Analysis
1. | Yes, reaction times increased when entering the telephone number. |
2. | Other distractions could include loading a CD; looking for something in the console, a purse, or a backpack; and shoulder checking. |
Identifying Alternatives/Perspectives
1. | Answers will vary. A sample answer is given.
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Researching the Issue
2. | Answers will vary. A sample answer is given.
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Analyzing the Issue
3. | Answers will vary. A sample answer is given.
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Reactions of three of the stakeholders to the lists of risks and benefits are as follows:
Taking a Stand and Defending Your Position
4. | Answers will vary. A sample answer is given.
HID headlights should be banned until the technology has been refined. The present circumstance is that the benefits to the driver using HID headlights is outweighed by the risks created for all the drivers in the stream of oncoming traffic. It does not make sense to make night-time driving safer for some drivers and more dangerous for many other drivers. A second point is that the drivers of vehicles equipped with HID headlights are not mutually exclusive from the drivers who face these lights from oncoming traffic. It is possible that a driver using HID headlights could face another vehicle with HID headlights; then nearly all the benefits afforded by this technology are nullified by the glare from the oncoming HID headlight user. |
Evaluation
5. | Answers will vary. A sample answer is given.
Other students made the argument that one of the problems with headlights in general is that they are mounted too high on some vehicles, especially SUVs and trucks, causing these lights to create a greater hazard for oncoming passenger cars. If headlights were mounted at the same height above the roadway, the glare problem would be reduced. Another point was that it is possible to modify these lights so they emit less blue light, thus reducing the effect of HID headlights on drivers in oncoming vehicles. Although these are good points, I still maintain that HID headlights should be banned until these changes and other improvements have been incorporated so that the risks for drivers in oncoming vehicles have been reduced. |
Procedure
1. and 2. | Note that the average velocity values reveal that within a reasonable amount of experimental error, the velocity was constant for this experiment.
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3. | ![]() |
4. | ![]() |
5. | ![]() |
6. | ![]() |
Analysis
7. | a. | The best-fit line of the position–time graph is a straight line with a constant slope of 30.0 cm/s. |
b. | The average velocity is determined by calculating the slope of this line. | |
c. | The total displacement is the point on the line corresponding to 4.50 s, which is 135 cm. | |
8. | a. | The best-fit line of the average velocity–time graph is a straight, horizontal line with a constant slope of 0. |
b. | The average velocity is where the best-fit line intercepts the vertical axis, which is 30.0 cm/s. | |
c. | The displacement is the area under this graph. Since the shape of this area is a rectangle, the area is equal to the length times the width. This means that the displacement is 30.0 cm/s × 4.50 s = 135 cm. |
Procedure and Analysis
1. | a. | Yes, this is uniform motion because the object moves in a straight line at a constant speed. |
b. | Yes, the shape of the best-fit line is the same. Also, this line has the same endpoints. | |
c. | The slope value of 0.30 m/s (30.0 cm/s) is the average velocity for this motion. This is consistent with the work done previously. | |
d. | The velocity–time graph is a horizontal, straight line with a y-intercept of 0.30 m/s (30.0 cm/s). This value is consistent with the work done previously. | |
e. | The area under the graph corresponds to the displacement of the object. This value, 1.35 m (135 cm), is consistent with the work done previously. |
Check for Understanding
2. | a. | ![]() |
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The rider is 90 m [N] from base camp. |
b. | ![]() |
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c. | The applet confirms the shapes of these graphs. |
d. | ![]() Using the Area tool, the displacement value is + 90 m. This is the same as in the answer to question 2.a. |
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3. | a. | Since the rider is travelling south, the velocity and displacement values will be negative. |
b. | ![]() |
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The rider is 54 m [S] from base camp. |
c. | ![]() |
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d. | The applet confirms the shapes of these graphs. | |
e. | Using the Area tool, the displacement is –54 m or 54 m [S]. This is the same as in the answer to question 3.b. |
Analysis
1. | Data will vary. Use the sample data provided in the table to calculate the average velcoity and midpoint times.
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2. | ![]() |
3. | ![]() |
4. | The values for acceleration will depend upon how steep the ramp is. The following calculation matches the graph in the answer to question 3.
