Conservation of Energy AP Physics C Mechanics FRQ – Exam Style Questions etc.
Conservation of Energy AP Physics C Mechanics FRQ
Unit 3: Work, Energy, and Power
Weightage : 15-25%
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
The variable x represents the position of particle A in a two-particle system. Particle B remains at rest. The above graph shows the potential energy U ( x) of the system as a function of position x of particle A, which has a mass of 0.40 kg. Particle A is released from rest at the position x = 2.0 m. Assume positive to be toward the right.
(a) Calculate the speed of particle A when it reaches position x = 14.0 m.
(b) i. Calculate the magnitude of the acceleration of particle A when it reaches position x= 3.0 m.
ii. What is the direction of the acceleration of particle A when it reaches position x= 3.0 m? Left Right
(c) Determine the magnitude of the acceleration of particle A when it reaches position x = 7.0 m.
(d) Particle A is again released from rest at the position x = 2.0 m.
i. Calculate the elapsed time for particle A to travel from position x = 2.0 m to position x = 6.0 m.
ii. Calculate the elapsed time for particle A to travel from position x = 6.0 m to position x = 8.0 m.
iii. Calculate the elapsed time for particle A to travel from position x = 8.0 m to position x = 14 m.
(e) Particle A is now placed at position x= 7.0 m. In order for particle A to reach the position x = 16.0 m and to be moving at a speed of 1.0 m s, what initial speed should particle A be given at its new initial position of x = 7.0 m?
(f) On the grid below, carefully draw a graph of the force F acting on particle A as a function of x for the range 0 < x < 16 m.
Answer/Explanation
(a) For correctly using conservation of energy \(U_1+K_1=U_2+K_2\) \(U_1+0=U_2+\frac{1}{2}mv^2\) For correctly substituting values from the graph into the equation above
\((8j)=(6j)+\left ( \frac{1}{2} \right )(0.40kg)v^2\)
\(v=3.16m/s\)
(b)
i) For correctly relating the slope of the line to the force exerted on the particle \(F=-slope=-\frac{(U_2-U_1)}{(X_2-X_1)}=-\frac{(12-0)J}{(0-6)M}=2.0N\)
For using Newton’s second law to calculate the acceleration of the particle \(a=\frac{F}{m}=\frac{(2.0N)}{(0.40kg)}=5.0/s^2\)
Note: point is awarded for \(5.0 m/ s^2\)
(b)
i) Alternate Solution
Solving for two velocities between 2.0 m and 6.0 m\( v_1 = 0\)
ii) For selecting answer consistent with part (b)(i)
(c) For a correct answer \(F=-slope=0\therefore a=0\) Note: credit is given even if no work is shown
(d)
i) For using a correct kinematic equation to calculate the time
\(d=v_1t+\frac{1}{2}at^2=0+\frac{1}{2}at^2\)
For correctly substituting acceleration from part (b)(i) into equation above
\(t=\sqrt{\frac{2d}{a}}=\frac{(2)(6.0-2.0)m}{(5.0m/s^2)}=1.26s\)
ii) For using a correct kinematic equation to determine the speed at x = 6.0 m or using the velocity at 6.0 m from alternate solutions part (b)(i) or (d)(i)
\(v^2_2=v^2_1+2ad=0+2ad
v_2=\sqrt{2ad}=\sqrt{(2)(5.0m/s^2)(6.0-2.0)m} v_{2}=6.32m/s
For using a kinematic equation for zero acceleration to calculate the time d =vt
\(t=\frac{d}{v}=\frac{(8.0-6.0)m}{(6.32m/s)}\) t=0.32s
Alternate Solution
For correctly using the energy equation to solve for the speed at 6.0 m (or using the value from alternate solution part (b)(i)
\(U_{1}+K_{1}=U_{2}+K_{2} \therefore (8J) +0+\frac{1}{2}(0.40kg)v^{2}_{2} v_{2}=6.32m/s\)
For using a kinematic equation with zero acceleration to calculate the time d=vt\( t=d/v=(8.0-6.0)m/(6.32m/s) t=0.32s\)
(d)
iii) For using a correct kinematic equation to calculate the time
\(d=\frac{1}{2}(v_1+v_2)t\)
\(t=\frac{2d}{(v_1+v_2)}=\frac{(2)(14-8.0)m}{(6.32m/s+3.16m/s)} t=1.26s\)
(e) For correctly using conservation of energy to calculate the speed
\(U_1+K_1=U_2+K_2\)
\(0+\frac{1}{2}m_1v^2_1\)
\(0+\frac{1}{2}m_1v^2_1=U_2+\frac{1}{2}m_2v^2_2\)
\(\left ( \frac{1}{2} \right )(0.40kg)v^2_1=(6j)+\left ( \frac{1}{2} \right )(0.40kg)(1.0m/s^2)
v=5.57m/s\)
(f) For a horizontal line at F = 2 N from x = 0 m to x = 6 m
For a horizontal line at F = 0 from x = 6 m to x = 8 m and from x = 14 m to x = 16 m
For a horizontal line at F = –1 N from x = 8 m to x = 14 m
Note: Two points earned if shape and magnitudes are correct but graph is inverted