Pre-Lab for velocity and acceleration-2

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Linh Nguyen Janet Kang Physics AL 01/23/24 Pre-Lab for velocity and acceleration Equation for position at constant velocity: y(t)=v0t +y0 a= acceleration t=time Exercise 1: a) The question states that the ball “perfectly reflects after the collision maintaining the same speed” this tells us that if the initial velocity is 7.5 m/s than the final velocity will equal -7.5 m/s. We can see a change in the balls acceleration through the equation (Vfinal - Vinitial = at). -7.5 m/s - 7.5 m/s= -15 m/s The equation demonstrates a change in the ball’s acceleration which indicates that a force was acting on the ball. b) velocity= 7.5 m/s The question states that the ball is 20m from the wall which means the total distance the ball covers when moving from the wall can be represented by (x+20); x representing the initial distance from the wall. The question also tells us that the time it takes for the ball to reach the wall is approximately 4 secs. To find x the ball traveled we can use the equation: V0 = displacement/time Math: 7.5 m/s= (x+20)/4 7.5 x 4= (x+20) 30=(x+20) 30-20=x x=10 m So the ball was initially 10m from the wall
c) Equation for velocity under constant acceleration: V(t) = at + V0 1. For the ball to reach the ground again, it first has to rise to its highest point and then fall back down. Since the ball starts at the ground and comes back to the ground, the total time in the air is twice the time it takes to reach its highest point. By using the equation of motion, v = u + at, where v is the final velocity (0 m/s at the highest point), u is the initial velocity (5 m/s), and a is the acceleration (-9.81 m/s^2), we find the time it takes to reach its highest point is 5/9.81 = 0.51 seconds. Therefore, the total time to reach the ground again would be 2*0.51 = 1.02 seconds. 2. As calculated in the previous part, the time taken to reach the highest point of the trajectory is 0.51 seconds. 3. The maximum height can be calculated using the equation h = ut + 0.5*a*t^2, where h is the height, u is the initial velocity, a is the acceleration (which is negative since it's acting downwards), and t is the time it took to reach the highest point. Plugging the values, we get h = 5*0.51 + 0.5*(-9.81)*0.51^2 = approximately 1.28 meters. Exercise 3 a) ( y0 - 0) = 0 * tf + 1/2 * 9.8 * tf 2 y0= (½) × 9.8×tf^2 b)
1. y0 = 1m = 1 = 1/2 * 9.8 * t2 So tf = 0.452 second 2. y0 = 5m = 5 = 1/2 * 9.8 * t2 So tf = 1.01 second 3. y0 = 20m = 20 = 1/2 * 9.8 * t2 So tf = 2.02 second 4. y0 = 50m = 50 = 1/2 * 9.8 * t2 So tf = 3.194 second 5. y0 = 0m = 0 = 1/2 * 9.8 * 6. So tf = 0 second c) Exercise 4 In the previous Part we discussed the ideal behavior of a ball under constant acceleration. Brainstorm three possible reasons why the time it takes for a ball to reach the ground after being dropped in our real-life experiment may not match the ideal theoretical predictions. For each possible reason, indicate whether the cause would result in a faster-than-ideal drop time or a slower-than-ideal drop time and why. a) Windy conditions can either accelerate or decelerate the ideal drop time, depending on the wind's direction and strength. This variation in wind conditions can either prolong or shorten the time of descent since wind acts as an additional force impacting the object's velocity. b) Regarding the initial speed of the ball, a slight downward throw can also extend the time of descent. For an object to be considered in free fall, it must descend solely under the
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influence of gravity without additional forces. Even a minor increase in the initial speed of the ball can have an impact on its descent time. c) Considering gravitational pull, as the ball falls from a greater height, the gravitational force strengthens, intensifying the pull on the ball. Consequently, this change in gravity reduces the time it takes for the ball to drop. Exercise 5: a) The distance from the drop zone to the rail on the second floor is: 5.68 m b) The distance from the drop zone to the rail on the third floor is: 9.26 m c) The distance from the drop zone to the rail on the fourth floor is: 12.93 m Exercise 6: Exercise 7: a) The object that