AP Physics 1 : Motion in Two Dimensions

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #1 : Motion In Two Dimensions

An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?

Possible Answers:

70m

420m

140m

280m

Correct answer:

140m

Explanation:

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use \(\displaystyle v_y = v_o sin\theta\).

\(\displaystyle v_y = (75 \frac{m}{s})sin 45^o = 53 \frac{m}{s}\)

Next we find how long it takes to reach the top of its trajectory using \(\displaystyle v_f = v_o +at\).

\(\displaystyle 0 \frac{m}{s} = 53 \frac{m}{s} + (-10 \frac{m}{s^2})t\)

 t = 5.3s

Finally, find how high the object goes with \(\displaystyle d = v_ot +\frac{1}{2}at^{2}\).

\(\displaystyle d = (53\frac{m}{s})5.3s + \frac{1}{2}(-10\frac{m}{s^2})(5.3s)^{2} = 140 m\) 

Example Question #2 : Motion In Two Dimensions

An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?

Possible Answers:

1350m

62.5m

675m

250m

Correct answer:

1350m

Explanation:

First, find the horizontal (x) and vertical (y) components of the velocity

\(\displaystyle v_x = v_o cos\theta\)

\(\displaystyle v_x = (125 \frac{m}{s}) cos(30^o) = 108\frac{m}{s}\)

\(\displaystyle v_y = v_o sin\theta\)

\(\displaystyle v_y = (125\frac{m}{s})sin(30^o) = 62.5\frac{m}{s}\)

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

\(\displaystyle v_f = v_o + at\)

\(\displaystyle 0 \frac{m}{s} = 62.5 \frac{m}{s} + (-10 \frac{m}{s^2})t\)

t = 6.25s 

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

\(\displaystyle d_x = v_x(t)\)

\(\displaystyle d_x = 108 \frac{m}{s} (12.5s) = 1,350 m\)

Example Question #3 : Motion In Two Dimensions

A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

What is the velocity of the box just before it hits the ground?

Possible Answers:

\(\displaystyle \small 0\frac{m}{s}.\)

\(\displaystyle \small 22.8\frac{m}{s}.\)

\(\displaystyle \small 37.2\frac{m}{s}.\)

\(\displaystyle \small 32.2\frac{m}{s}.\)

Correct answer:

\(\displaystyle \small 32.2\frac{m}{s}.\)

Explanation:

We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.

\(\displaystyle \frac{1}{2}=\frac{30m}{x}\rightarrow x=60m\)

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation \(\displaystyle \small gsin\Theta\).

\(\displaystyle gsin\Theta=(10)sin(60)=8.66\frac{m}{s^2}\)

We can plug these values into the following distance equation and solve for time.

\(\displaystyle \small x = x_{0} - \frac{1}{2}(gsin\Theta )t^{2}\)

\(\displaystyle \small 0 = 60 - \frac{1}{2}(8.66)t^{2}\)

\(\displaystyle \small \small t = 3.72 s\)

Now that we know the acceleration on the box and the time of travel, we can use the equation \(\displaystyle \small v = v_{0} + at\) to solve for the velocity.

\(\displaystyle \small v = 0 + (8.66\frac{m}{s^{2}})(3.72 s) = 32.2 \frac{m}{s}\)

Example Question #1 : Motion In Two Dimensions

A ball gets pushed off a \(\displaystyle 1.2m\) high table with a horizontal speed of \(\displaystyle 9\frac{m}{s}\). How far does the ball travel horizontally before hitting the ground?

1.2m_table

Possible Answers:

\(\displaystyle x=5.0\ \text{m}\)

\(\displaystyle x=2.9\ \text{m}\)

\(\displaystyle x=0.3\ \text{m}\)

\(\displaystyle x=4.4\ \text{m}\)

\(\displaystyle x=11.7\ \text{m}\)

Correct answer:

\(\displaystyle x=4.4\ \text{m}\)

Explanation:

This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.

\(\displaystyle y=y_o+v_{oy}t+\frac{1}{2}a_yt^2\)

Choosing the ground to be the zero height, we have \(\displaystyle y=0\) and \(\displaystyle y_o=1.2\ \text{m}\).

Also, knowing that the initial vertical velocity is zero, we know that \(\displaystyle v_{oy}=0\).

The kinematic equation simplifies using these values.

