Reference Frames and Relative Motion AP Physics C Mechanics FRQ – Exam Style Questions etc.
Reference Frames and Relative Motion AP Physics C Mechanics FRQ
Unit: 1. Kinematics
Weightage : 10-15%
Question
A projectile is launched from the back of a cart of mass m that is held at rest, as shown above. At time t = 0, the projectile leaves the cart with speed \(v_0\) at an angle θ above the horizontal. The projectile lands at point P. Assume that the starting height of the projectile above the ground is negligible compared to the maximum height reached by the projectile and the horizontal distance traveled.
(a) Derive an expression for the time\( t_p\) at which the projectile reaches point P. Express your answer in terms of \(v_0\) , θ , and physical constants, as appropriate.
(b) On the axes below, sketch the horizontal component \(v_x\) and the vertical component \(v_y\) of the velocity of the projectile as a function of time t from t = 0 until t = \( t_p\) . Explicitly label the vertical intercepts with algebraic expressions.
The projectile is again launched from the same position, but with the cart traveling to the right with speed \(v_1\) relative to the ground, as shown above. The projectile again leaves the cart with speed v0 relative to the cart at an angle q above the horizontal, and the projectile lands at point Q, which is a horizontal distance D from the launching point. Express your answers in terms of \(v_0\) , q , and physical constants, as appropriate.
(c) Give a physical reason why the projectile lands at point Q, which is not as far from the launch position as point P is, and explain how that physical reason affects the flight of the projectile.
(d) Derive an expression for \(v_1 \). Express your answer in terms of \(v_0\) , q , D, and physical constants, as appropriate. After the launch, the cart’s speed is \(v_2\) . Beginning at time t = 0, the cart experiences a braking force of F = -bv , where b is a positive constant with units of kg s and v is the speed of the cart. Express your answers to the following in terms of m, b, \(v_2\) , and physical constants, as appropriate.
(e) i. Using Newton’s second law, write but DO NOT solve a differential equation that represents the motion of the cart while it experiences the braking force.
ii. Show that the speed v ( t) of the cart as a function of time is given by the equation \(v(t)=v_2e^{-6t/m}\)
iii. Derive an expression for the distance the cart travels from t = 0 until the time it comes to a stop.
Answer/Explanation
(a)
- For using an appropriate kinematics equation to calculate the time to the highest point of the flight \(v_{x}=v_{x0}+a_{x}t\therefore v_{y}=v_{y0}+a_{y}t_{top}\)
- For substituting into the equation above and doubling the time\( 0 = v_o (sin\theta) – gt_{top}\therefore t_{top}=\frac{v_0(sin\theta)}{g} t =2t_{top}= \frac{2v_0(sin\theta}{g} \)
Alternate solution
- For using an appropriate kinematics equation to calculate the time of flight\( x = x_0 + V_{xo}t+\frac{1}{2}a_xt^2 \therefore \Delta y=v_{1y}t+\frac{1}{2}a_{y}t^2 \)
- For substituting into the equation above \(0 = v_o (sin\theta )t — \frac{1}{2} gt^2\therefore =\frac{2v_0(sin\theta)}{g}\)
(b)
- For a straight horizontal line with positive values on the \(v_x\) graph
- For correctly indicating the y-intercept on the \(v_x\) graph
- For a straight line with an initially positive value on the \(v_y\) graph
- For a line with negative slope that crosses the horizontal axis on the \(v_y\) graph
- For correctly indicating the y-intercept on the \(v_y\) graph
(c) For an appropriate physical reason and correct explanation
Claim: The projectile will not travel as far as the stationary case.
Evidence: The rightward component of velocity causes the initial horizontal launch velocity of the projectile with respect to the ground to be less than when the cart was stationary.
Reasoning: The projectile had a component of velocity to the right with respect to the ground at the time of launch.
(d) For a correct expression for the horizontal component of the velocity of the projectile \(v_x=v_0(cos\theta)-v_{1}\)
For correctly substituting into the equation for constant speed\( \Delta x=v_xt\therefore D=(v_0(cos\theta)-v_1)(2v_0(sin\theta)/g)\)
\(\frac{D}{2v_0(sin\theta)/g}=v_0(cos\theta)-v_1\therefore v_1=v_0(cos\theta)-\frac{gD}{2v_0(sin\theta)}\)
(e)
(i) For an appropriate differential equation
\(F= ma\)
\(-bv=m\frac{dv}{dt}\)
(ii) For a correct separation of variable in the above differential equation
\(-\frac{b}{m}dt=\frac{1}{v}dv\)
For integrating with appropriate limits or constant of integration
\(\frac{b}{m}=(t-0)=ln(v(t))-ln(v_2)=ln\left ( \frac{v(t)}{v_2} \right )\)
\(e^{-bt/m}=\frac{v(t)}{v_2}\therefore v(t)=v_2e^{-bt/m}\)
(iii) For indicating that the distance traveled is the integration of the above equation\( \Delta=\int vdt=\int v_{2}e^{-bt/m}dt\) For integrating with appropriate limits or constant of integration
