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--- courses:phy101l:1 [2023/06/03 06:35] – [4.1 Time and velocity] asad
+++ courses:phy101l:1 [2023/09/30 20:02] (current) – asad
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 ====== 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 60: / Line 62: @@
 Using the five measurements you will have to calculate the following quantities.
-==== - Time and velocity ====
+==== - Velocity ====
 First, calculate the time for the ball to leave the ramp and reach point A which is $t = \sqrt{2y/g}$.
 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}$.
-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 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 $.
+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 $.
-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 $v' = \sqrt{v_x^2+v_y'^2}$.
+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 $v' = \sqrt{v_x'^2+v_y'^2}$.
 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')$.
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 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 ====
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 ==== - 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>