Newton’s First Law AP Physics C Mechanics FRQ – Exam Style Questions etc.
Newton’s First Law AP Physics C Mechanics FRQ
Unit 2: Force and Translational Dynamics
Weightage : 20-15%
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
Question
A 0.50 kg fan cart is placed on a level, horizontal track of negligible friction, as shown. The fan is turned on, and the fan cart is released from rest and moves to the right. The cart travels along the horizontal track and then down an incline. Motion detector 1 measures the acceleration a of the cart from time t = 0 to t = 3 s. At t = 3 s, the cart makes a smooth transition to the incline, and motion detector 2 measures the acceleration of the cart after t = 3 s. The fan exerts the same magnitude of force on the cart during the entire motion. The graphs below show a as functions of t. For each motion detector, the positive direction is away from the detector.
(a) On the dots below that represent the cart at two different locations, draw and label the forces (not components) that act on the cart at each location. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
(b) Calculate the magnitude of the net force exerted on the fan cart when it is on the horizontal track.
(c) Calculate the angle θ of the incline.
(d) Suppose careful measurement determines the angle of the incline to be 3° larger than that calculated in part (c). Consider the following explanation.
“The scale used to measure the mass of the fan cart was not calibrated properly before the measurement, and this could account for the observed difference in the angle.”
Does the explanation sufficiently account for the observed discrepancy?
_____ Yes _____ No
Justify your answer.
The experiment is repeated for several trials, each with a different angle for the incline. The acceleration of the cart down the incline is measured for each angle. The graph below shows the plot of the acceleration a of the cart as a function of the sine of the angle sin θ.
(e)
i. Draw a best-fit line for the data.
ii. Using the straight line, calculate an experimental value for the acceleration due to gravity g.
(f) If the cart were replaced with a second cart of mass 1.0 kg that has a fan that exerts the same magnitude of force as the original fan, explain how the graph given in part (e) would change.
Answer/Explanation
Ans:
(a)
(b)
\(F = m\cdot a = 0.5 kg\cdot \frac{4}{5}m/s^{2}= \frac{2}{5}N\)
(c)
\(\frac{12}{5}=\frac{4}{5}+9sin\theta . \frac{8}{5}= 10sin\theta . s-h^{-1}=9.2^{0}\)
(d)
No
m does not appear in the calculations for acceleration and so nos no effect on the measured angle
(e)
\(a = c + 9 sin\theta . \frac{da}{vt}= \frac{9dsin\theta }{0t}\Rightarrow \frac{da}{dsin\theta }=9. \Rightarrow 9= 10 m/s^{2}\)
(f)
The slope would remain constant, but the y-intercept would halve.