Remember, you must choose two points on the best-fit line, not any of the plotted points. In this case, points (1.00 s, 14.0 cm/s) and (5.00 s, 39.0 cm/s) are used. The slope of the best-fit line is 6.25 cm/s2. |
5. | The position–time graph shows that the cart travelled further in each time interval than it did in the previous interval. Since the cart was travelling farther in each subsequent time interval, it must have been speeding up. |
6. | The velocity–time graph shows that the velocity values are increasing. This means that the cart was speeding up. |
Evaluation
7. | You should compare the actual shapes of the graphs with their predicted shapes. |
Design
Designs will vary. Your design may be similar to that used in the “Accelerated Motion—Speeding Up” investigation. For this investigation, however, the cart is made to travel from the lower end of the table towards the higher end. The trigger-released, spring-loaded piston fires from blocks of wood that the timer rests upon. The blocks of wood are secured against a wall. This end of the table is the lower end, while the far end of the table is raised with blocks of wood, as in the previous investigation (Part A).
Your procedure should indicate that you only analyze the motion of the cart after it has become free of the spring—that’s when the cart starts to slow down and the differences in position values (in the data) become smaller.
It is important to ensure that your design discusses safety issues and promotes the responsible use of the equipment. In your design, you should have accounted for the fact that the trigger-released, spring-loaded piston cannot “fire” into the timing device. Your design should include some safeguards so that the cart does not fall to the floor.
Analysis
3. | Data will vary. Sample data is provided.
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4. | ![]() |
5. | ![]() |
6. | Answers will vary. The following calculation of the slope of the velocity–time graph matches the sample graphs and the sample data.
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7. | The position–time graph shows that the cart travels less distance in each subsequent time interval. This means that the cart is slowing down. |
8. | The velocity–time graph shows that the velocity is decreasing. This means that the cart is slowing down. |
Evaluation
9. | Answers will vary. You should compare the shape of your predicted graph with the shape of your actual graph. |
Process
step 1: | You should have identified questions like the following that need to be investigated in the field study:
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step 2: | Methods and tools for collecting data might include the following:
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step 3: | Sample processes for data analysis might include the following:
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step 4: | Sample safety precautions might include the following:
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Part A: Varying the Mass of the Vehicle
Procedure
Part A: Varying the Mass of the Vehicle
Constant Values
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Analysis
1. | As the mass of the vehicle increases and the force of friction kept constant, the magnitude of the acceleration decreases. |
2. | In each of these trials, the only force acting is the force of friction. Since this force is the only force, it is also the sum of all the forces, or the net force. |
3. | Both the acceleration and the net force are pointing in the negative direction. |
Part B: Varying the Net Force
Procedure
Part B: Varying the Net Force
Constant Values
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Analysis
4. | As the magnitude of the net force decreases, the magnitude of the acceleration decreases. | |
5. | a. | The net force can be zero if all the forces acting on the vehicle cancel out. |
b. | The force of friction on a vehicle could be minimal if the truck were travelling at a very low speed (to reduce air resistance) and if the truck were moving on an extremely low-friction surface, such as a thin layer of water or ice (to reduce rolling resistance). | |
6. | The net force and the acceleration are both travelling in the negative direction. |
Part A: Causing the Vehicle to Speed Up
Procedure
Part A: Causing the Vehicle to Speed Up
Constant Values
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Analysis
1. | The net force was 1000 N less than the applied force in each case because the force of friction of 1000 N was acting in the negative direction. |
2. | Yes, these results are consistent with Newton’s second law because, in each case, the acceleration value is equal to the net force divided by the mass. |
Check Your Understanding
3. | a. | The applet returns a value of 0.500 m/s2 for this scenario. |
b. | ![]() |
First, determine the net force. | Next, determine the acceleration. | ||
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The acceleration of the vehicle is +0.5000 m/s2. |
Part B: A Special Case
Procedure
Part B: A Special Case
Constant Values
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Analysis
4. | Newton’s second law explains that the acceleration values are equal to the net force divided by the mass. If the net force value is zero, then so is the acceleration. |
5. | The animated vehicle on the screen also indicated that the acceleration was zero because it moved with uniform motion across the screen. |
