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AP Physics 1 : Newtonian Mechanics

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Example Questions

Example Question #21 : Motion In Two Dimensions

A 100 N force pushes a 15 kg block, from rest, 60 cm across a frictionless table that is 85 cm above the floor. If the force stops acting on the block when it leaves the table, calculate how far $\left(\Delta x \right )$ the block lands from the table.

Possible Answers:

.49 m

0.42 m

2.83 m

0.83 m

1.19 m

Correct answer:

1.19 m

Explanation:

This is a two-part problem. First, we need to know how fast the block is moving in the horizontal direction when it leaves the table. This can be calculated by using a kinematic equation, combined with Newton's second law.

$V_f=\sqrt{V_i^2+2a\Delta x}$

$V_f=\sqrt{0+2\left(\frac{F}{m} \right ) \Delta x}$

$V_f=\sqrt{\frac{2(100N)(.60m)}{15kg}}=2.83\frac{m}{s}$

Now that we know how fast the block is moving horizontally, we need to know how long the block will be in air before hitting the ground. This is found by considering the equations of motion for an object falling from rest:

$0=-\frac{1}{2}gt^2+h \Rightarrow t=\sqrt{\frac{2h}{g}}$

$t=\sqrt{\frac{2(.85m)}{9.81\frac{m}{s^2}}}=0.42s$

Lastly, we find the horizontal distance by:

$\Delta x = vt = 2.83 \frac{m}{s}.0.42s=1.19 m$

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Example Question #21 : Motion In Two Dimensions

A ball is thrown at a velocity $\vec{v}=\left (10\hat{i}+6\hat{j} \right )\frac{m}{s}$ . What is the speed of the ball?

Possible Answers:

$10\frac{m}{s}$

$4\frac{m}{s}$

$11.66\frac{m}{s}$

$6\frac{m}{s}$

Correct answer:

$11.66\frac{m}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector $\vec{v}=\left (10\hat{i}+6\hat{j} \right )\frac{m}{s}$ , the speed of the ball is:

$\left |\vec{v} \right |=\sqrt{10^{2}+6^{2}}=\sqrt{136}\approx11.66\frac{m}{s}$

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Example Question #111 : Linear Motion And Momentum

An ant is digging at a velocity $\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$ into the Earth. What is the speed of the ant?

Possible Answers:

$7\frac{mm}{s}$

$5\frac{mm}{s}$

$8.60\frac{mm}{s}$

$2\frac{mm}{s}$

Correct answer:

$8.60\frac{mm}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector:

$\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$

The speed of the ant is:

$\left |\vec{v} \right |=\sqrt{7^{2}+(-5)^{2}}=\sqrt{49+25}=\sqrt{74}\approx8.60\frac{mm}{s}$

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Example Question #111 : Linear Motion And Momentum

An ant is digging at a velocity $\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$ in the Earth. In what direction is the ant digging?

Possible Answers:

$59^{\circ}$ right of vertical

$59^{\circ}$ left of vertical

$54.5^{\circ}$ below the horizon

$54.5^{\circ}$ above the horizon

Correct answer:

$54.5^{\circ}$ below the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{-7}{5}\right )\approx-54.5^{\circ}$

The ant is digging at an angle $54.5^{\circ}$ below the horizon.

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Example Question #111 : Linear Motion And Momentum

A rollercoaster car is sliding down an incline at a velocity $\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$ . What is the speed of the rollercoaster car?

Possible Answers:

$15\frac{m}{s}$

$25\frac{m}{s}$

$29.15\frac{m}{s}$

$10\frac{m}{s}$

Correct answer:

$29.15\frac{m}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector:

$\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$

The speed of the rollercoaster car is:

$\left |\vec{v} \right |=\sqrt{(-15)^{2}+(-25)^{2}}=\sqrt{225+625}=\sqrt{850}\approx29.15\frac{m}{s}$

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Example Question #21 : Motion In Two Dimensions

A car travels east, south, then west. What is the magnitude of the displacement of the car?

