Question:
1 (a) Students are expected to use the definition of power to show that its base SI units are kg m2 s–3. [2 marks]
(b) Students have to use a mathematical expression for electrical power to determine the base SI units of potential difference. [2 marks]
2 (a) Students are expected to provide definitions of speed and velocity and make use these definitions to explain why speed is a scalar and velocity is a vector. [2 marks]
(b) A ball is released from rest. The ball falls vertically, hits the ground and rebounds vertically. Students may assume there is no air resistance.
(i) Students have to describe the variation of the velocity of the ball with time from t = 0 to t = 2.1 s. [3 marks]
(ii) Students have to calculate the acceleration of the ball after it rebounds from the ground. [3 marks]
(iii) Students have to calculate the distance and displacement from the initial position moved by the ball from t = 0 to t = 2.1 s. [5 marks]
(iii) Students have to calculate the distance and displacement from the initial position moved by the ball from t = 0 to t = 2.1 s. [5 marks]
(iv) Students have to plot a graph that shows the variation of the speed of the ball with time. [2 marks]
Mark Scheme:
1 (a) power = work / time = (force × distance) / time [1 mark]
[power] = (kg m s–2 × m) s–1 = kg m2 s–3 [1 mark]
[power] = (kg m s–2 × m) s–1 = kg m2 s–3 [1 mark]
(b) power = VI [1 mark]
[V] = kg m2 s–3 A–1 [1 mark]
[V] = kg m2 s–3 A–1 [1 mark]
2 (a) speed = distance / time and velocity = displacement / time [1 mark]
speed is a scalar as distance has no direction and velocity is a vector as displacement has direction [1 mark]
(b) (i) Indicate that the ball has a constant acceleration or linear/uniform increase in velocity until 1.1 s [1 mark]
Indicate that the ball rebounds or bounces or changes direction [1 mark]
Indicate that the ball decelerates to zero velocity having the same acceleration as the initial value [1 mark]
(ii) Use a = (v – u) / t or gradient of the graph [1 mark]
Student’s working shows that (8.8 + 8.8) / 1.8 or appropriate values from line or (8.6 + 8.6) / 1.8 [1 mark]
Student’s working shows that 9.8 (9.78) m s–2 or = 9.6 m s–2 [1 mark]
(iii) 1. distance = first area above graph + second area below graph [1 mark]
First area = (1.1 × 10.8) / 2 + Second area = (0.9 × 8.8) / 2.
Total area = 5.94 + 3.96 [1 mark]
Distance = 9.9 m [1 mark]
2. Displacement = first area above graph – second area below graph [1 mark]
Displacement = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m [1 mark]
(iv) The graph has a correct shape with straight lines and all lines are above the time axis [1 mark]
The graph shows correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds (10.8 m s–1 at 1.1 s and 8.8 m s–1 at 1.2 s and 3.0 s) [1 mark]
Possible answers:
1 (a) power = work / time = (force × distance) / time.
Base units of power = (kg m s–2 × m) s–1 = kg m2 s–3
(b) power = VI
Base SI units of VI = kg m2 s–3 (as shown above)
Base SI units of V = kg m2 s–3 / A = kg m2 s–3 A–1
2. (a) Speed is the rate of change of distance travelled with respect to time.
Velocity is the rate of change of displacement with respect to time.
Speed is a scalar quantity because distance has no direction, whereas velocity is a vector quantity because displacement has a direction.
Alternative answers:
Speed is a scalar quantity defined by the word equation = distance traveled/time taken. Velocity is a vector quantity defined by the word equation = change of displacement/time taken.
Speed is a scalar that is always positive because the distance is the numerator and velocity is a vector that can be positive or negative because the displacement is the numerator.
(b) (i) The graph has a constant acceleration from 0.0 sec to 1.1 sec. Then, it changes direction and decelerates to zero velocity having the same magnitude of acceleration as the initial value.
(ii) Using a = (v – u) / t = (8.8 + 8.8) / 1.8 = 9.78) m s–2
(iii) 1. Distance travelled = first area above graph + second area below graph
= (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 = 5.94 + 3.96 = 9.9 m
2. Displacement from the initial position = first area above graph – second area below graph = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m
(iv) Graph (variation of the speed of the ball with time)
Feynman’s insights or goofs?:
The mark scheme simply states that “speed = distance/time.” However, Feynman elaborates that “we can define the speed in this way: We ask, how far do we go in a very short time? We divide the distance by the time, and that gives the speed. But the time should be made as short as possible, the shorter the better… (Feynman et al., 1963, section 8–2 Speed).” To explain the subtleties in a definition of speed, Feynman provides the following joke: A lady driving a car is caught for high-speed driving, and a cop says, “Lady, you were going 60 miles an hour!” She argues that this is impossible because she was traveling for only seven minutes. Thus, the cop explains that “lady, if you kept on going the same speed in one hour you would go 60 miles.” Interestingly, her answer was, “Well, the car was slowing down and it would not go 60 miles.” Alternatively, she could argue that if the car moves at the same speed, then she would run into a wall! More important, we need to explain the concept of instantaneous speed and how speed can be measured by using an accurate speedometer.
The mark scheme states that velocity = displacement/time instead of velocity = (change of displacement)/(time taken). Nevertheless, Feynman provides a mathematical definition of velocity as follows: “[l]et us try to define velocity a little better. Suppose that in a short time, ϵ, the car or other body goes a short distance x; then the velocity, v, is defined as v = x/ϵ, an approximation that becomes better and better as the ϵ is taken smaller and smaller (Feynman et al., 1963, section 8–2 Speed). This definition of velocity is based on the ratio of an infinitesimal distance to the corresponding infinitesimal time. Theoretically speaking, we imagine what happens to the ratio as the time we use is shorter and shorter. In other words, we take a limit of the displacement traveled by an object divided by the time elapsed, as the time taken is assumed to be shorter and shorter, ad infinitum. This idea was independently invented by Newton and Leibnitz and it is now known as calculus.
Importantly, Feynman clarifies that “[o]rdinarily we think of speed and velocity as being the same, and in ordinary language they are the same. But in physics, we have taken advantage of the fact that there are two words and have chosen to use them to distinguish two ideas. We carefully distinguish velocity, which has both magnitude and direction, from speed, which we choose to mean the magnitude of the velocity, but which does not include the direction (Feynman et al., 1963, section 9–2 Speed and velocity).” In short, velocity is speed in a specified direction. Thus, we can also define velocity by describing how the x-, y-, and z-coordinates of an object change with time, as well as write vx = Δx/Δt, vy = Δy/Δt and vz = Δz/Δt.
Reference:
Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