\(\displaystyle 0=y_o+\frac{1}{2}at^2\)

Rearrange the equation to isolate time.

\(\displaystyle t=\sqrt{\frac{-2y_o}{a_y}}\)

We know that \(\displaystyle a_y\) is the acceleration due to gravity: \(\displaystyle a_y=-9.8\frac{m}{s^2}\). Plug in the values to solve for time.

\(\displaystyle t=\sqrt{\frac{-2(1.2\ \text{m})}{-9.8\ \frac{\text{m}}{\text{s}^2}}\)

\(\displaystyle t=0.49\ \text{s}\)

We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation \(\displaystyle x=x_0+v_xt\).

We know that \(\displaystyle v_x=9\ \frac{\text{m}}{\text{s}}\) and that \(\displaystyle x_o=0\). Using these values and the time, we can solve for the horizontal distance travelled.

\(\displaystyle x=0+(9\ \frac{\text{m}}{\text{s}})(0.49\ \text{s})\)

\(\displaystyle x=4.4\ \text{m}\)

Example Question #2 : Motion In Two Dimensions

A car drives north at \(\displaystyle \small 25 \frac{m}{s}\) for \(\displaystyle \small 60s\), then turns east and drives at \(\displaystyle \small 30 \frac{m}{s}\) for \(\displaystyle \small 120 s\). What is the magnitude and direction of the average velocity for the trip?

Possible Answers:

\(\displaystyle 21.7\frac{m}{s}\ \text{due east}\)

\(\displaystyle 27.5\frac{m}{s}\ \text{due north}\)

\(\displaystyle 21.7\frac{m}{s}\ \text{at}\ 22.6^o\ \text{above the horizontal}\)

\(\displaystyle 27.5\frac{m}{s}\ \text{at}\ 25^o\ \text{above the horizontal}\)

\(\displaystyle 25\frac{m}{s}\ \text{at}\ 22.6^o\ \text{above the horizontal}\)

Correct answer:

\(\displaystyle 21.7\frac{m}{s}\ \text{at}\ 22.6^o\ \text{above the horizontal}\)

Explanation:

First, determine how far the car travels in each direction:

\(\displaystyle d=v*t\)

\(\displaystyle (25\frac{m}{s})*(60s)=1500m\ N\)

\(\displaystyle (30\frac{m}{s})*(120s)=3600m\ E\)

Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.

\(\displaystyle \sqrt{1500^2+3600^2}=3900m\)

Find the average velocity by dividing the total displacement by the total time.

\(\displaystyle v=\frac{\Delta x}{\Delta t}\)

\(\displaystyle v= \frac{3900m}{60s+120s} =21.7\frac{m}{s}\)

Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.

\(\displaystyle \tan\theta=\frac{opp}{adj}=\frac{\Delta x_N}{\Delta x_E}\)

Use the north and the east directional displacements to find the angle.

\(\displaystyle \tan\theta=\frac{1500m}{3600m}\)

\(\displaystyle \theta=\tan^{-1}(\frac{1500m}{3600m})=22.6^o\)

Our final answer will be:

\(\displaystyle 21.7\frac{m}{s}\ \text{at}\ 22.6^o\ \text{above the horizontal}\)

Example Question #6 : Motion In Two Dimensions

A ball is launched at an angle of \(\displaystyle 30^o\) above the horizontal with an initial velocity of \(\displaystyle 8\frac{m}{s}\).  At what time is its vertical velocity \(\displaystyle 0\frac{m}{s}\)?

Possible Answers:

\(\displaystyle 0.90s\)

\(\displaystyle 1.64s\)

\(\displaystyle 0.5s\)

\(\displaystyle 0.41s\)

\(\displaystyle 2.45s\)

Correct answer:

\(\displaystyle 0.41s\)

Explanation:

Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.

\(\displaystyle v_f = v_i + at\)

We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.

\(\displaystyle v_{iy}=v_o\sin(\theta)\)

\(\displaystyle v_{iy}=(8\frac{m}{s})\sin(30^o)=4\frac{m}{s}\)

Using this in our kinematics formula, we solve for the time.

\(\displaystyle v_f = v_i + at\)

\(\displaystyle 0\frac{m}{s} = (4\frac{m}{s}) + (-9.8\frac{m}{s^2})(t)\)

\(\displaystyle t=\frac{-4\frac{m}{s}}{-9.8\frac{m}{s^2}}\)

\(\displaystyle t=0.41s\)

Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.

