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--- courses:phy101l:1 [2023/06/03 06:19] – [4.1 Time and velocity] asad
+++ courses:phy101l:1 [2023/09/30 20:02] (current) – asad
@@ Line 1: / Line 1: @@
 ====== 1. Trajectory of a projectile ======
+[[https://colab.research.google.com/drive/1Edmdk8mft0K5_MSEcSW7qc31sz7jlubL?usp=sharing|Report sample in Google Colab]].
 The report should have the following five sections. Each section must be written by a separate student. The whole report must be in a single Google Colab file and attached from Google Drive into the associated Google Classroom assignment.
@@ Line 39: / Line 41: @@
 You will measure different parameters of this trajectory, for example, the height of the ramp, the positions of the bounces and the maximum height of the second part of the trajectory. Using these parameters, you will calculate different velocity components and the energy, momentum and impulse related to this projectile. The aim is to get a complete description of a projectile including its impact with a horizontal surface.
 ===== - Experimental Setup =====
 For this experiment you have to use the following apparatus: a table, a bench, a ramp, some marble balls, 3 white recording papers, 3 carbon papers, some adhesive tape for fixing the papers on the table or bench, a long meter scale to measure long heights and lengths, a short meter scale for measuring short distances on the recording papers, and a weight scale to measure the mass of a ball.
-{{:courses:phy101l:m1.1.png?nolink|}}
+{{:courses:phy101l:phy101l-1.jpg?nolink|}}
-The setup is briefly sketched through the following points (elaborate in the report).
+Describe how you set up the apparatus on the table.
-  - A bench is placed on a table. There must be enough space in front of the bench for the balls to bounce.
-  - A ramp is placed on the bench.
 ===== - Method and Data =====
 ^ Quantity ^ Symbol ^ Value ^ Error ($\delta$) |
@@ Line 54: / Line 55: @@
 | Position of the first bounce at A | $s$ |  cm | $\delta s=$ cm |
 | Position of the second bounce at B | $s'$ |  cm | $\delta s'=$ cm |
-| Maximum height between A and B | $h$ |  cm | $\delta h' = $ cm |
+| Maximum height between A and B | $h$ |  cm | $\delta h = $ cm |
-===== - Analysis and Results =====
-Using the six measurements you will have to calculate the following quantities.
-==== - Time and velocity ====
+Describe how collected these data.
-Time for the ball to leave the ramp and reach point A
-$$ t = \sqrt{\frac{2y}{g}} $$
+===== - Analysis and Results =====
+Using the five measurements you will have to calculate the following quantities.
-where gravitational acceleration $g=981\pm 1$ cm s$^{-2}$.
-Horizontal component of the acceleration of the ball $a_x=?$
-The unchanging horizontal velocity of the ball $ v_x = s/t$.
-Vertical velocity of the ball just before it hits the table at point A: $ v_y = -gt = -\sqrt{2gh}$.
-The time the ball takes to reach point B from point A would be $ t' = (s'-s)/v_x$.
-Vertical component of the velocity of the ball after it leaves point A is $ v_y' = g t' / 2 $.
-Net velocity of the ball before it impacts at A is $v = \sqrt{v_x^2+v_y^2}$.
-Net velocity of the ball after it impacts at A is $v' = \sqrt{v_x^2+v_y'^2}$.
-==== - Angle and height ====
-Maximum height the ball reaches between points A and B
-$$ h' = \frac{g}{2}\left(\frac{t_{AB}}{2}\right)^2. $$
+==== - Velocity ====
+First, calculate the time for the ball to leave the ramp and reach point A which is $t = \sqrt{2y/g}$.
-The angle of the trajectory before reaching A
+Horizontal component of the acceleration of the ball $a_x=0$, so the horizontal velocity of the ball $ v_x = s/t$. The vertical velocity of the ball just before it hits the table at point A would be $ v_y = -gt = -\sqrt{2gh}$.
-$$ \theta = \arctan{\frac{v_y}{v_x}} $$
+From the maximum height $h$, you can calculate the time it took for the marble to go from points A to B ($t'$). We know $h=gt'^2/4$ and therefore $t' = \sqrt{4h/g}$. So the horizontal velocity of the ball after leaving A would be $ v_x' = (s'-s)/t'$. And the vertical component of the velocity of the ball after it leaves point A is $ v_y' = g t' / 2 $.
-and after leaving A
+So the net velocity of the ball before it impacts at A is $v = \sqrt{v_x^2+v_y^2}$. And the net velocity of the ball after it impacts at A is $v' = \sqrt{v_x'^2+v_y'^2}$.
-$$ \theta' = \arctan{\frac{v_y'}{v_x}}. $$
+The angle of the trajectory before reaching A is $ \theta = \tan^{-1}(v_y/v_x) $ and after leaving A $ \theta' = \tan^{-1}(v_y'/v_x')$.
 ==== - Kinetic energy ====
@@ Line 103: / Line 84: @@
 Vertical component of the momentum before reaching A $P_y=mv_y$ and after leaving A $P'_y = mv'_y$.
-Horizontal component of the momentum before reaching A $P_x=mv_x$.
+Vertical component of the momentum before reaching A $P_x'=mv_x'$ and after leaving A $P'_x = mv'_x$.
-Net momentum before reaching A $p = mv$.
+So the net momentum before reaching A $p = mv$ and the net momentum after leaving A is $p' = mv'$.
-Net momentum after leaving A $p' = mv'$.
 ==== - Impulse ====
@@ Line 117: / Line 96: @@
 ==== - Discussion ====
+Answer the following questions:
+  - What is the difference between systematic and stochastic error? Which errors in this experiment are systematic and which are stochastic?
+  - Why is $\delta s' > \delta h > \delta s$?
+  - Which velocity is greater, before reaching point A or after leaving A?
+  - Is the impulse positive or negative? Why?
 ==== - Conclusion ====
-===== Report sample =====
-<html>
-<script src="https://gist.github.com/kmbasad/907b3932678f9117b41f079025e26cb3.js"></script>
-</html>