LAB QUEST
STATION 1 - The Mass Question
Do more massive objects fall faster than lighter ones?
This is a question that was the subject of great debate for a very long time. Around the year 1600, Galileo was the first to truly understand that freely falling objects fall with a constant acceleration of -9.8 m/s2. This applies to all objects, regardless of mass.
(well, this is not exactly true, but we will learn why it varies later in the year)
With the advent of high speed filming, we can determine this with relative ease. Watch the videos and then answer the following questions:
To find the video, use paer.rutgers.edu/pt3 then click Motion then Pisa, simultaneous.
- Do more massive (‘heavier’) objects fall faster or slower than less massive (‘lighter’) objects? Explain how you know this.
The ‘freefall’ acceleration rate is the same for any object, independent of mass! We know this from watching the video frame by frame and by testing it out in station 3. We remember too that ‘freefall’ means only influenced by gravity. No air resistance, lift, etc.
- What is constant regardless of mass for a falling object (neglecting air resistance)?
an acceleration rate of ~9.81 m/s2 in the downward direction.
- Why is m/s2 the unit of acceleration? What does the seconds squared mean? Why is it different from the unit of velocity, which is m/s? Explain.
m/s2 is another way of writing m/s/s. Since avg. acceleration is change in velocity (measured in m/s) divided by a time interval (measured in s), then the unit of acceleration is (m/s) / s which reduces to m/s2. Avg. velocity is change in position divided by a time interval, measured in m/s. Velocity and acceleration are totally different things, and are not synonymous.
- If the acceleration due to gravity is constant, regardless of mass, why do you think a feather takes longer to fall than a rock, when dropped from the same height?
Acceleration is constant only for freefall. I’ll guess that the impact of air resistance causes the feather to fall at a reduced acceleration than the rock.
- How can we test your idea from question 4? What kind of experiment would you set up?
We could drop a rock and a feather in an airless environment (a vacuum or on the moon even) and we would expect their acceleration rates to be the same!
BONUS QUESTION
- How could Galileo convince those people who thought heavier objects fell faster that he was right and they were wrong? In his time, people thought an object twice as heavy would fall twice as fast. Can you think of an experiment that would refute the claims of the people of Galileo’s time?
(Remember, let’s pretend there was no video back in his day. A ridiculous and frightening thought, I know.)
Dropping two different masses connected by a chain would be one experiment to try. If one object was 10x the mass of the other, it would be easy to see that the more massive object does not come close to having a 10x greater acceleration.
STATION 2 - Gravity Check
What’s all this acceleration is constant business? -9.8 m/s2 ? If you still don’t buy it, I don’t blame you. Because in the real world, falling things don’t seem to accelerate constantly because of air resistance. But under controlled conditions, where air resistance is small, we may catch sight of the underlying truth….We’ll drop a ball from a height above the motion detector and see what results, but first:
1. DRAW PREDICTIONS of the x vs t, v vs t and a vs t graphs for the drop.
2. Then set up the motion detector and open Expt 06 – Ball Toss and do the actual experiment. You may need to repeat it a few times to get clear data.
3. Interpret the graphs made by the motion detector and determine the acceleration due to gravity. Do they match your predictions? Explain!
4. Next, predict what the graphs will look like when you take the ball and toss it gently in the air straight above the motion detector. You do this by drawing out your predicted x, v and a graphs vs time.
5. Then do the actual experiment. You will almost certainly need to do it a few times to get clear data. It takes a bit of practice.
6. Interpret the graphs made by the motion detector and determine the acceleration due to gravity. Do they match your predictions? Explain!
Drop
UpwardToss
Answer the following questions:
- The accepted value for the acceleration due to gravity is 9.8 m/s2 and is pointed towards the center of the earth. What did you get for the acceleration due to gravity (also known as ‘g’ )? The same! The a vs. t graph shows ~10 m/s2!
- Did the value of acceleration due to gravity ( ‘g’ ) change depending upon whether you dropped the ball or tossed it up? When you tossed the ball straight up, it stopped for an instant at its peak. What was its acceleration at the peak, when v=0 !? Check the slope of the v vs. t graph and the a vs. t graph! Still -9.8 m/s2 ! Explain how you can have an acceleration at a given instant even if you have no velocity at a given instant.
To get avg. acceleration you need TWO velocity values with a time interval in between.
since Δv = vf – vi , two velocity values are required. If you pick the peak as vi then vf will be negative, yielding a – Δv, and thus a – acceleration. If you pick the peak as vf, then vi will be positive (as the ball is on the way up to the peak) and Δv will be negative and thus acceleration will be too, since the time interval you divide by is always positive.
In BOTH cases, acceleration is negative. Remember, a = Δv /t not v/t !
- Why do we say g = negative 9.8 m/s2? What does the negative sign indicate?
it indicates that the acceleration is always in the downward direction.
10. Do you think the presence of air causes the ball to speed up a little or slow down a little? If we reduced the effect of air by dressing the ball up in one of those Olympic skin tight outfits, do you think it would accelerate downward a little faster or a little slower? What are your reasons for thinking this?
If it is the pull of the earth downward that causes the downward acceleration, then I’ll suppose that the push of the air upwards on the ball as it falls would reduce the acceleration rate. (slow down?)
