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AP Physics 1 : AP Physics 1

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7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #52 : Motion In One Dimension

Consider the following scenario:

Sledder

A sledder of mass is at the stop of a sledding hill at height with a slope of angle .

A sledder is accelerating down the hill that has a slope $20^{\circ}$ at a rate of $4\frac{m}{s^2}$ . If the sledder has an initial velocity of $2\frac{m}{s}$ , how far has the sledder traveled along the hill if her final velocity is $8\frac{m}{s}$ ?

Possible Answers:

7.5m

16m

12m

32m

Correct answer:

7.5m

Explanation:

We can use the follow kinematics equation to solve this problem:

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

Rearranging for change in x, we get:

$\Delta x = \frac{v_f^2-v_i^2}{2a}$

Now plugging in values for each variable:

$\Delta x = \frac{\left ( 8\frac{m}{s}\right )^2-\left ( 2\frac{m}{s}\right )^2}{2\left ( 4\frac{m}{s^2}\right )}$

$\Delta x = 7.5m$

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Example Question #53 : Motion In One Dimension

Consider the following scenario:

Sledder

A sledder of mass is at the stop of a sledding hill at height with a slope of angle .

If the sledder is traveling down a hill which has a slope $\angle 40^{\circ}$ and has a constant horizontal velocity of $v_x = 6\frac{m}{s}$ , how much time does it take for the sledder to drop a vertical distance of 12m .

Possible Answers:

0.3s

3.6s

1.1s

5.8s

2.4s

Correct answer:

2.4s

Explanation:

To solve this problem we will go in the following steps:

1. Calculate linear velocity parallel with slope

2. Calculate vertical component of velocity

3. Calculate time it takes to drop 12m

So first, we will use the cosine function:

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

Rearranging:

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

Plugging in our value:

$v = \frac{\left ( 6\frac{m}{s}\right )}{cos(40^{\circ})} = 7.83\frac{m}{s}$

Now to calculate the vertical component, we will use the sine function:

$sin(\theta)=\frac{v_y}{v}$

Rearranging

$v_y = vsin(\theta)$

$v_y = (7.83)sin(40^{\circ}) = 5.03\frac{m}{s}$

Now using this to calculate the time it takes to drop 12m :

$\Delta y = v_yt$

$t = \frac{\Delta y}{v_y} = \frac{12m}{5.03\frac{m}{s}}$

t = 2.4s

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Example Question #57 : Linear Motion And Momentum

An arrow is shot straight up in the air with an initial velocity of $35\frac{m}{s}$ from a height of. How long does it take for the arrow to hit the ground? Neglect air resistance.

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

Possible Answers:

7.1s

5.8s

6.4s

7.0s

2.0s

Correct answer:

7.1s

Explanation:

Since the arrow is shot straight up in the air, we can calculate how long it takes for the arrow to reach its high point (where it will have a velocity of 0):

v_f = v_i +at

$0\frac{m}{s}= 35\frac{m}{s}+\left ( -10\frac{m}{s^2}\right )t$

t = 3.5 s

Furthermore, this will be the same amount of time it takes to get to its original position. Then we just need to calculate how long it takes the arrow to travel from its original position to the ground, using the follow kinematics equation:

$\Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

Plugging in our values:

$-2m = \left ( -35\frac{m}{s}\right )t+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )t^2$

Rearranging:

$\left ( 5\frac{m}{s^2}\right )t^2+\left ( 35\frac{m}{s}\right )t-2m=0$

Using the quadratic equation, we get:

t = 0.057s or -7.06s

Since we can't have a negative time, the first option is the correct one. Adding this to our other time, we get:

t = 7.1s

Alternatively, we could have solved the equation by putting the initial conditions in the kinematic equation.

$-2m = \left ( 35\frac{m}{s}\right )t+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )t^2$

However, the first method emphasizes the importance of being able to solve a problem multiple different ways and incorporates multiple concepts surrounding projectile motion.

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Example Question #51 : Linear Motion And Momentum

A physicist over-pressurized a water heater that exploded and shot into the air. The water heater stayed in the air for $12 \textup{ s}$ before it came crashing back to earth. How far above the ground did the water heater reach at it's tallest point? (Ignore air resistance.)

Possible Answers:

$60\textup{ m}$

$720\textup{ m}$

$360\textup{ m}$

$120\textup{ m}$

$180\textup{ m}$

Correct answer:

$180\textup{ m}$

Explanation:

We can solve these problems using kinematic equations. First we start with:

$v = v_{0} + at$

$0= v_{0} + (-10\frac{m}{s^{2}})(6s))$

$v_{0} = 60\frac{m}{s}$

We know that the final velocity is $0\frac{m}{s}$ because at the top of an arc the water heater doesn't have vertical movement. The acceleration is due to gravity, and the time is only half the time that was given in the problem (half the time is the water heater going up, the other half is it falling back to the ground).

