) the work plus energy of an object must remain constant
d) work and energy are inversely related
Chapter 10: Collisions 1 – The ideal case
1.Would you consider momentum (m × v) a vector or scalar quantity?
a) scalar because mass is a scalar unit
b) scalar because velocity is a scalar unit
c) vector because mass is a vector quantity, so momentum must have a direction
d) vector because velocity is a vector quantity, so momentum must have a direction
2.According to the law of conservation of momentum, in an ideal collision:
a) any loss of mass from the colliding objects will result in a decrease in their velocity
b) the velocities of each individual object must be the same before and after the collision
c) the product of mass and velocity of all objects is the same before and after the collision
d) all of the above
3.A 150 g ball dropped from a height hits the stationary Earth, weighing 6 × 1024 kg, at 10 m·s-1. After an ideal collision of the ball with the ground what is the rebound momentum of the ball?
a) 1.5 kg·m·s-1
b) 1500 kg·m·s-1
c) 9 × 1028 kg·m·s-1
d) 9 × 1031 kg·m·s-1
4.A 60 kg volleyballer can produce enough force to gain a momentum of 840 kg·m·s-1. If they lost 3 kg in body mass, how much would their jump velocity change?
a) it would increase by 0.7 m·s-1
b) it would decrease by 0.7 m·s-1
c) it would increase by 47 866 m·s-1
d) it would increase by 52 906 m·s-1
5.Two players accidentally collide on a pitch. Player one was 80 kg and travelling to the left at 6 m·s-1 and player two was 100 kg and travelling to the right at 5 m·s-1. What is the speed and direction of the two collided players immediately after their collision?
a) 0.11 m·s-1 in the direction of player two
b) 5.4 m·s-1 in the direction of player one
c) 20 m·s-1 in the direction of player two
d) it cannot be determined from the information given
Chapter 11: Collisions 2 – The coefficient of restitution
1.If a collision is associated with a higher coefficient of restitution:
a) the combined masses of the colliding objects must be higher
b) the combined speeds of the colliding objects must be higher
c) more energy is lost by the objects during the collision
d) more energy is retained in the objects after the collision
2.A ball hits a wall at a velocity of 15 m·s-1. If the coefficient of restitution of the collision is 0.77, what is the rebound velocity of the ball?
a) 0.77 m·s-1
b) 11.55 m·s-1
c) 19.48 m·s-1
d) this cannot be determined from the information given
3.A ball is dropped from a height of 1 m and rebounds 0.8 m. What is the coefficient of restitution of the collision of the ball with the ground?
a) 0.8
b) 0.89
c) 1.12
d) 1.25
4.One way to increase the coefficient of restitution in a direct collision is to:
a) increase the masses of the objects
b) increase the speed of the objects
c) decrease the speed of the objects
d) there is no way to alter the coefficient of restitution for two objects
5.A home run is more likely to be hit over right field in baseball because:
a) the greater angle of incidence in the bat–ball collision increases the coefficient of restitution
b) the lesser angle of incidence in the bat–ball collision increases the coefficient of restitution
c) balls pitched wide of the batter are usually faster
d) swinging later allows the ball to increase speed more
1.The friction force that opposes motion when an object is sliding is referred to as:
a) static friction
b) sliding friction
c) kinetic friction
d) rolling friction
2.When measured on a force platform, the coefficient of friction of the object–platform interface is equal to:
a) the ratio of the horizontal force to the vertical (normal) force
b) the ratio of the vertical (normal) to horizontal force
c) the horizontal force
d) the vertical force multiplied by the horizontal force
3.An object’s friction on a given surface will increase if:
a) its velocity increases
b) its mass increases
c) its surface area increases
d) its surface area decreases
4.What is the friction force developed between a football boot and the ground when a 100 kg player stands stationary on a surface where the coefficient of friction is 2.2?