If there is air resistance, then by removing it, we should achieve an acceleration closer to the freefall acceleration, and thus the downward acceleration should be a bit greater in value.
STATION 3 - Quick Draw
How fast are YOU, hombre?
I am sure you have heard the words ‘reaction time’ before. But now, knowing the acceleration due to gravity here near the earth, we can determine YOUR reaction time!
You need a partner for this experiment.
1. Hold a meter stick steadily between your partners’ unpinched fingers. Note the centimeter mark right at the finger level.
2. Do not tell your partner when you are going to do it, but at some point drop the meter stick. Make sure your partner knows they are supposed to catch it by pinching their fingers as quickly as they can after you drop it. Note the centimeter mark where they caught the stick.
What did you just find? You found ∆x!
But what direction is the ∆x in? Is it upward or is it downward ?
*How does the direction of the displacement affect the sign of ∆x ? This is critical!!!
You know a and you know vi so with these three variables known, find the time the stick was in the air. This is your partner’s reaction time! Pretty awesome huh?
Each person does this three times.
3. Calculate your average reaction time three times.
Then get your average reaction time by adding your three reaction times up and dividing by three.
Be sure your times fall between 0.1 and 0.5 seconds. It is likely that any other time values are indicative of a problem with units, signs or calculations. Do not neglect signs. If you are attempting to take the square root of a negative value, this also indicates that a closer look is warranted. Hint: Consider the sign of the displacement.
STATION 4 - Droppin’ a Dime
Take a dime and stand on top of a chair. Or the table but be careful! Raise the dime up as high as you can. Have partners measure the distance from the floor to the dime. DO NOT DROP THE DIME YET.
Come back down from the chair and as a group, PREDICT, USING THE EQUATIONS OF CONSTANT ACCELERATION, how long the dime will take to reach the ground once dropped!
***Remember, the ∆x the dime has to cover is negative displacement because it is in the downward direction!!!
When you have a prediction in hand, do the drop, use the stopwatches and see how close you were.
Repeat the experiment above with a shoe or a ball or some other object big enough to not be greatly affected by air resistance.
Answer the following (THINK ABOUT THESE BEFORE YOU ANSWER):
- Did using -9.8 m/s2 work for the heavier object too? What can you conclude about the dependence of g the acceleration due to gravity, on the mass of an object in freefall? Does it matter how massive the object is?
Amazingly, the accelerations were the same regardless of mass. We observed it!
- Let’s say that because of your reaction time, your experimental time is a bit longer than your predicted time. If you were using this time to calculate g would your g values be greater than 9.8 m/s2 or less than 9.8 m/s2 ? Explain your reasoning.
- Imagine dropping a dime down a bottomless well. One second later, you drop a penny. Describe what happens to the distance between the dime and the penny as they fall. Be sure to explain your answer in detail….provide supporting statements for your thoughts!
The dime would remain ahead of the penny, but the distance gap would continue to increase!
This may be hard to see at first. But realizing that the dime will always be faster than the penny might help. After 1 second, the dime is moving at 10 m/s while the penny has v=0 since it just was dropped. At 2 seconds the dime is at 20 m/s but the penny is at 10 m/s. The dime is still faster, thus the dime is moving farther away from the penny.
Interestingly, the time gap between the two remains at 1 second!
STATION 5 - Muzzle Velocity (a.k.a. the starving dude)
In a few weeks we will use this launcher for another lab. However, when we do, we will NEED to know how fast it can launch the metal ball.
How can we figure this out?
1. Describe IN DETAIL every step we could use in an experiment to figure this out. Explain the math expressions you would use, and what you would be solving for.
When you are done with your description, do the experiment and come up with a value for the speed at which the ball leaves the launcher. This speed is also referred to as the muzzle velocity of the ball.
By shooting the ball upwards and measuring the peak height, we’ll have Δ x. Acceleration from the moment it leaves the muzzle to its peak is -9.8 m/s2. And vf is zero at the peak! With these three values we can calculate the vi that the ball had upon emergence from the muzzle! Which turned out to be approximately 4.9 m/s.
2. You are part of a rescue effort where there is a person stranded on a high cliff. You and your team will launch a sandwich to them, so they don’t starve. DESCRIBE IN DETAIL why there is no mayo on the sangwidge.
It’s a PB&J sandwich. Mayo on that would be gross.
3. The launcher uses explosives. For every 5 grams of powder explosive you use, you will impart 1 m/s of initial velocity to the sammich. If the starving dude is on a 110m high cliff, how much explosive should you use? Assume we are shooting straight up, and we want to just barely reach the dude.
Based on a Δ x of 110 meters at peak height, we calculated the required vi to be 46.9 m/s.
And if 5g of powder will provide 1 m/s, the total amount of powder was 234.5 grams
4. How long does it take the sandwhich to get to the dude?
approximately 4.7 seconds
5. What velocity (speed and direction) does the samitch have after 1.5 seconds?
when vi = 46.9 m/s and t =1.5s under the freefall acceleration, vf= 31.9 m/s upward.
6. How high up the cliff is it then??
Δ x = 59.1 m