6. | In all these cases, the vehicle will not accelerate because the forces are balanced, so the velocity will remain a constant value of zero. |
Check Your Understanding
7. | a. | If the vehicle is given an initial velocity of 4.00 m/s when the force of friction was set to be zero, the value that must be entered into the applet for the applied force is zero in order for the vehicle to maintain its velocity and not accelerate. |
b. | According to Newton’s second law, in order for a vehicle not to accelerate and move with uniform motion, the net force must be zero. In other words, the applied force and the force of friction must cancel each other out. Since the force of friction is zero, the applied force must also be zero. |
Procedure and Analysis
1. | According to Newton’s first law of motion, in the absence of a net force, a body at rest tends to remain at rest. If no head restraint is used, the seat is unable to exert a net force on the ball. Since there is no net force acting on the ball, the ball tends to remain at rest as the seat is shot out from underneath it. |
2. | The reasoning is identical to that used with the ball in question 1. Without the head restraint, the seat is unable to exert a net force on the head, and so the head tends to remain at rest as the rest of the body is accelerated out from underneath it. |
3. | If no head restraint is used, the head tends to remain at rest while the body is accelerated out from underneath it. Since the body and the head are still connected by the tissues in the neck, these tissues experience significant stresses and strains as the body is rapidly accelerated out from underneath the head. Eventually, the displacement of the head from the body reaches its limit, and the head is suddenly pulled forward by the hyperextended tissues in the neck. This creates an effect like the snap from the end of a whip, as the head is now pulled forward to catch up with the body. The neck is now hyperextended in the forward direction as the head moves ahead of the body. The rapid hyperextension in two different directions—first backwards then forwards—causes the neck injuries known as whiplash. |
4. | Newton’s first law of motion says that a body will remain at rest only if there is no net force acting on it. To prevent the head from remaining at rest while the rest of the body accelerates, it is necessary to exert a net force on the head so that it accelerates forward with the rest of the body. The head restraint allows the driver’s seat to provide a net force on the head at the same time that it applies a net force to the rest of the body. |
Procedure and Analysis
1. | According to Newton’s first law of motion, an object in motion will tend to maintain its velocity in the absence of a net force. If no seat belt is used when the seat strikes the barrier, there is no net force that can be exerted on the ball. In this circumstance, the ball maintains its velocity and continues to travel beyond the seat, even though the seat has stopped. |
2. | The explanation is identical to that given for the ball that had no seat belt in question 1. Since there is no seat belt to provide a net force on the crash test dummy, the dummy maintains its velocity and continues to move forward until it collides with the interior of the car or the windshield. |
3. | If no seat belt is used, the person’s body maintains its velocity and continues to move forward. While the person’s abdomen and trunk collide with the lower parts of the dashboard, the person’s head can collide with the upper parts of the dashboard and the windshield. Passengers in the back seat can collide with the back of the front seat. The driver can collide with the steering wheel. |
4. | Newton’s first law indicates that injuries can be reduced if the occupants inside a vehicle are not permitted to maintain their velocity when a front-end crash occurs. The seat belt provides a net force to decelerate the occupants as the vehicle decelerates. |
Procedure
Although the results will vary greatly with the steepness of the ramp, the following results are considered typical.
Number of Marbles Released |
Release Point from Bend (cm) |
Number of Marbles Ejected After Collision | Distance Cup Moves (cm) |
1 | 10 | 1 | 3.5 |
1 | 20 | 1 | 9.5 |
1 | 30 | 1 | 13.0 |
2 | 10 | 2 | 7.0 |
2 | 20 | 2 | 15.0 |
2 | 30 | 2 | 26.0 |
Analysis
1. | Increasing the height increases the impacting velocity of the released marble as it reaches the four target marbles. |
2. | As the height of the released marble increases, the velocity of the marble ejected after the collision also increases. Since the ejected marble has a greater velocity, it is able to move the cup a greater distance. |
3. | When the number of released marbles is doubled, the total mass of the incoming marbles is doubled. |
4. | When the number of released marbles is doubled, the mass of the ejected marbles also doubles. (Two marbles are now sent toward the cup.) The cup moves roughly twice as far as it did for a single marble released from the same height. |
Evaluation
5. | Based on the results, two significant characteristics of an object in motion are mass and velocity. |
Background Information
1. | The cup experienced the greatest resistance to its motion when it was pushed on the textured paper towel. The force of friction must have been greatest in this case. The cup experienced the least resistance on the aluminium foil. The force of friction must have been the least value in this case. |
Procedure
Data will vary. Sample data is given. Calculate the average stopping distance for each trial.