a) 0.002 N
b) 22.4 N
c) 220 N
d) 2158 N
5.One way to slide a heavy object (e.g. heavy opponent in rugby) is to:
a) push briefly with less force
b) push the object slightly downwards and horizontally
c) push the object slightly upwards and horizontally
d) push perfectly horizontally
Chapter 13: Fluid dynamics – drag
1.Turbulent flow is characterised by:
a) smooth, parallel layers of fluid flow with minimum energy
b) smooth, parallel layers of fluid that take energy away from an object
c) the mixing of adjacent layers of fluid that helps a moving object retain energy
d) the mixing of adjacent layers of fluid that takes energy away from an object
2.Form drag is influenced by three parameters, which are the object’s:
a) form drag coefficient (Cdρ), frontal surface area and relative velocity (object vs. fluid)
b) form drag coefficient (Cdρ), frontal surface area and the squared relative velocity (object vs. fluid)
c) form drag coefficient (Cdρ), frontal surface area and relative mass
d) form drag coefficient (Cdρ), frontal surface area, relative velocity and relative mass
3.Surface drag is strongly influenced by:
a) the roughness (macro- and microscopically) of the object’s surface
b) the total surface area of the object
c) the viscosity of the fluid
d) all of the above
4.What is the form drag of a 70 kg cyclist riding at 50 km·h-1, if they have a frontal surface area of 0.6 m2 and their measured coefficient of drag in air is 0.8?
a) 6.7 N
b) 24 N
c) 92.6 N
d) 1200 N
5.You measure the friction force on a cyclist on six occasions to determine the reliability of your measurements. Your values are 67, 69, 80, 63, 66 and 70 N. What is the coefficient of variation (as a percentage) for these tests?
a) 5.9%
b) 8.5%
c) 11.8%
d) 69.2%
Chapter 14: Hydrodynamics 1 – drag
1.Wave drag is:
a) present at the interface of the water and air in swimming
c) increased with greater up-down (i.e. bobbing) movements of a swimmer
b) the biggest source of drag in fast crawl-stroke swimming
d) all of the above
2.The wave that forms at the head of a swimmer is commonly called the:
a) bow wave
b) stern wave
c) anterior wave
d) prominent wave
3.While it’s still not clear which factors influence wave drag the most, it is likely that the following will factor strongly (choose the most correct answer):
a) swimming speed, body roll, body mass
b) body roll, body mass, kick amplitude
c) swimming speed, body roll, vertical position of the swimmer in the water
d) none of the above are correct
4.One benefit of the small-amplitude flutter kick, which is used by crawl-stroke swimmers, is that it:
a) provides substantial propulsion
b) increases turbulence and thus minimises form drag
c) increases turbulence and thus minimises stern wave formation
d) prevents wave assistance for other swimmers
5.Increasing yaw of the body in swimming will likely:
a) decrease form drag and improve swimming performance
c) decrease surface drag and improve swimming performance
b) increase form drag and reduce swimming performance
d) decrease skin drag and improve performance
Chapter 15: Hydrodynamics 2 – propulsion
1.Propulsive efficiency is a measure of the ability of a swimmer to fully utilise which physical law?
a) Bernoulli’s law
b) Newton’s Third Law
c) Law of hydrodynamic efficiency
d) Law of cosines
2.Most of the propulsion in swimming comes from the presence of two forces. They are:
a) drag and braking
b) lift and braking
c) drag and friction
d) drag and lift
3.Bernoulli’s theorem is best described as:
a) increases in fluid flow velocity cause decreases in fluid pressure
c) changes in fluid flow velocity are associated with changes in fluid pressure
b) decreases in fluid pressure cause increases in fluid flow velocity
d) the kinetic energy of a fluid must remain constant
4.Research has shown that increases in crawl-stroke swimming speeds are associated with:
a) increases in ventral hand pressures
b) increases in dorsal hand pressures
c) increases in ventral hand pressures and decreases in dorsal hand pressures
d) decreases in both ventral and dorsal hand pressures
5.Lift forces around an aerofoil (such as the hand in swimming) can be best explained by which of the following mechanisms?
