Simple and Physical Pendulums AP Physics C Mechanics FRQ – Exam Style Questions etc.
Simple and Physical Pendulums AP Physics C Mechanics FRQ
Unit 7: Oscillations
Weightage : 10-15%
Question
Block A and Block B of masses m and 3m, respectively, are arranged in a setup consisting of an ideal spring
with spring constant k and a horizontal surface. Friction between the surface and the blocks is negligible except in
a region of length D, where the coefficient of kinetic friction between Block A and the surface is m. Block B is
attached to a string of length and negligible mass, as shown in Figure 1. Block A is held against the spring,
compressing the spring a distance xc.
At time=0, Block A is located at position x=xο and is released from rest. After the block is released, the following occurs.
• At time $t = t_{1}$, Block A is at$ x =x_{1}$ after traveling a distance xc. Block A moves with speed v, and the spring is at its equilibrium position.
• At time $t = t_{1}$, the left side of Block A is at$ x = x_{1}$ after passing through a distance D across the region with nonnegligible friction.
• At time $t = t_{1}$, Block A is at x = $t_{3}$, and Block A collides with and sticks to Block B.
(a) For parts (a)(i) and (a)(ii), express your answer in terms of m, k , D, m, xc, and physical constants, as appropriate.
i. Derive an expression for the speed v of Block A at time $t_{1}$.
ii. Derive an expression for the speed vA,B of the two-block system immediately after the collision at time $t_{3}$.
(b) i. On the following axes, sketch a graph of the kinetic energy K of Block A as a function of time t from
time $t = 0$ to time $t_{3}$.
ii. Use principles of work and energy to justify the graph drawn in part (b)(i) for the time interval t = 0 to t = t1.
Explicitly reference features of the shape of the graph you drew in part (b)(i).
After the collision, the two-block system instantaneously comes to rest at time $t_{1}$, which occurs when the string
makes a small angle θmax with the vertical, as shown in Figure 2. For times $t > t_{1}$, the system oscillates with frequency f.
The support holding the string is raised, and the procedure is then repeated using a new string of length 2.
(c) Indicate how the new frequency of oscillation f2 of the system on the new string of length 2 will compare to the frequency of oscillation f from the original procedure.
_____ f > f 2 _____ f < f 2 _____ f = f 2
Briefly justify your answer.
▶️Answer/Explanation
Ans:-
(a)(i) For a multi-step derivation with an application of the conservation of mechanical energy that indicates that all of the energy of the system is initially Us.
$E_{initial}=E_{final}$
$U_{s} = K$
$\frac{1}{2} kx_{c}^{2}=\frac{1}{2}mv^{2}$
$v=\sqrt{\frac{kx_{c}^{2}}{m}}$
$v=x_{c} \sqrt{\frac{k}{m}}$
(a)(ii) For a derivation to solve for the speed at $x_{2}$, that includes one of the following:
• An appropriate application of the conservation of energy
• An appropriate kinematics equation.
For one of the following that is consistent with the previous point in the response for part (a)(ii):
• A correct expression for the energy dissipated by friction
• A correct expression for the acceleration of the block in the region with nonnegligible friction
For attempting to derive an expression for $v_{A,B}$ by using the conservation of momentum
For substituting the expression for the speed at $x_{2}$ that is consistent with the first point of
the response in part (a)(ii) and substituting the correct masses into an expression for conservation of momentum
(b)(i) For a nonlinear sketch that begins at zero and increases for the entire time interval $0 ≤ t ≤t_{1}$
For a sketch that decreases for the entire time interval $t_{1} ≤ t ≤ t_{2}$ but does not go to zero
For a sketch that is concave up for the time interval t_{1}≤ t≤t_{2}
For a continuous function for the time interval $t_{1} ≤ t ≤ t_{3} $that has a horizontal line that is greater than zero for the time interval $t_{2}≤t ≤t_{3}$.
