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AP Physics 1 : Motion in Two Dimensions

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Example Question #411 : Newtonian Mechanics

Find the vector going from point A to point B, shown on the graph below.

Vectors 1

Possible Answers:

<5,2>

<2,3>

< 2,5>

<0,5>

<-2,5>

Correct answer:

< 2,5>

Explanation:

The correct answer is <2,5> . This vector indicates that in order to get to point B from point A you must be 2 units to the right - the first number in vector notation indication horizontal movement or movement in the x-direction - and 5 units upward - the second number in vector notation indicating vertical movement or movement in the y-direction.

Vectors 2

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Example Question #460 : Ap Physics 1

A remote control car slides off a curb .2 meters high and lands 1 meter from the base of the curb. Select the initial horizontal velocity of the car.

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

Possible Answers:

$14 \frac{m}{s}$

$5 \frac{m}{s}$

$8\frac{m}{s}$

$10\frac{m}{s}$

$4\frac{m}{s}$

Correct answer:

$5 \frac{m}{s}$

Explanation:

To solve this problem you must consider (separately) the x and y components of an object that is both moving horizontally and vertically.

For the vertical component we find the time that the car was in the air:

$\Delta y=V_{0y}t+.5a_{y}t^2$

Keep in mind the acceleration vertically is just gravity.

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

*Make gravity negative because it pulls downward and the height negative because the object is falling. You could also just remember that time can't be negative and ignore the signs.

Solve for t:

$\frac{-.2m}{-5\frac{m}{s^2}}=t^2$

t=.2s

For the horizontal component we use:

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

Acceleration in the horizontal is 0.

$1m=V_{0x}(.2s)$

$\mathbf{V_{0x}=5\frac{m}{s}}$

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Example Question #421 : Newtonian Mechanics

Object X has initial velocity $<5,5>\frac{m}{s}$ and mass 50kg . Object Y has initial velocity $<-5,5>\frac{m}{s}$ and mass 40kg . These objects collide in deep space. Determine their final velocity if they stick together after the collision.

Possible Answers:

$<0,4.44>\frac{m}{s}$

None of these

$<5.56,.444>\frac{m}{s}$

$<.556,5>\frac{m}{s}$

$<5,10>\frac{m}{s}$

Correct answer:

$<.556,5>\frac{m}{s}$

Explanation:

$\overrightarrow{P}_{Total}=\overrightarrow{P}_1+\overrightarrow{P}_2...$

Definition of momentum:

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

Combine equations:

$\overrightarrow{P}_{Total}=m_1\overrightarrow{v}_1+m_2\overrightarrow{v}_2$

Plug in values:

$\overrightarrow{P}_{Total}=50*<5,5>+40*<-5,5>$

$\overrightarrow{P}_{Total}=<250,250>+<-200,200>$

$\frac{\overrightarrow{P}_{Total}}{m}=\frac{<50,450>kg\frac{m}{s}}{90kg}$

$\overrightarrow{v}=<.556,5>\frac{m}{s}$

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AP Physics 1 : Motion in Two Dimensions

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #411 : Newtonian Mechanics

Example Question #460 : Ap Physics 1

Example Question #421 : Newtonian Mechanics

All AP Physics 1 Resources

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