a) the fluid flows faster over the top surface than the bottom because it has to travel further, and fluid separating at the leading edge of the hand must reach the trailing edge at the same time; this causes a pressure differential and lift
b) fluid striking the underside of the aerofoil (or hand) causes an upward pressure on the hand, and thus a lift force
c) fluid flowing over the top surface of the aerofoil (or hand) accelerates to move through a smaller space, as the fluid further above acts as a ‘lid’; the faster flow is associated with lower pressure (Bernoulli’s theorem) so the aerofoil is forced upwards
d) the angle of the aerofoil (or hand) causes a turning of the oncoming fluid and change in fluid direction must occur with an opposite movement of the aerofoil according to Newton’s Second Law (i.e. action–reaction or conservation of momentum)
1.The lift force created about a spinning object can be best explained by:
a) Newton’s First Law
b) the Magnus effect
c) Bernoulli’s Effect
d) none of the above
2.The air closest to a ball, which tends to rotate with a spinning ball, is called:
a) the boundary layer
b) the Bernoulli layer
c) the Magnus layer
d) the vortex
3.You see a soccer ball flying through the air and notice that it is spinning back on itself (i.e. backspin). This ball will have a tendency to:
a) swing sideways as it travels
b) dip quickly as it travels
c) swing upwards, or at least tend to ‘hang’ in the air
d) follow a normal parabolic flight path
4.A golf ball spinning faster will:
a) swing further than a ball spinning slower
b) swing less than a ball spinning slower
c) swing the same as a ball spinning slower
d) always swing six times as much as a ball spinning at half the rate
5.In order for a soccer or volleyball to move in a near-random trajectory along a near-parabolic path, it is best to:
a) put topspin on the ball as it is kicked or hit
b) put backspin on the ball as it is kicked or hit
c) kick or hit the ball with no spin at all
d) kick or hit the ball slower
Chapter 17: The kinetic chain
1.A push-like pattern is one in which:
a) all involved joints extend in a sequential order from proximal to distal
b) all involved joints extend in a sequential order from distal to proximal
c) all involved joints extend simultaneously
d) there is no clear sequence of joint extension
2.A movement pattern that is ideal for tasks requiring very high forces is the:
a) push-like pattern
b) throw-like pattern
c) sequential pattern
d) summation pattern
3.The high movement speeds accomplished using throw-like patterns result from either or both of two mechanisms:
a) the transfer of momentum from distal to proximal segments and the re-use of stored elastic energy
b) the transfer of momentum from distal to proximal segments and inertial forces inherent in moving limbs
c) the transfer of momentum from proximal to distal segments and the re-use of stored elastic energy
d) none of the above
4.The learning of complex, throw-like patterns usually:
a) progresses from push-like to throw-like with practice and learning
b) occurs rapidly, even in inexperienced movers
c) cannot be accomplished in children of any age
d) can only occur with specific and detailed movement practice programmes
5.Which of the following would be least likely to be performed with a throw-like movement pattern?
a) a baseball bat swing
b) a soccer kick for maximum distance
c) a fast tennis serve (e.g. first serve)
d) a shot put by a young child
Chapter 18: Gait: Walking and running
1.In walking, the last phase in which the body is in double support is called?
