AP Physics C Mechanics Unit 1: Kinematics

Build calculus-based models of motion using vectors, displacement, velocity, acceleration, representations, and component-by-component kinematics.

Covered syllabus points

1.1 Scalars and Vectors 1.2 Displacement, Velocity, and Acceleration 1.3 Representing Motion 1.4 Reference Frames and Relative Motion 1.5 Motion in Two or Three Dimensions

[Maximum number: 1]

A small sled slides across a rough horizontal table with an initial velocity $v_{0}$ . The coefficient of kinetic friction between the sled and the table is $\mu_{k}$ . A string connects the sled to a device on the ground. The device maintains constant tension $F_{T}$ in the string by unwinding the string as the sled slides to the right. The total mass of the sled is m. The string is attached to the device at x=0 and at a height of y, as shown in Figure 1. The horizontal position of the sled is represented by x, as shown in Figure 2. Express all algebraic answers in terms of $m, \mu_{k}$ , $F_{T}, x, y$ , and physical constants, as appropriate.

Determine an expression for the angle $\theta$ that the string makes with the vertical when the sled has traveled a horizontal distance x.

For any correct trigonometric expression for $\theta$ in terms of the given quantities 1 point

\sin \theta=\frac{x}{\sqrt{y^{2}+x^{2}}} \quad \therefore \theta=\sin ^{-1}\left(\frac{x}{\sqrt{y^{2}+x^{2}}}\right)

\cos \theta=\frac{y}{\sqrt{y^{2}+x^{2}}} \quad \therefore \theta=\cos ^{-1} \frac{y}{\sqrt{y^{2}+x^{2}}}

\tan \theta=\frac{x}{y} \quad \therefore \theta=\tan ^{-1} \frac{x}{y}

[Maximum number: 15]

Table for Question 1 Kinematics question 2 — AP Physics C: Mechanics

Mech. 1. A new sports car is undergoing acceleration tests to determine its specifications. The following data on speed v versus time t are recorded for the car as it accelerates from rest along a straight track.

(a)

Table for Question (a) — AP Physics C: Mechanics

(a) On the axes below, plot v as a function of t and sketch a curve that best represents the data.

Figure for Question (a) — AP Physics C: Mechanics

[ 3 ]

(b)

Table for Question (b) — AP Physics C: Mechanics

(b) Answer each of the following for the time period $t=0 \mathrm{~s}$ to $t=10 \mathrm{~s}$ .

[ 2 ]

(i)

Table for Question (i) — AP Physics C: Mechanics

i. Does the speed of the car increase, decrease, or stay the same? Increase Decrease Stay the same Justify your answer.

[ 1 ]

(ii)

Table for Question (ii) — AP Physics C: Mechanics

ii. Does the acceleration of the car increase, decrease, or stay the same? Increase Decrease Stay the same Justify your answer.

[ 1 ]

(c)

Table for Question (c) — AP Physics C: Mechanics

(c) Explain how you would use your graph in part (a) to find the distance traveled by the car between $t=2 \mathrm{~s}$ and $t=8 \mathrm{~s}$ .

[ 2 ]

(c) 2 points For mentioning area under the curve 1 point For specifically describing the area between $t=2 \mathrm{~s}$ and $t=8 \mathrm{~s}$ 1 point Example: To determine the distance traveled from a velocity-time graph, one should find the area under the curve. For time $t=2 \mathrm{~s}$ to $t=8 \mathrm{~s}$ , one can approximate the area using a trapezoid that approximately fills the area under the curve.

(d)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

Table for Question (d) — AP Physics C: Mechanics

(d) Derive an expression for the acceleration of the car a(t) as a function of time t.

[ 2 ]

(d) 2 points For beginning with an indication that the acceleration is the derivative of velocity $a(t)=\frac{d v}{d t}$ For correctly evaluating the derivative $a(t)=\frac{d}{d t}\left(-0.3 t^{2}+7 t\right)=-0.6 t+7$

(e)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

Table for Question (e) — AP Physics C: Mechanics

(e) Calculate the position of the car as a function of time, x(t), assuming that the car starts from rest at the origin of a coordinate system.

[ 3 ]

(e) 3 points For an indication that the displacement is the integral of the velocity $\Delta x=\int_{t_{0}}^{t} v(t) d t$ For attempting to integrate with appropriate limits or constant of integration $\Delta x=\int_{0}^{t}\left(-0.3 t^{2}+7 t\right) d t$ For the correct answer $x(t)=-0.1 t^{3}+3.5 t^{2}$

(f)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

Table for Question (f) — AP Physics C: Mechanics

(f) Calculate the distance traveled by the car between $t=2 \mathrm{~s}$ and $t=8 \mathrm{~s}$ .

[ 3 ]

(f) 3 points For recognizing that the distance is the change in position $\Delta x=x(8)-x(2)$ For using the expression from part (e) to calculate x(2) and x(8) $x(2)=13.2 \mathrm{~m}$ $x(8)=172.8 \mathrm{~m}$ For the correct answer with units $\Delta x=159.6 \mathrm{~m}$ Alternate solution For using the integral of the speed and recognizing that the distance is the change in position For using the correct limits or constant of integration $\Delta x=\int_{2}^{8}\left(-0.3 t^{2}+7 t\right) d t$ $\Delta x=\frac{-0.3 t^{3}}{3}+\left.\frac{7 t^{2}}{2}\right|_{2} ^{8}=-51.2+224-(-0.8+14)$ For the correct answer with units $\Delta x=159.6 \mathrm{~m}$

[Maximum number: 8]

Scientists have created a new type of lightweight foam and are performing experiments to investigate the properties of the foam. The mass of Cart A is 1000 kg and the mass of Cart B is 2000 kg. A piece of foam with negligible mass is attached to the front of Cart A, as shown. Cart A moves with a constant speed toward Cart B, which is initially at rest. At time $t=0 \mathrm{~s}$ , the foam connected to Cart A makes contact with Cart B. The foam remains in contact with Cart B for 0.5 s, after which the carts separate and both carts move with constant velocities.

