Unit 1: Energy & Motion

 

Supervisor’s Manual

Contents

Answers to Questions

Lesson 1.5: Inertia

Lessons 1.6: Vectors

Lesson 1.7: Components of Vectors

Lesson 1.8: Pressure

 

Laboratory Classes

Lab 1.5 Integrated Forces & Bows

Lab 1.6 Acceleration: Ticker Tape Timer

Lab 1.7 Vector Frog Racing

Lab 1.8 Manometer

 

Unit 1 – Energy & Motion

Answers to Questions

 

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Lesson 1.5 Inertia

  1. What is the difference between inertia and momentum?
  2. How is inertia defined? What is the SI unit of inertia?
  3. What is the difference between mass and volume?
  4. What is the difference between mass and weight? What is the mass of 1 kg on the moon?
  5. State Newton’s first law of motion.
  6. What does the gravitational force between two objects depend upon?
  7. What is the weight of 1 kg on the Moon?
  8. If two astronauts decide to practice throwing and catching a baseball while outside their spaceship, what difficulties might they encounter?

 

  1. No speed is needed for an object to have inertia. Momentum is the product of an object’s mass and velocity.
  2. Inertia is a measure of an objects mass or tendency to oppose acceleration. The SI unit of inertia is the kilogram (kg).
  3. Mass is a measure of the amount of matter. Volume is a measure of the amount of space an object or body of matter occupies.
  4. Weight is the force that gravity exerts on a certain amount of mass. 1 kg has a mass of 1 kg anywhere but on the moon it will have a weight far less than it would have on earth where the force of gravity is much greater.
  5. Newton’s First Law: An object that is at rest will tend to stay at rest and an object that is in motion will tend to continue its motion in the same direction with the same speed unless a force or set of unbalanced forces acts upon it.
  6. Both of their masses and their distance apart.
  7. 1 kg weighs roughly 1.6 N on the moon.
  8. Every time they throw the ball, they will accelerate in the opposite direction with nothing to stop them – ever.

 

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Lesson 1.6 Vectors

 

Review Questions

  1. What is the difference between speed and velocity?
  2. If an airplane is flying in a crosswind that blows at 90º to the plane’s direction through the air, how does the wind affect the path of the plane relative to the ground?
  3. On the diagram below, one arrow represents the speed of the plane relative to the air (it’s airspeed). The second arrow represents the velocity of the wind relative to the ground. Draw an arrow on the diagram that represents the velocity of the plane relative to the ground.
  4.  
  5. Use the diagram below to determine the resultant of 2 forces acting at right angles to each other. In terms of the scale used, Force 1 has a magnitude of 4 Newtons and has a length of 4 graduations on the scale. Force 2 has a magnitude of 7 N and has a length of 7 graduations. Use a ruler to measure the length of the resultant and estimate its magnitude.
  6.  
  7. Use the diagram below to determine the resultant of 2 forces acting at less than 90º to each other. Force 1 has a magnitude of 5 Newtons and force 2 has a magnitude of 7 N. Complete the parallelogram, use a ruler to measure the length of the resultant and estimate the magnitude of the resultant.

Lesson 1.6 Vectors

  1. Speed is a scalar quantity. No direction needs to be specified. Velocity is speed in a given direction. Velocity is a vector quantity.
  2. The path that the plane follows relative to the ground is a combination of the air speed of the plane and the wind speed.
  3.  

 

4. On the diagram 55mm = 7 N. The length of the resultant is 63mm.
This is equivalent to 63/55 x 7 = 8 N.
We can measure the angle or calculate it to be tan-1 (4/7) = 29.7º.



5 On the diagram 58mm = 7 N. The length of the resultant is 90mm.
This is equivalent to 90/58 x 7 = 10.86 N.
We can measure the angle or calculate it to be tan-1 (2.7/7) = 21.1º.