a) loading response
b) midstance
c) toe-off or pre-swing
d) swing
2.In walking the push-off from one phase to the next is largely accomplished by:
a) ankle joint extension (plantar flexion)
b) knee extension
c) simultaneous hip and knee extension
d) appropriately-timed arm swing
3.To walk at a constant velocity, energy lost or dissipated through foot-ground collision or dissipated from muscles is replaced:
a) by generation of arm swing
b) by active muscle work
c) completely through the recovery of elastic (potential) energy
d) through arm swing and recovery of elastic (potential) energy
4.Running allows us to generate rapid forces against the ground so it is a good way to move quickly. But running also helps to reduce energy cost compared to walking at higher speeds because:
a) there is less coactivation of muscles, and therefore less loss of energy from muscle contraction
b) the arms and legs are shortened (joints are flexe
d) in much of the stride, which reduces limb moments of inertia
c) there is no double support phase so the negative (braking) ground reaction forces are smaller
d) none of the above are correct
5.During running the spring-like behaviour of the leg is thought to contribute to improved movement economy (compared to movement without spring-like behaviour) because:
a) some gravitational energy and energy that might be lost through foot-ground collision and other mechanisms can be stored in elastic tissues (e.g. tendons) and reused later in the propulsive phase
b) all the energy that might be lost through foot-ground collision and other mechanisms can be stored in elastic tissues (e.g. tendons) and reused later in the propulsive phase
c) as discussed in Chapter 17 (The Kinetic Chain), tendon recoil is a better mechanism through which to provide efficient muscle-tendon power, so storage and release of tendon elastic (potential) energy is useful in higher-speed gaits such as running
d) both a and c are probably correct
Chapter 1 1. a 2. d 3. d 4. b 5. a
Working for Q3:
v = d/t
To calculate distance we can use Pythagoras’ theorem, where we calculate the hypotenuse
(C2) = A2 + B2 = 36 + 100
Therefore 
Thus v = 11.66/5.2 = 2.24 m·s-1
To find the direction from the start point to end point we can use the sine rule:
Sin θ = opposite/hypotenuse
Sin θ = 10/11.66 = 0.86
θ = inverse sin of 0.86 = 59° (1.03 radians)
Working for Q4:
a = ∆v/∆t = 4.8/2 = 2.4 m·s-2
Chapter 2 1. d 2. c 3. b 4. b 5. b
Working for Q3:
ω = θ/t, but we need to convert degrees to radians so the angle is 260/57.3 = 4.54 radians
ω = 4.54/0.16 = 28.4 rad·s-1
Working for Q4:
α = ∆ω/∆t = (ω2-ω1)/t
α = (40-800)/0.3
α = -760/0.3 = -2533.3°·s-2
But we need the answer in rad·s-2 so we divide by 57.3 = -44.2 rad·s-2 (you could convert degrees to radians in step 1)
Working for Q5:
First we convert 100 cm to metres (divide by 100 = 1 m)
v = rω = 1 × 3.2 = 3.2 m·s-1
Chapter 3 1. b 2. a 3. a 4. c 5. c
Working for Q4:
v = u + at (remember, the ball velocity is zero at the top of the trajectory)
0 = u + -9.81 × 1.4
-u = -13.7 m·s-1 (or throw velocity = 13.7 m·s-1)
Chapter 4 1. c. 2. d 3. c 4. b 5. a
Working for Q4:
F = ma, so a = F/m
a = 1869/80 = 23.4 m·s-2
Chapter 5 1. c 2. b 3. b 4. a 5. d
Working for Q1:
Momentum = m × v, so the largest is 80 kg × 8 m·s-1 = 640 kg·m·s-1
Working for Q2:
The mass of the runner is inconsequential because impulse = F × t
Impulse (J) = 1100 × 0.18 = 198 Ns
Working for Q3:
Ft = ∆mv, and since the mass does not change, the change in velocity is equal to the impulse
v = Ft/m = 460 × 0.22 / 7.26
v = 13.9 m·s-1
Chapter 6 1. c 2. d 3. d 4. c 5. a
Working for Q2:
