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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #41 : Motion In Two Dimensions

An archer has a bow, which can be modeled as a spring. The bow has a spring constant of $700\frac{N}{m}$ (that is, for each meter that the string of the bow is pulled back from equilibrium, 700N of force is applied). If the archer pulls back the bow 0.4m with an arrow of mass 0.15kg and releases it at an angle of $25^{\circ}$ above the horizontal, what is the range of the arrow? Neglect air resistance.

$g = 10\frac{m}{s^2}$

Possible Answers:

23.0m

2.1m

44.4m

57.2m

36.7m

Correct answer:

57.2m

Explanation:

Since the problem statement tells us that the bow can be modeled as a spring, we can use the following expression to determined how much energy is stored in the bow:

$U_{spring}=\frac{1}{2}kx^2$

Where is the spring constant, and is the distance the string of the bow is pulled from equillibrium. Plugging in our values, we get:

$U_{spring}=\frac{1}{2}\left ( 700\frac{N}{m}\right )(0.4m)^2$

$U_{spring}= 56J$

Then we can assume that all of this energy is transferred to the arrow as kinetic energy:

$U_{spring}=K_{arrow_{o}}$

$\frac{1}{2}mv_o^2=U_{spring}$

Rearranging for initial velocity:

$v_o=\sqrt{\frac{2U_{spring}}{m}}$

Plugging in our values:

$v_o = \sqrt{\frac{2\left ( 56J\right )}{(0.15kg)}}$

$v_o = 27.3\frac{m}{s}$

Then we can use the range equation to calculate the range of the arrow. Since we were given no initial or final heights in the problem statement, we will assume that the initial and final heights are the same; thus, the range equation is applicable:

$R = \frac{v_o^2sin(2\theta)}{g}$

$R = \frac{\left ( 27.3\frac{m}{s}\right )^2sin(50^{\circ})}{\left ( 10\frac{m}{s^2}\right )}$

R= 57.2m

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Example Question #42 : Motion In Two Dimensions

An arrow is shot from a height of and at an angle of $50^{\circ}$ above the horizontal. If the arrow reaches a maximum height of 40m , what was the arrows initial velocity? Neglect air resistance.

$g = 10\frac{m}{s^2}$

Possible Answers:

$2.4\frac{m}{s}$

$36.5\frac{m}{s}$

$13.9\frac{m}{s}$

$22.7\frac{m}{s}$

$24.8\frac{m}{s}$

Correct answer:

$36.5\frac{m}{s}$

Explanation:

We can use the following kinematic equation to determine the arrow's initial vertical velocity:

$v_f^2 = v_i^2 +2a\Delta y$

Rearranging for initial velocity, and eliminating final velocity:

$v_i=\sqrt{-2a\Delta y}$

Substituting in values:

$v_i = \sqrt{-2\left ( -10\frac{m}{s^2}\right )(39m)}$

$v_{y_{i}} = 27.9\frac{m}{s}$

Then using the following equation to determine total initial velocity:

$sin(\theta)=\frac{v_{y_{i}}}{v_i}$

Rearranging for initial velocity:

$v_i = \frac{v_{y_{i}}}{sin(\theta)}$

Plugging in our values:

$v_i = \frac{\left ( 27.9\frac{m}{s}\right )}{sin(50^{\circ})}$

$v_i = 36.5\frac{m}{s}$

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Example Question #41 : Motion In Two Dimensions

If an arrow is shot with a horizontal velocity of $25\frac{m}{s^2}$ from a height of and travels 150m before hitting the ground, at what angle to the horizontal was the arrow shot at? Neglect air resistance.

$g= 10\frac{m}{s^2}$

Possible Answers:

$40.7^{\circ}$

$50.2^{\circ}$

$68.1^{\circ}$

$23.6^{\circ}$

$15.9^{\circ}$

Correct answer:

$50.2^{\circ}$

Explanation:

We can use the range equation to solve this problem:

$R = \frac{v_o^2sin(2\theta)}{g}$

Now let's start to rearrange for theta:

$v_o^2sin(2\theta) = Rg$

Let's stop here and develop an expression for initial velocity in terms that we know"

$v_x = v_ocos(\theta)$

Rearranging:

$v_o = \frac{v_x}{cos(\theta)}$

Now let's plug this back into the range equation:

$\left (\frac{v_x^2}{cos^2(\theta)}\right )sin(2\theta)=Rg$

Then we can use the following trig identity:

