Mock Exams AP Physics – 1 – FRQ Set 1

Ans:

(a) Using conservation of energy, we set the potential energy at the top of the ramp equal to the kinetic energy at the bottom to get \(mgh=\frac{1}{2}mv^{2}\). Solving for v gives \(v=\sqrt{2gh}=\sqrt{(2)(10)(20)}=\sqrt{400}=20m/s\).

(b) Throughout the entire loop, there are only two forces acting on the cart: \(F_{g}\) and \(F_{N}\). The loop can be considered as a case of uniform circular motion, so the net force will always be directed toward the center and will be equal to \(\frac{mv^{2}}{r}\). At the bottom of the loop, that means \(\frac{mv^{2}}{r}=F_{N}-F_{g}\), so \(F_{N}=\frac{mv^{2}}{r-F_{g}}\). At the top of the loop, \(\frac{mv^{2}}{r}=F_{N}+F_{g}\), so \(F_{N}=\frac{mv^{2}}{r-F_{g}}\). Thus, the normal force must always be greater at the bottom of the loop.

(c) Mechanical energy will be constant throughout since the track is frictionless, so set the energy at the beginning equal to the energy at the top of the loop. This gives \(mgh=mg(2r)+\frac{1}{2}mv^{2}\), where r is the radius of the circle and v is the speed at the top of the loop.

Next, recall that in order for the car to complete the loop, it must be going fast enough to create some normal force at the top. In general, this equation would look like \(\frac{mv^{2}}{r}=F_{N}+F_{g}\). Maximizing the loop means setting \(F_{N}=0\), so \(\frac{mv^{2}}{r}=mg\). Moving the r to the other side gives \(mv^{2}=mgr\). Substituting this value into the first equation gives \(mgh=mg(2r)+\frac{1}{2}(mgr)\). You can cancel out m and g since they appear in every term, so \(h=2r+\frac{1}{2}r=2.5r\). The problem shows h = 20, so solving for r gives r = h/(2.5) = (20)/(2.5) = 8 m.

(d) Conservation of energy can be used to find the speed of the car as it enters the final segment. Set the potential energy of the initial position equal to the potential and kinetic energies as the cart enters the final stretch. This gives \(mgh_{1}=mgh_{2}+\frac{1}{2}mv^{2}\). First, realize m can be dropped since it appears in every term. Solving for v then gives \(v=\sqrt{2g}(h_{1}-h_{2})=\sqrt{(2)(10)(20-10)}=\sqrt{200}=10\sqrt{2}m/s\).

In order for the friction to bring the cart to a stop, it must do work to the cart. We know \(W=\Delta KE\), so \(Fd\cos \theta=\frac{1}{2}mv{_{0}}^{2}\). The first KE term can be dropped since the final speed will be 0. We can substitute \(F_{f}=\mu F_{N}=\mu mg\) for F term since friction is the force doing the work. Additionally, θ = 180° because the frictional force will act in a direction opposite the cart’s motion. All of this gives \((\mu mg)d\cos180^{\circ}=\frac{1}{2}mv{_{0}}^{2}\). Canceling the m on each side and solving for d then gives \(d=-\frac{1}{2}v{_{0}}^{2}/(\mu g\cos180^{\circ})=-\frac{\frac{1}{2}(10\sqrt{2})^{2}}{(0.2)(10)(-1)}=50m\).