added ap-physics1-exam

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diff --git a/ap-physics1-exam/q1.txt b/ap-physics1-exam/q1.txt
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+WY5V8181
+HR
+
+(a)
+
+Graph B is correct because the magnitude of acceleration is constant in
+simple circular motion, so the magnitude of net force is proportional
+and thus constant (F = ma).
+
+(b)
+
+The magnitude of tension on the string is greater than the centripetal
+force on the sphere. Tension has two components, the opposite force to
+gravity because the sphere is in equilibrium vertically, and the
+centripetal force. The resultant of two nonzero forces is larger than
+either component, so tension > centripetal force.
+
+(c)
+
+The inward acceleration needn't be as large for a very small tangential
+speed to keep the sphere in constant circular motion. This means that
+the centripetal force can be smaller. If the centripetal force can be
+smaller but the vertical force of tension counteracting gravity is the
+same, then the angle of the string and the pole is smaller.
+
+(d)
+
+This equation is consistent with part (c) because, for small values of
+theta (string nearly vertical), tan(theta)*sin(theta) ~ 0, so v^2 ~ 0,
+so v ~ 0 (small).
+
+(e)
+
+v^2 = gL*tan(theta)*sin(theta)
+=> g = v^2 / (tan(theta) * sin(theta)).
+
+If v^2 were graphed on the y-axis of a graph and
+(tan(theta) * sin(theta)) on the x-axis, the slope of the line of best
+fit would give an experimental value of g.
+
+(f)
+
+As theta decreases, the distance of the sphere to the rod decreases.
+The masses of all objects remains const, I = mr^2 for a point-like
+object, and the distances from the pivot (and thus inertia) remain the
+same for the rod and the platform. Therefore, as r -> 0, I -> 0.
diff --git a/ap-physics1-exam/q2.tex b/ap-physics1-exam/q2.tex
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+{\obeylines%
+WY5V8181
+HR
+}
+\noindent(a)
+
+Speed is absolute velocity (i.e. regardless of direction). Both A and B
+have increasing absolute distance from the x-axis at every time shown on
+the graph, so neither has decreasing speed.
+
+\noindent(b)
+
+At $t=0$, $t=2$, A and B are moving at the same speed because their
+lines on the graph intersect.
+
+\noindent(c)
+
+Object A.
+
+\noindent(d)
+
+$$W = \Delta E.$$
+The initial velocities of both objects are zero, so their initial
+energies are zero, and $$W = E_f = {1\over2}mv^2.$$ Because
+$m_A = m_B,$ and $$|v_A| > |v_B| \to v_A^2 > v_B^2\hbox{ at $t=4$},$$
+object A has more energy and has thus had more net work done to it.
+
+\bye