Figure for Question 1 Kinematics question 3 — AP Physics C: Mechanics

(a)

The graph shows the velocity v of Cart A as a function of time t for the time interval when the foam and Cart B are in contact.

What feature(s) of the graph could be used to estimate the displacement of Cart A during the collision?

Using the information shown in the graph, determine the speed of Cart B at $t=0.5 \mathrm{~s}$ .

On the following grid, draw a smooth curve of the velocity of Cart B as a function of time.

[ 1 ]

(i)

For stating that the area bounded by the curve is the displacement

1 point

(a)(ii)

For indicating momentum is conserved

1 point

Example Response

$\Sigma p_0=\Sigma p_f$

For the correct answer for speed with units (2 m/s)

1 point

Example Solution

$m_{\mathrm A}v_{\mathrm{A0}}+m_{\mathrm B}v_{\mathrm{B0}}=m_{\mathrm A}v_{\mathrm{Af}}+m_{\mathrm B}v_{\mathrm{Bf}}$ $m_{\mathrm A}v_{\mathrm{A0}}-m_{\mathrm A}v_{\mathrm{Af}}=m_{\mathrm B}v_{\mathrm{Bf}}$ $v_{\mathrm{Bf}}=\frac{m_{\mathrm A}}{m_{\mathrm B}}\left(v_{\mathrm{A0}}-v_{\mathrm{Af}}\right)$ $v_{\mathrm{Bf}}=\frac{1000\,\mathrm{kg}}{2000\,\mathrm{kg}}(5\,\mathrm{m/s}-1\,\mathrm{m/s})$ $\therefore v_{\mathrm{Bf}}=2\,\mathrm{m/s}$

For a graph that starts at (0,0) and ends at (0.5,2) or value consistent with (a)(ii)

1 point

For a smooth, continuous curve that transitions from concave up to concave down

1 point

Example Solution

$v(\mathrm{m/s})$ $v(\mathrm{m/s})$

Total for part (a)

5 points

(b)

For $0 \leq t \leq 0.50 \mathrm{~s}$ , the velocity v of Cart A can be described by the function $v(t)=64 t^{3}-48 t^{2}+5$ .

Calculate the magnitude of the maximum net force acting on Cart A during this interval.

On the following grid, draw a smooth curve of the magnitude of the force acting on Cart A as a function of time. Clearly indicate the value of the maximum force on the vertical axis.

Figure for Question (b) — AP Physics C: Mechanics

The foam is removed from the front of Cart A and the experiment is repeated. The carts collide, with both Cart A and Cart B having the same initial and final velocities as in the original collision. The time intervals during which the carts are in contact are different in the collision with the foam and the collision without the foam. In the collision without the foam, Cart A is in contact with Cart B for a shorter duration than in the original collision, when the foam was present.

For the original collision when the foam is present, the magnitude of the average net force exerted on Cart B is $F_{1}$ . For the collision without the foam, the magnitude of the average net force exerted on Cart B is $F_{2}$ .

[ 7 ]

$\begin{aligned} a(t)&=\frac{dv}{dt}=192t^2-96t,\\ \frac{da}{dt}&=384t-96=0 \quad\Longrightarrow\quad t=0.25\ \mathrm{s}. \end{aligned}$

At (t=0.25\ \mathrm{s}), (a=-12\ \mathrm{m\,s^{-2}}). With (m=1000\ \mathrm{kg}),
$F_{\max}=m|a|=(1000)(12)=12\,000\ \mathrm{N}.$

For the graph,
$|F(t)|=1000\,|192t^2-96t|=96\,000t(1-2t), \qquad 0\le t\le0.50,$
so it is a smooth downward-opening curve through (0\ \mathrm N) at (t=0) and (t=0.50\ \mathrm s), with maximum (12\,000\ \mathrm N) at (t=0.25\ \mathrm s).

[Maximum number: 2]

A block of mass m is placed on top of an ideal spring of spring constant k. The block is pushed against the spring, compressing the spring a distance $\Delta x$ . The block is released from rest, leaves the spring at the position shown in the figure, travels upward, and enters a track with a constant radius of curvature R that has negligible friction. The block enters the track at point A, maintains contact with the track, and exits horizontally at point B, a distance 3 R above the point the block was released. The block then falls to the ground and lands a horizontal distance D from the end of the track. Express all algebraic answers in terms of $m, k, \Delta x, R$ , and physical constants, as appropriate. The size of the block is much smaller than the radius of curvature of the track.

Calculate the distance D that the block travels.

For correctly relating the height of fall to the time of fall
$y=y_{0}+v_{o y} t+\frac{1}{2} a_{y} t^{2} \therefore H=0+0+\frac{1}{2} g t^{2} \therefore t=\sqrt{\frac{(2)(4 R)}{g}}=\sqrt{\frac{8 R}{g}}$

1 point

For correctly substituting into the equation for constant velocity consistent with part (b)(i)
$D=v_{x} t=\left(\sqrt{\frac{k}{m}(\Delta x)^{2}-6 g R}\right)\left(\sqrt{\frac{8 R}{g}}\right)$

1 point

Total for part (d)

2 points

All question bank results loaded

Covered syllabus points

1 Kinematics question 1

1 Kinematics question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)

Question (e)

Question (f)

1 Kinematics question 3

Question (a)

Question (b)

1 Kinematics question 4