Change in Momentum and Impulse AP Physics C Mechanics FRQ – Exam Style Questions etc.
Change in Momentum and Impulse AP Physics C Mechanics FRQ
Unit 4: Linear Momentum
Weightage : 15-25%
Question
A student sets up an experiment with a cart on a level horizontal track. The cart is attached with an elastic cord to a force sensor that is fixed in place on the left end of the track. A motion sensor is at the right end of the track, as shown in the figure above. The cart is given an initial speed of v0 = 2.0 m s and moves with this constant speed until the elastic cord exerts a force on the cart. The motion of the cart is measured with the motion detector, and the force the elastic cord exerts on the cart is measured with the force sensor. Both sensors are set up so that the positive direction is to the left. The data recorded by both sensors are shown in the graphs of velocity as a function of time and force as a function of time below.
(a) Calculate the mass m of the cart. For time period from 0.50 s to 0.75 s, the force F the elastic cord exerts on the cart is given as a function of time t by the equation F A = sin(ωt) , where A = 6.3 N and ω = 12.6 rad/s.
(b) Using the given equation, show that the area under the graph above is 1.0 N .s
(c) The experiment is repeated using a different cord that exerts a larger average force on the cart. The cart starts and ends with the same speeds as those in the original experiment. Will the area for the graph of force as a function of time for the new cord be greater than, less than, or equal to the area for that of the original cord? ____ Greater than ____ Less than ____ Equal to Justify your answer. The elastic cord from the original experiment can be modeled as an ideal spring with force constant k .
(d) Derive an expression for the maximum change in length \(x_{MAX}\) for the cord. Express your answer in terms of m, k, \(v_0\) , and physical constants, as appropriate.
The student performs several trials of the experiment. For the first trial, the cart is empty. In each succeeding trial, a block is added to the cart. In all trials, the cart has an initial speed of 2.0 m s to the right, the cart rebounds to the left with a speed of 2.0 m s , and the maximum change in length of the elastic cord is measured. The total mass M of the cart and the maximum change in length of the cord in each trial are recorded in the table below.
(e) Indicate below which quantities should be graphed to yield a straight line with a slope that could be used to calculate a numerical value for the force constant of the elastic cord k. Vertical axis:
Horizontal axis: Use the remaining columns in the table on the previous page, as needed, to record any quantities that you indicated that are not given.
(f) Plot the data points for the quantities indicated in part (e) on the graph below. Clearly scale and label all axes, including units as appropriate. Draw a straight line that best represents the data.
(g) Use the best-fit line to calculate the force constant k of the elastic cord.
Answer/Explanation
(a) For correctly using the area under the graph to calculate the mass of the cart j=Area = \(\Delta p=m(v_2-V_1)\) For a correct answer with units \(m=\frac{Area}{(v_2-v_1)}\)=\(\frac{(1.0N.s)}{(2.0m/s-(-2.0m/s))}\) m = 0.25 kg
(b) For relating the time integral of given equation to the area under the curve For applying the correct limits of integration
\(j=Area=\int_{0.50}^{0.75}Asin(\omega t)dt\)
\(j=-\frac{A}{\omega }[cos\omega t]_{0.50}^{0.75}=-\frac{(6.3N)}{(12.6rad/s)}[cos(12.6t)]_{0.50}^{0.75}\)
\(j=-(0.50)(cos[12.6(0.75)])-cos[(12.6)(0.50)]\)
\(j=1.0N.s\)
(c) For selecting “Equal to” with an attempt at a relevant justification For a correct justification Example: Since the cart starts and ends at the same velocity, the impulse on the cart will be the same so the area under the graph will be the same. Note: If the wrong selection is made, the justification is ignored.
(d) For correctly using conservation of energy
\(U_{MAX}=K_{MAX}\)
\(\frac{1}{2}kx^{2}_{MAX}=\frac{1}{2}mv_{0}v^2\)
\(x_{MAX}=v_{0}\sqrt{\frac{m}{k}}\)
(e) For correctly indicating two variables that will yield a straight line that could be used to determine a value for Example: Vertical Axis: \(x_{MAX}\) Horizontal Axis:\(\sqrt{m}\)
Note: Student earns full credit if axes are reversed or if they use another acceptable combination
(f) For a correct scale that uses more than half the grid 1 point For correctly labeling the axis with variables and units consistent with part (e) Note: Student earns full credit if axes are reversed
For correctly plotting data For drawing a straight line consistent with the plotted data
(g) For correctly calculating the slope from the best-fit line and not the data points
Example: \(slope=\frac{\Delta x_{MAX}}{\Delta \sqrt{m}}=\frac{(0.40-0.20)m}{(1.32-0.70)\sqrt{kg}}=0.34m/\sqrt{kg} 0.33m/\sqrt{kg}\)
\(\frac{v_{0}}{\sqrt{k}}\therefore k=\frac{v^2_{0}}{(slope)^2}=\frac{(2.0m/s)^2}{(0.34m/\sqrt{kg})^2}\) k=34.6N/m
(linear reression k=36.7N/m)