First we convert 35 cm to metres (divide by 100 = 0.35 m)
Then we convert 10 kilograms to newtons of force (multiply by 9.81 = 98.1 N)
Torque = F × d = 98.1 × 0.35
Torque = 34.3 Nm
Working for Q3:
First we convert 5 cm to metres (divide by 100 = 0.05 m)
Torque (τ) = F × d, so F = τ/d
F = 10/0.05 = 200 N
Chapter 7 1. c 2. b 3. b 4. a 5. d
Working for Q3:
Torque (τ) = Iα, so τ = mk2 × α
τ = 2 × 0.82 × 18 = 23 Nm
Chapter 8 1. a 2. d 3. a 4. b 5. b
Working for Q3:
Angular momentum (H) = Iω so if I is reduced by 10% then ω must increase by 10% to keep momentum constant
Working for Q4:
H = mk2ω (remember, I = mk2), so if the arm is brought closer the overall radius of gyration (k) is reduced and ω must increase in order to keep angular momentum (H) constant
Chapter 9 1. c 2. c 3. c 4. d 5. b
Working for Q1:
Given that the load is moved at constant velocity (i.e. no force is required to accelerate it) then the force is only large enough to keep the object moving. Therefore ‘force’ is equal to exactly that needed to overcome the 40 kg load, so we simply convert the kilogram load to newtons (multiply by 9.81 = 40 × 9.81 = 392.4 N)
W = F × d = 392.4 × 0.6
W = 235.4 J
Working for Q2:
Power = F × d/t = 1400 × 0.6/0.25
Power = 3360 W
Working for Q3:
KE = ½mv2 so if velocity increases by three times then KE increases by 32 = 9 times
Working for Q4:
First we convert 100 g to kilograms (divide by 1000 = 0.1 kg)
Total energy (TE) = kinetic energy + potential energy
TE = KE + PE = ½mv2 + mgh
TE = (½ × 0.1 × 202) + (0.1 × 9.81 × 5)
TE = 24.9 J
Chapter 10 1. d 2. c 3. a 4. a 5. a
Working for Q3:
The mass of the Earth is inconsequential
First we convert 150 g to kilograms (divide by 1000 = 0.15 kg)
Momentum (p) = mv = 0.15 × 10 = 1.5 kg·m·s-1
Momentum (p) = mv
So, the velocity at 60 kg = p/m = 840/60 = 14 m·s-1
Velocity after weight loss = 840/57 = 14.7 m·s-1
Change in velocity = vafter – vbefore = 14.7 – 14 = 0.7 m·s-1 (i.e. an increase in velocity)
Working for Q5:
Total momentum (ptot) of players after the collision = m1v1 + m2v2 (where 1 and 2 refer to players 1 and 2)
ptot = (80 × 6) + (100 × -5) = -20 kg·m·s-1 (I chose player two to be running in the negative direction)
If their combined mass (masstot) = 80 + 100 kg = 180 kg, then the resulting velocity = ptot/masstot
Resulting velocity = -0.11 m·s-1 (i.e. in direction of player two)
Chapter 11 1. d 2. b 3. b 4. c 5. a
Working for Q2
Only 0.77 (77%) of the energy is retained in the collision, and since the mass won’t change the velocity must decrease to 77% of its value = 0.77 × 15 = 11.55 m·s-1
We could also use the equation v1 – v2 = -e (u1 – u2). If the ball is object 1 and the wall is object 2 we can write:
v1 – 0 = -e (u1 – 0), v1 = -0.77 × 15 = -11.55 m·s-1 (the negative sign denotes the ‘rebound’ direction of the ball, so the rebound velocity is 11.55 m·s-1)
Working for Q3:
Chapter 12 1. b 2. a 3. b 4. d 5. c
Working for Q4:
First we convert 100 kg to newtons, which is the weight force (normal reaction force, R) acting straight down (multiply by 9.81 = 981 N)
Ff = µR = 2.2 × 981 = 2158 N
Chapter 13 1. d 2. b 3. d 4. c 5. b
Working for Q4:
The mass of the cyclist is inconsequential
First we convert 50 km·h-1 to metres per second (× 1000/3600 = 13.9 m·s-1)
Fd = kAv2 = 0.8 × 0.6 × (13.92) = 92.6 N
Mean (X–) of the measurements = 69.16 N
Standard deviation (SD) of the measurements = 5.85 N
CV (%) = SD/X– × 100 = 5.85/69.16 × 100 = 8.45% (or 8.5%)
Chapter 14 1. d 2. a 3. c 4. c 5. b
Chapter 15 1. b 2. d 3. c 4. d 5. d
Chapter 16 1. b 2. a 3. c 4. a 5. c
Chapter 17 1. c 2. a 3. c 4. a 5. d
Chapter 18 1. c 2. a 3. b 4. b 5. d