Example Question #3 : Motion In Two Dimensions

A 2kg box is at the top of a frictionless ramp at an angle of \(\displaystyle 60^o\). The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

When the box is released, how long will it take the box to reach the ground?

Possible Answers:

\(\displaystyle 2.45s\)

\(\displaystyle 3.46s\)

\(\displaystyle 2.63s\)

\(\displaystyle 4.21s\)

\(\displaystyle 3.72s\)

Correct answer:

\(\displaystyle 3.72s\)

Explanation:

We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation \(\displaystyle \small cos(60^{\circ}) = \frac{30}{x}\).

\(\displaystyle \frac{1}{2}=\frac{30m}{x}\rightarrow x=60m\)

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of \(\displaystyle \small gsin\Theta\). Since the angle is 60o, the acceleration on the box is \(\displaystyle \small 8.66 \frac{m}{s^{2}}.\) 

\(\displaystyle gsin\Theta=(10)sin(60)=8.66\frac{m}{s^2}\)

Finally, we can plug these values into the following distance equation and solve for time.

\(\displaystyle \small x = x_{0} - \frac{1}{2}(gsin\Theta )t^{2}\)

\(\displaystyle \small 0 = 60 - \frac{1}{2}(8.66)t^{2}\)

\(\displaystyle \small \small t = 3.72 s\)

 

Example Question #4 : Motion In Two Dimensions

A plane is traveling from Portland to Seattle, which is 100 miles due north of Portland. There is a constant wind traveling southeast at 30 mph. If the plane needs to get to Seattle in one hour while flying due north, at what speed (relative to the wind) and angle should the pilot fly?

Possible Answers:

\(\displaystyle 123\text{mph}\ 9.9^o\ \text{north of west}\)

None of the other answers

\(\displaystyle 142\text{mph}\ 14.2^o\ \text{west of north}\)

\(\displaystyle 123\text{mph}\ 9.9^o\ \text{west of north}\)

\(\displaystyle 142\text{mph}\ 14.2^o\ \text{north of west}\)

Correct answer:

\(\displaystyle 123\text{mph}\ 9.9^o\ \text{west of north}\)

Explanation:

First we need to find out at what the speed relative to the ground the plane needs to fly. The plane needs to cover 100 miles in 1 hour, so it's simply 100 mph due north.

Now we need to calculate its speed relative to the wind and its angle. We know the sum of the wind vector and plane vector needs to equal 100 mph due north. Therefore, all east/west movement must cancel out and all north/south movement must add to 100mph.

We can separate the wind velocity into its components:

\(\displaystyle w_{x} = 30cos(45^{\circ})\)

\(\displaystyle w_{y} = 30sin(45^{\circ})\)

We can also represent the components of the plane's velocity:

\(\displaystyle p_x = p\cdot sin(\Theta )\)

\(\displaystyle p_y = p\cdot cos(\Theta )\)

If the trig functions seem reversed, this is because the angle in question is between the y-coordinate and a vector pointing slightly west of north.

We can also represent the components of the velocity relative to the ground:

\(\displaystyle v_x = 0\)

\(\displaystyle v_y=100\)

Since the horizontal velocity is equal to 0, we can set the x-components of the wind and plane vectors equal to each other:

\(\displaystyle 30cos(45^{\circ}) = p\cdot sin(\theta)\)

The sum of the y-components of the wind and plane vectors must equal 100:

\(\displaystyle p\cdot cos(\theta) - 30sin(45^{\circ}) = 100\)

The wind vector is subtracted because it is in the opposite direction of the plane vector.

Now we just need to isolate a variable and substitute one equation into another. We will isolate the total plane velocity in the first equation:

\(\displaystyle p = \frac{30cos(45^{\circ})}{sin(\theta)}\)

Substituting this into the second equation, we get:

\(\displaystyle \frac{30cos(45^{\circ})cos(\theta)}{sin{\theta}} - 30sin(45^{\circ}) = 100\)

Solving for theta, we get \(\displaystyle \theta = 9.9^{\circ}\) (west of north)

We can plug this into the first equation to get:

\(\displaystyle p = \frac{30cos(45^{\circ})}{sin(9.9^{\circ})} = 123 mph\)

Example Question #5 : Motion In Two Dimensions

A baseball is traveling with a velocity of \(\displaystyle 12\frac{m}{s}\) at an angle of \(\displaystyle 50^{\circ}\) above horizontal. What is the velocity of the ball after two seconds?