Answers:
Surface Under Cup | Stopping Distance (cm) | Average Stopping Distance (cm) | ||||
Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 5 | ||
textured paper towel | 13.5 | 12.5 | 14.0 | 13.0 | 13.0 | 13.2 |
loose-leaf paper | 16.0 | 15.0 | 14.5 | 15.0 | 15.0 | 15.1 |
aluminium foil | 17.0 | 15.0 | 19.0 | 16.0 | 17.0 | 16.8 |
Analysis
2. | Since the marble was released from the same height in each case, it acquired the same velocity at the bottom of the ramp for each trial. Also, since the marble came to rest in each case, the final velocity was also the same for each trial. |
3. | Since the cup travelled the longest distance on the aluminium foil, the foil provided the longest time interval. Similarly, since the cup travelled the shortest distance on the textured paper towel, it provided the shortest time interval. |
4. | According to the sample data, the largest force of friction was caused by the cup moving over the textured paper towel. The smallest force of friction was caused by the cup moving over the aluminium foil. |
5. | The change in momentum was the same in all cases. This is due to the fact that the cup and marble had the same mass and that the initial and final velocities were the same in all trials. |
Evaluation
6. | For a given change in momentum, as the force increases, the time interval decreases, and vice versa. In each case, the change in momentum is constant and equal to the force multiplied by the time interval. |
Procedure
1. | The impulse-change in momentum theory states that if the change in momentum is the same, then the force must decrease as time increases. As the following chart indicates, the shock-absorbing bumpers, crumple zones, padded interior, and seat belts will increase the time. Therefore, less force is exerted on the occupants of the vehicle. |
Automobile Technology of the Past |
Automobile Technology of the Present |
How the Change in Technology Helped Reduce Injury |
metal front and rear bumpers welded solidly to the main frame of the automobile | large plastic bumpers attached through shock absorbers to a collapsible frame |
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solid metal frame | collapsible frame with crumple zones |
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metal dashboard, hard plastic steering wheel solidly attached to steering mechanisms of the automobile | padded dashboard and interior with a collapsible, shock-absorbing steering wheel |
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no seat belts or air bags | seat belts and driver- and passenger-side air bags; often side-impact air bags |
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Analysis
2. | A brief history of seat-belt use and the reduction of injuries is as follows:
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Procedure
Newton’s Laws and the Interaction of Objects
Constant Values
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Analysis
1. | The force on Vehicle 1 and the force on Vehicle 2 have the same magnitude, but opposite direction. |
2. | The results from the table confirm the presence of an action force and a reaction force consistent with Newton’s third law. |
3. | The negative sign in the force on Vehicle 1 indicates that the force is directed to the left. |
4. | a. | ![]() |
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b. | ![]() |
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When m2 = 1.00 kg,
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When m2 = 2.00 kg,
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When m2 = 4.00 kg,
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When m2 = 5.00 kg,
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When m2 = 6.00 kg,
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When m2 = 8.00 kg,
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When m2 = 10.00 kg,
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c. | The negative acceleration values for Vehicle 1 indicate that Vehicle 1 is accelerating to the left. The positive acceleration values for Vehicle 2 indicate that Vehicle 2 is accelerating to the right. |
5. | According to Newton’s second law, the acceleration of a vehicle is equal to the net force on the vehicle divided by the vehicle’s mass. The magnitude of the acceleration values are sometimes different even though the net force is the same because the masses have different values. The acceleration values have different signs (positive or negative) because the forces have different signs. | ||||||||||||||||||||||||||||||||||||||||
6. | As the acceleration of the vehicles decreases, the velocity decreases proportionally. | ||||||||||||||||||||||||||||||||||||||||
7. | The vehicle with the smaller mass experienced the greater change in velocity. | ||||||||||||||||||||||||||||||||||||||||
8. |
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9. | Yes, the results verify that the smaller mass undergoes the greater change in velocity. This can be understood in terms of Newton’s second law because a vehicle with a smaller mass will experience a larger acceleration. It can also be understood in terms of momentum because if the momentum of the vehicle remains the same, a vehicle with a smaller mass would need a larger velocity to have the same momentum. |
10. | a. | Yes, the force is the same for both vehicles. |
b. | No, the acceleration is not the same for both vehicles. | |
c. | Yes, the time of interaction is the same for both vehicles. | |
d. | No, the final velocity is not the same for both vehicles. | |
e. | Yes, the impulse is the same for both vehicles. | |
f. | Yes, the momentum is the same for both vehicles. |
11. | Acceleration and final velocity are not the same. The larger the mass, the smaller the acceleration and the smaller the final velocity of each vehicle. |
Part A: Hit-and-Stick Collisions
Procedure
Part A: Hit-and-Stick Collisions
Constant Values
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Analysis