$sin(2\theta)=2sin(\theta)cos(\theta)$

Plugging this into our expression:

$\left (\frac{v_x^2}{cos^2(\theta)}\right )2sin(\theta)cos(\theta)=Rg$

Now we can cancel out a cos function to get:

$\left (\frac{v_x^2}{cos(\theta)}\right )2sin(\theta)=Rg$

And then:

$2v_x^2tan(\theta)=Rg$

Rearranging for theta:

$\theta=tan^{-1}\left ( \frac{Rg}{2v_x^2}\right )$

We have values for each of these variables, so time to plug and chug:

$\theta = tan^{-1}\left ( \frac{(150m)\left ( 10\frac{m}{s^2}\right )}{2\left ( 25\frac{m}{s}\right )^2}\right )$

$\theta = tan^{-1}(1.2)$

$\theta = 50.2^{\circ}$

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Example Question #44 : Motion In Two Dimensions

An object with mass slides down an incline plane with slope $\theta$ and height . What does the coefficent of friction $\mu$ have to be in order for the object to not gain velocity?

Possible Answers:

$\mu=\frac{1}{2}mh^{2}$

$\mu=mg(cos(\theta ))$

$\mu=mg$

$\mu=tan(\theta )$

$\mu=mg(sin(\theta))$

Correct answer:

$\mu=tan(\theta )$

Explanation:

In order for the object not to gain velocity, all the forces acting on it need to cancel out. The normal force $(F_{N})$ will already cancel with the perpendicular force due to gravity ( $(F_{\perp })$ ) by definition. In order to find the correct coefficient of friction, we set the magnitude of the force due to friction (F_ f) equal to that of the parallel force due to gravity $(F_{\parallel })$ :

$|F_f|=|F_{\parallel }|$

$F_{f} = \mu F_{N} = \mu mg(cos(\theta ))$

$F_{\parallel }=mg(sin(\theta ))$

$\mu mg(cos(\theta ))=mg(sin(\theta ))$

The mass and gravity cancel here.

$\mu=\frac{sin(\theta )}{cos(\theta )}$

$\mu=tan(\theta )$

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Example Question #41 : Motion In Two Dimensions

A ball near the surface of the earth has initial velocity $<10,10>\frac{m}{s}$ . Determine the velocity 2.1 seconds later.

Possible Answers:

$v_f=<10,-9.8>\frac{m}{s}$

None of these

$v_f=<10,-10.58>\frac{m}{s}$

$v_f=<10,10.58>\frac{m}{s}$

$v_f=<-10,-10.58>\frac{m}{s}$

Correct answer:

$v_f=<10,-10.58>\frac{m}{s}$

Explanation:

Using

v_i+a*t=v_f

<10,10>+<0,-9.8>*2.1=<10,-10.58>

$v_f=<10,-10.58>\frac{m}{s}$

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Example Question #46 : Motion In Two Dimensions

A car traveling at $100\ \frac{\textup{m}}{\textup{s}}$ launches off a ramp of height $2\textup{ m}$ and length $5 \textup{ m}$ . Determine the maximum height of the car. Ignore air resistance.

Possible Answers:

$142.45\textup{ m}$

$108.96\textup{ m}$

$121.15\textup{ m}$

None of these

$116.12\textup{ m}$

Correct answer:

$142.45\textup{ m}$

Explanation:

Finding the angle at which the car launches at:

y:2

x:5

$\theta=tan^{-1}(\frac{2}{5})$

$\theta=21.8^\circ$

Determining y-component of velocity:

$100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$

Using conservation of energy in a closed system:

E_i=E_f

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

.5mv_y^2+mgh_i=mgh_f

Solving for h_f

$\frac{v_y^2+gh_i}{g}=h_f$

Plugging in values:

$\frac{37.1^2+9.8*2}{9.8}=h_f$

h_f=142.45m

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Example Question #47 : Motion In Two Dimensions

A car traveling at $100\ \frac{\textup{m}}{\textup{s}}$ launches off a ramp of height $2\textup{ m}$ and length $5\textup{ m}$ . Determine the "hang time" of the car. Ignore air resistance.