When we know the initial velocity of the water heater we can then use:

$y = v_{0}t + \frac{1}{2}at^{2}$

$y=(60\frac{m}{s})(6s)+\frac{1}{2}(-10\frac{m}{s})(6s)^{2}$

y = 180m

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Example Question #59 : Linear Motion And Momentum

An object on frictionless wheels is being pulled with a force of 20N and accelerates at rate of $5 \frac{m}{s^{2}}$ . If a second object is also pulled at 20N but weighs twice as much as the first object, what would it's acceleration be?

Possible Answers:

$2.1 \frac{m}{s^{2}}$

$4.1 \frac{m}{s^{2}}$

$2.5 \frac{m}{s^{2}}$

$4.5 \frac{m}{s^{2}}$

Correct answer:

$2.5 \frac{m}{s^{2}}$

Explanation:

First we must find the original object's mass using the equation F=ma . . We then double the mass for the second object and again use . Solve for acceleration using Mass2 = 8kg and Force2 = 20N .

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Example Question #51 : Linear Motion And Momentum

A ball is dropped from a building and at one instant its speed is $25\frac{m}{s}$ . What is its approximate speed one second later?

Possible Answers:

$30\frac{m}{s}$

$35\frac{m}{s}$

$25\frac{m}{s}$

$40\frac{m}{s}$

Correct answer:

$35\frac{m}{s}$

Explanation:

The acceleration of an object due to gravity is $9.8\frac{m}{s^{2}}$ . If you multiply this value by one second, you can see that an object increases by a speed of $9.8\frac{m}{s}$ per second. After one second the ball is traveling $9.8\frac{m}{s}$ faster.

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Example Question #61 : Motion In One Dimension

A car traveling at $20 \frac{m}{s}$ sees a red light ahead and begins to slow down. The gas pedal is not touched until after the car has left the intersection. It takes to reach the light, which turns green again before the car comes to a complete stop. At this point the car rolls through the intersection at $5 \frac{m}{s}$ . What was the car's acceleration during this time?

Possible Answers:

$-.33 \frac{m}{s^{2}}$

$3 \frac{m}{s^{2}}$

$.33 \frac{m}{s^{2}}$

$-3 \frac{m}{s^{2}}$

Correct answer:

$-3 \frac{m}{s^{2}}$

Explanation:

The equation for acceleration is $\frac{\Delta v}{\Delta t}$ . The final velocity minus the initial velocity is $-15 \frac{m}{s}$ and the change in time is . When dividing the change in velocity by the change in time you receive the answer $-3 \frac{m}{s^{2}}$ which means that the car is decelerating due to the negative sign.

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Example Question #61 : Motion In One Dimension

While riding a bike down a steep hill at $9\frac{m}{s}$ you skid to a stop in 3.2 seconds. How far did the bike slide? Assume uniform acceleration.

Possible Answers:

$\Delta x=14.4m$

$\Delta x=1.4m$

$\Delta x=12.0m$

$\Delta x=10.4m$

$\Delta x=4.4m$

Correct answer:

$\Delta x=14.4m$

Explanation:

$\Delta x= (\frac{V_{i}+V_{f}}{2})*t=\frac{9\frac{m}{s}}{2}*3.2s=\mathbf{14.4m}$

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Example Question #332 : Ap Physics 1

A motorcycle accelerated from rest to $12\frac{m}{s}$ over a distance of 50 meters. What is the rider's acceleration?

Possible Answers:

$a=2\frac{m}{s^2}$

$a=1.44\frac{m}{s^2}$

None of these.

$a=.744\frac{m}{s^2}$

$a=2.2\frac{m}{s^2}$

Correct answer:

$a=1.44\frac{m}{s^2}$

Explanation:

To find the acceleration, use the kinematic equation $v^2=V_{0}^2+2a(x-x_{0})$

$V_{f}=12\frac{m}{s}$

$V_{0}=0$

$(12\frac{m}{s})^2=2a(50m)$

$a=\frac{144\frac{m^2}{s^2}}{100m}=\boldsymbol{1.44\frac{m}{s^2}}$

*the rider's acceleration will match that of the motorcycle.

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Example Question #301 : Newtonian Mechanics

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the maximum height of the football.

Possible Answers:

18.25m

21.23m

None of these

23.98m

16.81m

Correct answer:

21.23m

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

The ball will be at it's maximum height when the component of it's velocity, and thus that direction of kinetic energy, is zero.

The velocity will be decreased by the negative work done by gravity.

.5mv_y_i^2+mgh=.5mv_i_f^2

Plugging in values:

.5*20.4^2+(-9.8)h=.5*0^2

Solving for

h=21.23meters

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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 #52 : Motion In One Dimension

Example Question #53 : Motion In One Dimension

Example Question #57 : Linear Motion And Momentum

Example Question #51 : Linear Motion And Momentum

Example Question #59 : Linear Motion And Momentum

Example Question #51 : Linear Motion And Momentum

Example Question #61 : Motion In One Dimension

Example Question #61 : Motion In One Dimension

Example Question #332 : Ap Physics 1

Example Question #301 : Newtonian Mechanics

All AP Physics 1 Resources

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