Question
For a particular nonlinear spring, the relationship between the magnitude of the applied force, F, and the stretch of the spring, x, is given by the equation \(F = kx^{1.5}\). How much energy is stored in the spring when is it stretched a distance \(x_1\)
(A) \(\frac{2k_1^{2.5}}{5}\)
(B) \(\frac{kx_1^{1.5}}{5}\)
(C) \(kx_1^{2.5}\)
(D) \(\frac{1}{2}kx_1^2\)
(E) \(1.5 kx_1^{0.5}\)
▶️Answer/Explanation
Ans: A
For a spring that is not linear (i.e., does not obey Hooke’s law) the energy stored is not \(\frac{1}{2} kx^2\). The magnitude of the energy stored will be equal to the magnitude of the work done to stretch the spring to \(x_1\)/ the steps to calculate the work are shown below.
Question
The figure shows a view from above of two objects attached to the end of a rigid massless rod at rest on a frictionless table. When a force F is applied as shown, the resulting rotational acceleration of the rod about its center of mass is kF/(mL).
What is k?
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{5}{8}\)
(D) \(\frac{3}{4}\)
(E) \(\frac{5}{6}\)
▶️Answer/Explanation
Ans: C
The center of mass of the system is at a distance of \(y_{cm}=\frac{(m)(0) + (2m)(L)}{m +2m} = \frac{2}{3} L\) below the mass m. With respect to this point, the clockwise torque produced by the force F has magnitude
\(\tau = rF = (\frac{2}{3}L – \frac{1}{4} L) F = \frac{5}{12} LF\)
Since the rotational inertia of the system about its center of mass is
\(I = \sum m_i r_i^2 = m (\frac{2}{3}L)^2 + (2m) (\frac{1}{3}L)^2 = \frac{2}{3} mL^2\)
the equation \(\tau = Ia\) becomes
\(\frac{5}{12} LF = (\frac{2}{3}mL^2) \alpha \Rightarrow \alpha = \frac{5}{8} \frac{F}{mL}\)
Question
An astronaut lands on a planet whose mass and radius are each twice that of Earth. If the astronaut weighs 800 N on Earth, how much will he weigh on this planet?
(A) 200 N
(B) 400N
(c) 1,600 N
(D) 3,200 N
▶️Answer/Explanation
Ans: B
The value of g near the surface of a planet depends on the planet’s mass and radius:
\(mg = \frac{GMm}{r^2} \Rightarrow g = \frac{GM}{r^2}\)
Therefore, calling this Planet X, we find that
\(g_x = \frac{GM_x}{r^2_x} = \frac{G(2M_{Earth})}{(2r_{Earth})^2} = \frac{1}{2} \frac{GM_{Earth}}{r^2_{Earth}}=\frac{1}{2} g_{Earth}\)
Since g is half as much on the planet as it is on Earth, the astronaut’s weight (mg) will be half as much as well. The astronaut would weigh half of Boo N, which is 400 N, (B).
Alternatively, you can plug in some numbers, although understand that this can introduce new errors if you’re not careful with your calculations. This allows you to make some unreal assumptions with numbers that are easy to work with, in this case, that the Earth has a mass of 10 and a radius of 1. Now it’s a lot easier to solve for Earth, \(g = GM/R^2 = G. 10/1^2 = 10G\). Planet X has twice the mass and radius of Earth, so for Planet X, M = 20 and r = 2. This gives \(g = GM/R^2 = G.20 /2^2 = 5G\). Since g is half as much on the planet as it is on Earth, the astronaut’s weight will be half as much as well.
Question
The radius of a collapsing spinning star (assumed to be a uniform sphere with a constant mass) decreases to \(\frac{1}{16}\) of its initial value.
What is the ratio of the final rotational kinetic energy to the initial rotational kinetic energy?
(A) 4
(B) 16
(C) \(16^2\)
(D) \(16^3\)
(E) \(16^4\)
▶️Answer/Explanation
Ans: C
Since no external torques act on the star as it collapses (just like a skater when she pulls in her arms), angular momentum, Iw, is conserved, and the star’s rotational speed increases. The moment of inertia of a sphere of mass m and
radius r is given by the equation \(I = kmr^2 \) (with \(k = \frac{2}{5}\), but the actual value is irrelevant), so
we have:
\(\omega _f = \frac{I_i}{I_f} \omega _i = \frac{kmr^2_i}{kmr^2_f}\omega_i = \frac{r^2_i}{r^2_f} \omega _i = \frac{r_i^2}{(\frac{1}{16}r_i)^2} \omega _i = 16^2 \omega_i\)
Therefore,
Question
A toy car and a toy truck collide. If the toy truck’s mass is double the toy car’s mass, then, compared to the acceleration of the truck, the acceleration of the car during the collision will be
(A) double the magnitude and in the same direction
(B) double the magnitude and in the opposite direction
( C) half the magnitude and in the same direction
(D) half the magnitude and in the opposite direction
(E) dependent on the type of collision
▶️Answer/Explanation
Ans: B
We know that the forces during a collision are an example of an action-reaction pair. This means the force on the car will be equal in magnitude but opposite in direction compared to the force on the truck. Because the forces are in opposite directions, the accelerations will be as well. Eliminate (A) and (C). Because the forces are equal in magnitude, Newton’s Second Law tells us that the acceleration of the lighter object will be double the acceleration of the heavier object. Eliminate (D). Finally, these things will be true regardless of what type of collision takes place. Therefore, the answer is (B).