Possible Answers:

$3.95\textup{ s}$

$6.11\textup{ s}$

$7.61\textup{ s}$

$8.44\textup{ s}$

None of these

Correct answer:

$7.61\textup{ s}$

Explanation:

Finding the angle at which the car launches at:

y:2

x:5

$\theta=tan^{-1}(\frac{2}{5})$

$\theta=21.8^\circ$

Determining component of velocity:

$100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$

Using conservation of energy in a closed system:

E_i=E_f

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

.5mv_y^2+mgh_i=mgh_f

Solving for h_f

$\frac{v_y^2+gh_i}{g}=h_f$

Plugging in values:

$\frac{37.1^2+9.8*2}{9.8}=h_f$

h_f=142.45m

Determine the velocity in the y direction when the car hits the ground

.5mv_f^2=mgh_i

Solving for v_f

$v_f=\sqrt{gh_i}$

Plugging in values:

$v_f=\sqrt{9.8*142.45}$

$v_f=37.4\frac{m}{s}$

Now, breaking the jump into two parts, the "up" and the "down" and using v_i+at=v_f

Up:

$37.1\frac{m}{s}-9.8*t_{up}=0$

Down:

$0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$

$t_{up}=3.79s$

$t_{down}=3.82s$

$t_{up}+t_{down}=7.61s$

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Example Question #48 : Motion In Two Dimensions

A car traveling at $100\ \frac{\textup{m}}{\textup{s}}$ launches off a ramp of height $2\textup{ m}$ and length $5\textup{ m}$ . Determine the horizontal distance traveled while airborne. Ignore air resistance.

Possible Answers:

None of these

$740\textup{ m}$

$707\textup{ m}$

$626\textup{ m}$

$819\textup{ m}$

Correct answer:

$707\textup{ m}$

Explanation:

Finding the angle at which the car launches at:

y:2

x:5

$\theta=tan^{-1}(\frac{2}{5})$

$\theta=21.8^\circ$

Determining component of velocity:

$100\frac{m}{s}*sin(21.8^\circ)=37.1\frac{m}{s}$

Using conservation of energy in a closed system:

E_i=E_f

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

.5mv_y^2+mgh_i=mgh_f

Solving for h_f

$\frac{v_y^2+gh_i}{g}=h_f$

Plugging in values:

$\frac{37.1^2+9.8*2}{9.8}=h_f$

h_f=142.45m

Determine the velocity in the y direction when the car hits the ground

.5mv_f^2=mgh_i

Solving for v_f

$v_f=\sqrt{gh_i}$

Plugging in values:

$v_f=\sqrt{9.8*142.45}$

$v_f=37.4\frac{m}{s}$

Now, breaking the jump into two parts, the "up" and the "down" and using v_i+at=v_f

Up:

$37.1\frac{m}{s}-9.8*t_{up}=0$

Down:

$0\frac{m}{s}+9.8*t_{down}=37.4\frac{m}{s}$

$t_{up}=3.79s$

$t_{down}=3.82s$

$t_{up}+t_{down}=7.61s$

Multiplying by the horizontal velocity:

$7.61s*100\frac{m}{s}cos(21.8^\circ)=707m$

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Example Question #51 : Motion In Two Dimensions

Deep in space Object has mass 350kg and is initially traveling with velocity $<30,50>\frac{m}{s}$ . At t=0 , it collides with Object , which has mass 500kg and is initially motionless. The two objects stick together.

Assuming the collision took place at the origin, determine the location at t=55s , fifty five seconds after the collision.

Possible Answers:

<808,225>m

<626,395>m

<682,1133>m

<545,1138>m

None of these

Correct answer:

<682,1133>m

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$

Using

$r_i+\overrightarrow{v}*\Delta t=r_f$

Plugging in values

(0,0)+<12.4,20.6>*55=r_f

r_f=<682,1133>m

Report an Error

Example Question #51 : Motion In Two Dimensions

Assuming the collision took place at the origin, determine the location of object at t=-15s , fifteen seconds before the collision.

Possible Answers:

<-450,-750>m

<450,750>m

None of these

<-720,-360>m

<-456,-113>m

Correct answer:

<-450,-750>m

Explanation:

Using

$r_i+\overrightarrow{v}*\Delta t=r_f$

Plugging in values

r_i+<30,50>*15=(0,0)

r_i=<-450,-750>m

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #41 : Motion In Two Dimensions

Example Question #42 : Motion In Two Dimensions

Example Question #41 : Motion In Two Dimensions

Example Question #44 : Motion In Two Dimensions

Example Question #41 : Motion In Two Dimensions

Example Question #46 : Motion In Two Dimensions

Example Question #47 : Motion In Two Dimensions

Example Question #48 : Motion In Two Dimensions

Example Question #51 : Motion In Two Dimensions

Example Question #51 : Motion In Two Dimensions

All AP Physics 1 Resources

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