 

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Lesson 1.7 Components of Vectors

Review Questions

  1. If a force acts at 25º to the vertical, can a portion of the force act in the vertical direction? Is this the vertical component of the force? Does the force acting at 25º to the vertical also have a horizontal component?
  2.  
  3. The diagram at the right represents the velocity of an object moving at 10 m/s in a direction 36.87º to the horizontal. Use the diagram to measure
    a) The vertical component of its velocity and:
    b) The horizontal component of its velocity.
  4.  
  5. If a cannonball is fired at an angle of 36.87º to the horizontal with a velocity of 50 m/s, use the vector diagram below to estimate a) The vertical component of its velocity? and b) The horizontal component of its velocity?
  6.  
  7. Use Diagram 4.2 to determine the resultant of the vectors acting on point A in Diagram 4.1 by moving the vectors such that Vector 2 starts at the end of Vector 1 etc.
  8.  
  9. What is the sum of the vertical components of the forces shown in Diagram 5.1 below?
  10. What is the sum of the horizontal components of the forces shown in Diagram 1.5 above?
  11.  
  12. Determine the vertical and horizontal components of the 4 forces shown acting on point A in Diagram 7.1 below. Use Diagram 7.2 to indicate the sum of the vertical components and the sum of the horizontal components and to estimate the resultant of the 4 forces.

 

 

Lesson 1.7 Components of Vectors

  1. Yes. Yes. Yes.
  2. Vertical component = 6 grid marks. 1 grid mark = 1 m/s.
    Vertical component = 6 m/s.
    Horizontal component = 8 m/s
  3. Vertical component = 30 m/s, horizontal component = 40 m/s.
  4. The resultant is shown on the diagram.
    It’s length and direction are measured on the graph.
  5. The blue vector has a vertical component of 2, the red vector has a vertical component of 1 and the green vector has a vertical component of –2. The sum of the vertical components is thus 1.
  6. The horizontal components are 5, -3 and –1 which add up to 1.
  7. The vertical components are: 2, 2, -2 and –4. These add up to –2. The horizontal components are 3, -1, -5 an 4. These add up to 1.
    We can estimate the resultant by measuring its length and direction the diagram below:

 

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Lesson 1.8 Pressure

 

Review Questions

  1. What is the pressure exerted on the floor by a person who weighs 50 kg and rests all of her weight on a stiletto heel with an area of 1 square centimeter?
  2. Why is it a good idea to let the air out of the tires if a car gets stuck in soft sand?
  3. If a probe is used to measure the pressure at a certain depth in a water tank, would a more accurate reading be obtained if the probe was pointing upwards, sideways or downwards (at the same depth)?
  4. Why is it easier to compress gases than liquids?
  5. What is the pressure (above atmospheric pressure) at a depth of 30 meters in salt water with a density of 1100 kilograms per cubic meter?
  6. What is the pressure above atmospheric pressure at the bottom of a beaker containing a liquid with a density of 950 kg/m3 if the depth of the liquid is 8 cm?
  7. If nearly the earth’s entire atmosphere is located in a region that is roughly 30 kilometers deep, why is the pressure at 15 km above the earth’s surface not roughly half of that at the earth’s surface?
  8. A hydraulic jack has a small piston with a diameter of 1cm and a larger piston with a diameter of 3cm. How much pressure is exerted on the circular surface of the larger piston if the smaller piston exerts a pressure of 300 kPa on the oil connecting the circular surfaces of the two pistons? (Assume that the surfaces of the pistons are at the same depth in the oil)
  9. A hydraulic jack has a small piston with a diameter of 1cm and a larger piston with a diameter of 4cm. How much force is exerted on the circular surface of the larger piston if the smaller piston exerts a force of 300 N on the oil connecting the circular surfaces of the two pistons? (Assume that the surfaces of the pistons are at the same depth in the oil)

Lesson 1.8 Pressure

  1. 50 kg has a weight of 50 x 9.81 N = 490.5 N
    1 square cm has an area of 1/(100 x 100) = 0.0001 m2.
    The pressure is thus 490.5 / 0.0001 = 4,905,000 Pa.
  2. The weight of the car can be spread over a larger area and the pressure on the sand is reduced considerably.
  3. The pressure is the same in all directions. There would be no difference between the probe readings.
  4. I suppose that the easiest answer is that gases are compressible and liquids are usually regarded as incompressible. The reason for this is that the gas molecules are not held together with the same forces that hold liquid particles together and they can move about more freely and thus move closer to each other or further apart.
  5. 1100 x 9.81 x 30 = 323,730 Pa.
  6. 950kg/m3 x 9.81N/kg x 0.08m = 745.6 Pa.
  7. The density of the air drops with pressure. It varies as the air pressure varies.
  8. The pressure is the same: 300 kPa. (300,000 Pa.)
  9. Area of small piston in contact with the oil = p x (1/2)2 = 0.7854 cm2
    Area of larger piston in contact with the oil =     p x (4/2)2 = 12.57 cm2
    The force on the larger surface will be multiplied by the ratio of the areas:
    Force = (12.57/0.7854) x 300N = 4800 N

