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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #111 : Linear Motion And Momentum

A projectile from a hand gun is fired at $30^{\circ}$ above the horizontal at a speed of $200\frac{m}{s}$ . How many seconds is the projectile airborne? Assume $g=-10\frac{m}{s^2}$ and that the hand gun was fired a negligible distance from the ground.

Possible Answers:

t=20s

t=40s

None of these

t=25s

t=50s

Correct answer:

t=20s

Explanation:

Since the question is asking how long the projectile was airborne, we are dealing with motion in y-direction only. Find initial velocity in the y-direction and substitute into the kinematic equation:

$V=V_{0}+at$

$V_{0y}=V_{0}sin\theta=200*sin30^\circ=100\frac{m}{s}$

$-100\frac{m}{s}=100\frac{m}{s}-10\frac{m}{s^2}t$

$-200\frac{m}{s}=-10\frac{m}{s^2}t$

t=20s

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Example Question #12 : Motion In Two Dimensions

A medieval cannon fires its projectile horizontally from a castle wall 70 meters high at a speed of $80\frac{m}{s}$ . How long will the cannonball be airborne?

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

Possible Answers:

t=7.3s

t=4.4s

t=3.7s

t=4.2s

t=2.4s

Correct answer:

t=3.7s

Explanation:

Narrow the choices of kinematic equations by including only ones with distance in their formulas. Of those two, we want the one that includes time:

$\Delta x={\color{Red} V_{0}t}+.5at^2$

The red term becomes zero because our projectile is not moving at first. Plug in known values and solve.

$\70=.5(10\frac{m}{s^2})t^2$

$140=(10\frac{m}{s^2})t^2$

14=t^2

{t=3.7s}

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Example Question #21 : Motion In Two Dimensions

An archer fires an arrow horizontally off a 60 meter high castle wall at a speed of $200\frac{m}{s}$ . What is the closest you could stand to the castle wall without fear of being hit?

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

Possible Answers:

823m

540m

1000m

575m

693m

Correct answer:

693m

Explanation:

Find the time the object is airborne by solving for t in the equation:

$\Delta x=V_{0}t+.5at^2$

Then, multiply the time in the air by the initial velocity to get the range of the projectile. Note that the initial velocity is zero in the y-direction.

$t=\sqrt{\frac{2 \Delta y}{a}}=3.46s$

$Range= V_{0}*t=200\frac{m}{s}*3.46s=693m$

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Example Question #351 : Newtonian Mechanics

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:

0.83 m

2.83 m

0.42 m

.49 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 #21 : Motion In Two Dimensions

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:

$5\frac{mm}{s}$

$8.60\frac{mm}{s}$

$7\frac{mm}{s}$

$2\frac{mm}{s}$

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:

$25\frac{m}{s}$

$15\frac{m}{s}$

$10\frac{m}{s}$

$29.15\frac{m}{s}$

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 #352 : Newtonian Mechanics

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

Possible Answers:

11m

8.54m

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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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #111 : Linear Motion And Momentum

Example Question #12 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #351 : Newtonian Mechanics

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 #352 : Newtonian Mechanics

Example Question #22 : Motion In Two Dimensions

All AP Physics 1 Resources

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