1. | ![]() |
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2. | The total momentum before the collision equals the total momentum after the collision in each trial. |
3. | As two groups of freight cars travel toward each other, they are separate objects. Once they couple to form the train, they then travel as one object. This is equivalent to a hit-and-stick collision. |
Part B: Hit-and-Rebound Collisions
Procedure
Part B: Hit-and-Rebound Collisions
Constant Values
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Analysis
4. | ![]() |
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5. | The total momentum before the collision equals the total momentum after the collision in each trial. |
6. | When two billiard balls roll toward each other, they are separate objects. After they collide, they rebound as separate objects, each with its own individual momentum. This is the essence of a hit-and-rebound collision. |
Part C: Explosions
Procedure
Part C: Explosions
Constant Values
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Analysis
7. | For all pairs of mass values,
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8. | The total momentum before equals the total momentum after in each trial. Since neither object was moving before the explosion, the total momentum before was zero; and since the total momentum after equals this value, the total momentum after was also zero. |
9. | Prior to the trigger being squeezed, the bullet and the rifle are motionless, making the total momentum of the system zero. After the explosion within the barrel, the bullet leaves with an amount of momentum in one direction, while the rifle recoils with an equal amount of momentum in the opposite direction. Since the momentums after the interaction are equal in magnitude but opposite in direction, the total momentum after the interaction is zero. |
Evaluation
10. | In all cases, the total momentum before an interaction was equal to the total momentum after an interaction. |
11. | If the force of friction were present, the vehicles would slow down every time they moved. In this scenario, the momentum values before and after the interactions would be continually decreasing, making it very difficult to compare the total momentum before and after. |
Procedure
Data will vary. Sample data is given. Complete the table with your calculations
Trial | Egg Without Helmet | Egg with Helmet |
Mass of Egg, Helmet, and Clamp (kg) |
0.061 40 | 0.083 64 |
Height Above Reference Level (m) | 0.040 | 1.50 |
Gravitational Potential Energy at Release Point (J) |
0.0241 | 1.23 |
Kinetic Energy Just Before Impact (J) |
0.0241 | 1.23 |
Speed Just Before Impact (m/s) |
0.886 | 5.42 |
Momentum Just Before Impact (kg•m/s) |
+ 0.0544 | + 0.454 |
Approximate Change in Momentum Due to Impact (kg•m/s) |
– 0.0544 | – 0.454 |
Approximate Impulse Delivered During Impact (kg•m/s) |
– 0.0544 | – 0.454 |
Estimated Time of Impact (s) | 0.010 | 0.040 |
Approximate Acceleration During Impact (m/s2) |
– 89 | – 1.4 × 102 |
Approximate Force Exerted During Impact (N) | – 5.4 | – 11 |
Analysis
1. | a. | The unprotected egg made contact with the cement block in the middle of its face. This explains the pattern of cracking that starts in the middle of the egg’s face. |
b. | The cracking on a helmeted egg will vary depending upon the design of the helmet. For example, cracking across the eyes is likely due to the fact that the helmet incorporated a protruding section for both the forehead and the jaw of the egg test dummy. When the impact occurred, the forces transmitted to the forehead and jaw of the egg test dummy caused cracking along the eyes. | |
2. | Refer to the sample data for the completed calculations. | |
3. | Answers will vary. The helmeted egg in this case was able to hit the cement block approximately six times faster before damage occured. | |
4. | Answers will vary. In the sample provided, the helmeted egg was able to hit the cement block and withstand approximately twice the force as the egg without a helmet before damage occured. | |
5. | a. | Even though the helmeted egg struck the cement block with about six times the velocity as the egg without a helmet, the force was only twice as great because the time interval was four times larger. |
b. | The foam lining of the helmet caused the time interval of the impulse to be large enough to reduce the force. | |
6. | a. | The kinetic energy was used to do work in each collision. The helmeted egg was able to withstand about five times the kinetic energy as the egg without a helmet. This is because it took about five times as much work to damage the helmeted egg. |
b. | The helmeted egg was able to withstand about five times the kinetic energy as the egg without a helmet. This is because it took five times as much work to change the shape of the plastic shell and foam liner of the helmet before damage could be done to the egg. |
Evaluation
7. | Results will vary depending on the design. One major difference will be between groups that used some sort of chin protection (as in a motorcycle helmet) as opposed to those groups that did not use chin protection, exposing the chin area to the impact of the collision. |
8. | Results will vary depending on the design. Groups that incorporated a design that was more effective at spreading the force over a larger surface area and incorporated a lining that effectively absorbed the impact would have superior results. |
9. | Responses will vary. The original design and the testing will reveal unique defects. Therefore, what you would do differently would be based on your experience with the egg test dummy. These may include
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Analysis
1. | Typical results for overall trends in the data are as follows:
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Evaluation
2. | The results for this question will vary depending on the information used by others. |
Science 20 © 2006, Alberta Education