(b)(ii) For a statement about the change in kinetic energy that is consistent with the graph drawn in the response for part (b)(i)
Example Response
From $0< t<t_{1}$, the kinetic energy of Block A increases. The force exerted on the block by
the compressed spring transfers the elastic potential energy in the block-spring system to the
kinetic energy of the block. Because the force exerted by the spring is not applied at
a constant rate, the kinetic energy of the block does not increase at a constant rate.
(c) For selecting with an attempt at a relevant justification ,
For correctly applying an equation that relates the length of a pendulum to the period or frequency of the pendulum
Example Response
The period of a pendulum is calculated by using $T=2\pi \sqrt{\frac{l}{g}}$. Therefore, as the length is
increased, the period will also increase. Because frequency and period are inversely related,
an increase in period will result in a decrease in frequency.
Question
A small dart of mass 0.020 kg is launched at an angle of 30° above the horizontal with an initial speed of 10 m/s. At the moment it reaches the highest point in its path and is moving horizontally, it collides with and sticks to a wooden block of mass 0.10 kg that is suspended at the end of a massless string. The center of mass of the block is 1.2 m below the pivot point of the string. The block and dart then swing up until the string makes an angle θ with the vertical, as shown above. Air resistance is negligible.
(a) Determine the speed of the dart just before it strikes the block.
(b) Calculate the horizontal distance d between the launching point of the dart and a point on the floor directly below the block.
(c) Calculate the speed of the block just after the dart strikes.
(d) Calculate the angle θ through which the dart and block on the string will rise before coming momentarily to rest.
(e) The block then continues to swing as a simple pendulum. Calculate the time between when the dart collides with the block and when the block first returns to its original position.
(f) In a second experiment, a dart with more mass is launched at the same speed and angle. The dart collides with and sticks to the same wooden block.
i. Would the angle θ that the dart and block swing to increase, decrease, or stay the same?
_____ Increase _____ Decrease _____ Stay the same
Justify your answer.
ii. Would the period of oscillation after the collision increase, decrease, or stay the same?
_____ Increase _____ Decrease _____ Stay the same
Justify your answer.
Answer/Explanation
Ans:
(a)
Vtop = ?
Vtop, y = O m/s because top of protectory
Vtop,x only speed: Vo cosθ = 10 m/s (Cos 300) = 8,66 m/s
(b)
d = ? Vtop, y = O, V0, y = V0sin 300
d = Vx (t) \(\frac{V_{0}y}{g}=t\)
d = 8.66 m/s × 0.51s \(t = \frac{V_{0}sin30^{0}}{g}=0.51s\)
(c)
Vblock = ?
Cons. of angular momentum:
Li = Lf
rmv = Iw, \(w = \frac{v}{r}\) M = mdart + Mblock
\(rmv = MR^{2}(\frac{V}{R})\)
mdart (V0) = mdart + Mblock) (Vt)
10 m/s (OtoZ) = (0.oz + 0.1) (Vf)
Vf = 1.67 m/s
(d)
θ = ?
\(\frac{1}{2}mv^{2}=mgh\)
\(\frac{1}{2}(V_{f})^{2}=gh\) \(\frac{1}{2}(1.67 m/s)^{2}=9.81 m/s (h)\)
h = 0.14 m
\(cos\theta = \frac{a}{L}\)
\(cos\theta = \frac{1.2 m – h}{1.2 m}\)
\(cos\theta = \frac{1.2 – 0.14m}{1.2 m}\)
\(\theta = 28.2^{0}\)
(e)
Time = T/2
\(T = 2\pi \sqrt{\frac{L}{q}} \) simple pendulum
\(\frac{T}{2} = \pi \sqrt{\frac{L}{q}} \)
\(\frac{T}{2} = \pi \sqrt{\frac{1.2}{9.81}} \Rightarrow T/2 = 1.15\)
(f) i.
x Decrease
because the final velocity would be less and thus answer angle;
ii. x Stay the same
\(T = 2\pi \sqrt{\frac{L}{g} }\) the period would stay the same because the length and a stays the same despite change in mass