Possible Answers:

8.54m

11m

Correct answer:

8.54m

Explanation:

To find the magnitude of the displacement of the car, we must represent each velocity of the car as a vector. We will then add the vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

The car first travels east , represented as a vector: $\vec{r}=\left (7\hat{i} \right )m$

The car then travels south , represented as a vector: $\vec{r}=\left (-8\hat{j} \right )m$

The car finally travels west , represented as a vector: $\vec{r}=\left (-4\hat{i} \right )m$

Adding the vectors together gives us the displacement of the car:

$\vec{r}=\left (7\hat{i}-8\hat{j}-4\hat{i} \right )m=\left (3\hat{i}-8\hat{j} \right )m$

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , its magnitude, $\left | \vec{r} \right |$ , is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$

Given the displacement vector: $\vec{r}=\left (3\hat{i}-8\hat{j} \right )m$ , the magnitude of the displacement vector is:

$\left |\vec{r} \right |=\sqrt{3^{2}+(-8)^{2}}=\sqrt{9+64}=\sqrt{73}\approx8.54m$

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Example Question #22 : Motion In Two Dimensions

A car travels east, south, then west. What is the direction of the displacement of the car?

Possible Answers:

$20.6^{\circ}$ below horizontal

$20.6^{\circ}$ above horizontal

$20.6^{\circ}$ left of vertical

$20.6^{\circ}$ right of vertical

Correct answer:

$20.6^{\circ}$ below horizontal

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The angle of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem, the displacement vector is:

$\vec{r}=\left (3\hat{i}-8\hat{j} \right )m$

The angle of displacement is:

$\theta=tan^{-1}\left ( \frac{3}{-8}\right )\approx-20.6^{\circ}$

The car is traveling $20.6^{\circ}$ below horizontal.

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Example Question #22 : Motion In Two Dimensions

A boat is crossing a river is traveling east at $7\frac{m}{s}$ . The current of the river is traveling north at $3\frac{m}{s}$ . At what angle is the boat traveling?

Possible Answers:

$82.4^{\circ}$ below horizontal

$23.2^{\circ}$ above horizontal

$82.4^{\circ}$ above horizontal

$23.2^{\circ}$ below horizontal

Correct answer:

$23.2^{\circ}$ above horizontal

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The angle of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem, the velocity vector is: $\vec{v}=\left (7\hat{i}+3\hat{j} \right )\frac{m}{s}$

The angle of the boat is:

$\theta=tan^{-1}\left ( \frac{3}{7}\right )\approx23.2^{\circ}$

The boat is traveling $23.2^{\circ}$ above horizontal.

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Example Question #21 : Motion In Two Dimensions

A ball is thrown at a velocity $\vec{v}=\left (10\hat{i}+6\hat{j} \right)\frac{m}{s}$ . What is the direction of the ball (the angle at which the ball is thrown)?

Possible Answers:

$59^{\circ}$ above the horizon

$9^{\circ}$ right of vertical

$59^{\circ}$ below the horizon

$9^{\circ}$ left of vertical

Correct answer:

$59^{\circ}$ above the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{10}{6}\right )\approx59^{\circ}$

The ball is thrown $59^{\circ}$ above the horizon.

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Example Question #121 : Linear Motion And Momentum

A rollercoaster car is traveling along an incline at a velocity $\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$ . What is the direction of the rollercoaster car?

Possible Answers:

To the right at an angle $54.5^{\circ}$ above the horizon

To the right at an angle $54.5^{\circ}$ below the horizon

To the left at an angle $54.5^{\circ}$ above the horizon

To the left at an angle $54.5^{\circ}$ below the horizon

Correct answer:

To the right at an angle $54.5^{\circ}$ below the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{-25}{-15}\right )\approx59^{\circ}$

The rollercoaster car is traveling backwards at an angle $54.5^{\circ}$ below the horizon.

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AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #111 : Linear Motion And Momentum

Example Question #111 : Linear Motion And Momentum

Example Question #111 : Linear Motion And Momentum

Example Question #21 : Motion In Two Dimensions

Example Question #22 : Motion In Two Dimensions

Example Question #22 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #121 : Linear Motion And Momentum

All AP Physics 1 Resources

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