Differences

This shows you the differences between two versions of the page.

--- course_planning:183_projects:f17_project_15_solution [2017/12/07 00:04] – pawlakal
+++ course_planning:183_projects:f17_project_15_solution [2017/12/07 00:48] (current) – pawlakal
@@ Line 10: / Line 10: @@
  $w=7.42 rad/s$
-We can use v=rw to convert this to the linear speed that the steel ball leaves the catapult at:
+We can use $v=rw$ to convert this to the linear speed that the steel ball leaves the catapult with:
-$$v_{i, steel}=(7.42 rad/s)*(4 m)$$
+$$v_{i, steel}=(4 m)*(7.42 rad/s)$$
- $v_{i, steel}=29.68$
+$$v_{i, steel}=29.68 m/s$$
-Now we use energy conservation and linear momentum conservation to consider the collision between the steel ball and the shark. We take our system to be the shark and steel ball, and we assume that the shark starts at rest. For momentum conservation, we get:
+Now we use energy conservation and linear momentum conservation to consider the collision between the steel ball and the shark. We take our system to be the shark and steel ball, and we assume that the shark starts at rest. We also assume that the collision is elastic. For momentum conservation, we get:
   $p_{i, steel}+p_{i, shark}=p_{f, steel}+p_{f, shark}$
@@ Line 22: / Line 22: @@
 For energy conservation, we get:
-$$/frac{1}{2}m_{steel}v_{i, steel}^2=/frac{1}{2}m_{steel}v_{f, steel}^2+/frac{1}{2}m_{shark}v_{shark}^2$$
+$$KE_{i, steel}+KE_{i, shark}=KE_{f, steel}+KE_{f, shark}$$
+$$\frac{1}{2}m_{steel}v_{i, steel}^2=\frac{1}{2}m_{steel}v_{f, steel}^2+\frac{1}{2}m_{shark}v_{shark}^2$$
+Students will need to look up a mass for a shark. Using 900 kg, solving this system of equations for the final velocities of the steel ball and shark, we get:
+ $v_{f, steel}=-29.03 m/s$
+ $v_{shark}=0.65 m/s$
+So, the shark doesn't recoil much, but hopefully enough to scare it away!
+<WRAP tip>
+==Tutor Questions:==
+  * **Question:**  What did you choose for your system?
+  * **Expected Answer:**  For the clay ball-catapult collision, we included the clay ball, the steel ball, and the catapult. For the steel ball-shark collision, we included the steel ball and the shark.
+  * **Question:**  What was conserved in each collision?
+  * **Expected Answer:**  In the first collision, only angular momentum was conserved. Angular momentum is conserved because there were no external torques. The pivot point exerts an external force on the system, so linear momentum is not conserved. It is an inelastic collision, so energy is not conserved. In the second collision, linear momentum is conserved because there are no external forces, and energy is conserved because we assume the collision is elastic.
+  * **Question:**  What assumptions did you make?
+  * **Expected Answer:**  The water exerts no force resisting the shark's recoil. That gravity doesn't accelerate the steel ball its way from the catapult to the shark. That the steel ball experiences no air resistance. That the clay ball hits the catapult perpendicular to its arm at the very end of the arm. That the shark-steel ball collision is elastic.
+</WRAP>