most closely follows the theoretically predicted droptime would be the tennis ball b) The beach ball is the object which has the greatest percent difference with the theoretical drop time because it consistently is greater than ten percent. c) The beach ball in part b will have a percent difference greater for higher drop times. Exercise 8: a) The plot does look linear because the data points resemble a straight a line b) The slope of the line of best fit us 0.151 s/m c) The physical meaning of the line of best fit tells us that there is an external force affecting the recorded drop time because the data should give us an exponential relationship between the drop height and average drop time. Exercise 9 a) The average velocities indicate that there is an upward trend when the drop height increases. It indicates a positive relationship because as the height increases so does the velocity. b) In order to predict the average velocity for the beach ball to be dropped at a height of 500m we would have to implement the equation from our data points. The equation: 0.151x+ 0.379 By putting in 500m into the equation we can find that it would take the ball 75.879 seconds to reach the ground. This gives us an average velocity of 500 m/75.879 s= 6.589 m/s Exercise 10: a) We can see that the data indicates the beach ball was most affected by air resistance because it demonstrates the highest percent difference amongst the other balls/objects.
b) It does make the most sense that the beach ball has the greatest affect because it has a low volume/surface area. Exercise 11: a) I think that the multiple factors that lead to the discrepancies within the data is very prevalent such as the different reaction times (human error) can effect the average and also how the air resistance effected the time of the object significantly. This can be shown through our plot of datapoints because originally the data should be exponential but our experiment indicates a linear relationship. b) I think the systematic error wouldn’t have a significant bias because although it is true that the reaction times for the timers could be delayed or early we also must factor in the delay and earliness of the people who drop the objects. The human error on both sides of the experiment somehow offset each other which minimizes the bias. c) The environmental factors within the experiment would be the air resistance. The wind depending on which direction it blows could either prolong or shorten the time it take for an object to hit the ground which obviously has a significant impact on the data. Exercise 12: I think that the experiment does not hold the concept of the law of linear motion because our data points which indicate a linar relationship neither have a constant velocity and has differing net force acting on the object. In newton’s law the data should be linear only if it has a contant velocity but the objects in question perform a positive relationship between height and velocity as the velocity is shown to increase. As for the data within our plot it performs a linear graph which directly opposes newton’s law stating that objects with a inconsistent velocity and objects that are subjected to a net force should not be linear. Data link: https://docs.google.com/spreadsheets/d/1TdEkcBmAd7swTSJtD-xNRuKObYIN6tcZhFwUmbyg 73A/edit?usp=sharing
Ball Drop Height (m) Theoretical tf (s) Object 1 Avg. Measured tf (s) Object 1 Percent Different (%) Measured Avg. Velocity (m/s) 5.68 1.08 1.22 -12.7 4.65 9.26 1.37 1.80 -26.9 5.14 12.93 1.624 2.316 -35.16 5.582 Table 1.1: Processed Data (Beach Ball) Drop Height (m) Theoretical tf (s) Object 2 Avg. Measured tf (s) Object 2 Percent Different (%) Measured Avg. Velocity (m/s) 5.68 1.08 0.803 29.0 7.07 9.26 1.37 1.14 18.3 8.10 12.93 1.624 1.505 7.564 8.590 Table 1.2: Processed Data (Basket Ball) Drop Height (m) Theoretical tf (s) Object 3 Avg. Measured tf (s) Object 3 Percent Different (%) Measured Avg. Velocity (m/s) 5.68 1.08 0.982 9.13 5.78 9.26 1.37 1.36 1.06 6.81 12.93 1.624 1.671 -2.880 7.738 Table 1.3: Processed Data (Tennis Ball) Drop Height (m) Object 1 tf (s) Object 2 tf (s) Object 3 tf (s) 5.68 1.22 0.803 0.982 9.26 1.80 1.14 1.36 12.93 2.316 1.505 1.671 Table2: Raw data
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