 

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Physics Lab 1.5

Integrated Forces & Bows

Integrated Forces & Bows: Measure the force applied over a number of small distances as the string of a bow is pulled away from the bow. Multiply each small distance by the average force used to move that distance. The sum of these quantities represents the energy stored in the bow.

Marble Coaster: Design a marble coaster track (chute) that has as many loops, twists and turns as possible. Marbles and chutes will be used later to study projectiles and conservation of momentum in collisions.

Question

If a line graph is plotted with the force needed to move the string on the y-axis and the distance moved by the string on the x-axis, how would the area under the line be related to the energy stored in the bow?

Answer

Each pair of readings could be represented as a vertical bar on a graph. The area of the bar (Force x Distance) would equal the energy stored in the bow during the movement of the string through that particular distance.

The area under a line graph could be regarded as the combined area of an infinite number of bar graphs – each with an infinitesimally small width on the x-axis. The area under the graph represents the total energy or the integral of the forces with distance.

 

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Physics Lab 1.6 Measuring Acceleration

  1. A car covers a measured mile in 55 seconds. What is its average speed?
  2. The time taken for a toy car to travel between two marks on a table is 3.8 seconds. If the marks are 1.5 meters apart, what is the average speed of the car between the two marks on the table?
  3. A laboratory cart is attached to a strip of paper that passes through a timer as it accelerates. The timer makes dots on the strip of paper every 0.025 seconds. The distances between three successive dots are 7.4mm and 8.8mm.
    What is the average speed of the cart represented by the distance between the first two dots?
    At what rate is the cart accelerating?

 

Physics Lab 1.5 Measuring Acceleration

  1. 1 x 60/55 = 1.09 miles per minute or 65.45 miles per hour.
  2. 1.5 / 3.8 = 0.395 m/s
  3. Average speed between first two dots = 7.4 mm / .025 s = 296 mm/s = 0.296 m/s
    Average speed between the next two dots = 8.8 / .025 = 352 mm/s = 0.352 m/s
    Average increase in speed in 0.025 seconds = (0.352 – 0.296) = 0.056 m/s
    0.056 m/s increase in 0.025 seconds = 2.24 m/s2.

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Physics Lab 1.7 Vector Frog Racing

Questions for Discussion

  1. What tends to happen when the frog is lifted too quickly?
  2. Is there any advantage to standing up as opposed to sitting down while racing a frog? Why?
  3. The base of the frog moves forward when it is lifted as a result of an increase in tension in the string. Why does the string not slip through the hole in the frog when the tension in the string is increased?

Answers to questions

  1. The frog tends to flip past the vertical and will fall backwards when the tension in the string is released.
  2. There may be an advantage. Less tension will be needed to lift the frog and it may be easier to prevent the frog from flipping over backwards. However if the string is kept low, this will limit the maximum height to which the frog may rise during each cycle. Keeping the string low – while sitting down – may be a better way to ensure that the frog keeps moving forward.
  3. The friction between the string and the frog depends on the force between the two surfaces. When the string is released, the frog can flop forwards because the friction between the string and the plywood is very small and the friction between the base of the frog and the floor is relatively large. When there is tension in the string, the force between the plywood and the string increases. The process of lifting the frog also reduces the friction between the base of the frog and the floor. Increased friction between the frog and the string fixes the position of the upper portion of the frog. Reduced friction between the base of the frog and the floor allows the base to slide forward.

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Physics Lab 1.8 Simple Manometer

If water with a density of 1000 kg/m3 is used in the manometer indicated below, what pressure difference is it indicating. (The difference in height of fluid in the two legs is 50 cm.)

 

Answer:

Pressure = 1000 kg/m3 x 9.81 N/kg x 0.5 m = 4905 N/m2 (or Pa.)

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