\(\displaystyle g = 10\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 23.7\frac{m}{s}\)

\(\displaystyle 11.4\frac{m}{s}\)

\(\displaystyle 16.8\frac{m}{s}\)

\(\displaystyle 13.3\frac{m}{s}\)

\(\displaystyle 20.4\frac{m}{s}\)

Correct answer:

\(\displaystyle 13.3\frac{m}{s}\)

Explanation:

To solve this problem, we first need to split the velocity into its vector components.

Initial vertical velocity:

\(\displaystyle v_y = vsin(\theta)\)

\(\displaystyle v_y = 12sin(50^{\circ})=9.193\frac{m}{s}\)

Initial horizontal velocity:

\(\displaystyle v_x = vcos(\theta)\)

\(\displaystyle v_x =12cos(50^{\circ})=7.713\frac{m}{s}\)

Since we are neglecting air resistance, horizontal velocity does not change over time. We only need to calculate the new vertical velocity after two seconds, using kinematics:

\(\displaystyle v_{yf} = v_{yi} - at\)

\(\displaystyle v_{yf}= 9.913\frac{m}{s}-(10\frac{m}{s^2})(2s) = -10.81\frac{m}{s}\)

The negative sign simply means that the vertical velocity has changed direction, and is now pointed downward.

We can use the following equation to determine the total velocity, which will be the sum of the horizontal and vertical velocity vectors:

\(\displaystyle v^2 = v_y^2+v_x^2\)

\(\displaystyle v = \sqrt{v_y^2 + v_x^2}=\sqrt{(-10.81)^2+(7.71)^2} = 13.3\frac{m}{s}\)

Example Question #5 : Motion In Two Dimensions

Suppose that a golf ball is struck such that it travels at a speed of \(\displaystyle 50\: \frac{m}{s}\) at an angle \(\displaystyle 30^{o}\) to the horizontal. Neglecting air resistance, how long will the golf ball remain in the air before it touches the ground again?

\(\displaystyle g=9.8\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 5.10s\)

\(\displaystyle 2.26s\)

\(\displaystyle 1.60s\)

\(\displaystyle 2.55s\)

Correct answer:

\(\displaystyle 5.10s\)

Explanation:

We're told that the golf ball is starting its parabolic journey with a certain velocity at an angle to the ground. To solve for the time the golf ball stays airborne, we'll need to consider the x and y-components of the ball's trajectory.

First, we'll need an expression that considers the ball's velocity in the x-direction.

\(\displaystyle v_{xi}=v_{i}cos(\theta )\)

We'll also need an expression for the y-component of the velocity.

\(\displaystyle v_{yi}=v_{i}sin(\theta )\)

We'll need an expression that can relate the vertical distance traveled with the time spent in the air. Since the only acceleration occurring in this scenario is due to gravity, the result is that acceleration is constant. Therefore, we can make use of some of the kinematics equations:

\(\displaystyle \Delta y=v_{yi}t+\frac{1}{2}at^{2}\)

It's also important to note that once the ball lands back on the ground, only its horizontal displacement will have changed, while its vertical displacement will remain unchanged.

\(\displaystyle \Delta y=0=v_{yi}t+\frac{1}{2}at^{2}\)

\(\displaystyle v_{yi}t=-\frac{1}{2}at^{2}\)

We also have to remember that in this case, the source of acceleration is from gravity, which points downwards. If we define up as the positive y direction, then down must be the negative y direction. Therefore, we can write:

\(\displaystyle a=-g\)

\(\displaystyle v_{yi}t=\frac{1}{2}gt^{2}\)

Plug in the vertical component of velocity and solve for time.

\(\displaystyle vtsin(\Theta)=\frac{1}{2}gt^{2}\)

\(\displaystyle t=\frac{2vsin(\theta)}{g}\)

\(\displaystyle t=\frac{2* 50\frac{m}{s}* sin\left ( 30^{o}\right )}{9.8\frac{m}{s^{2}}}=5.10s\)

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