AP Physics C: Mechanics : Linear Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

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Example Questions

Example Question #21 : Motion

A guillotine blade weighing \(\displaystyle 400N\) is accelerated upward into position at a rate of \(\displaystyle 2.0\frac{m}{s^2}\) . 

What is the the approximate mass of the guillotine blade?

Possible Answers:

\(\displaystyle 3920kg\)

\(\displaystyle 800kg\)

\(\displaystyle 200kg\)

\(\displaystyle 41kg\)

\(\displaystyle 400kg\)

Correct answer:

\(\displaystyle 41kg\)

Explanation:

The force of gravity on the blade is \(\displaystyle 400N\), which is the same as \(\displaystyle 400\frac{kg\cdot m}{s^2}\)

This unit relationship comes from Newton's second law.

\(\displaystyle F=ma\) is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.

Solve the expression by plugging in known values.

\(\displaystyle 400 \frac{kg\cdot m}{s^2}=m \cdot 9.8\frac{m}{s^2}\)

\(\displaystyle \frac{400}{9.8}kg=mass\approx41kg\)

Example Question #1 : Interpreting Linear Motion Diagrams

A guillotine blade weighing \(\displaystyle 400N\) is accelerated upward into position at a rate of \(\displaystyle 2.0\frac{m}{s^2}\) . 

What is the tension on the rope pulling the blade, while it is accelerating into position?

Possible Answers:

\(\displaystyle 482N\)

\(\displaystyle 377N\)

\(\displaystyle 82N\)

\(\displaystyle 318N\)

\(\displaystyle 400N\)

Correct answer:

\(\displaystyle 482N\)

Explanation:

The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.

\(\displaystyle F_{T}-F_{g}=ma\)

Rearrange the equation. 

\(\displaystyle F_{T}=ma+F_{g}\)

Plug in known values.

\(\displaystyle F_T=41kg\cdot2\frac {m}{s^2} + 400N\)

\(\displaystyle F_T = 482N\)

Example Question #21 : Mechanics Exam

Two objects moving in one dimension created the following velocity vs. time graph:

Velocity time graph labeled

From the graph above, what is true about the two objects at time \(\displaystyle \small t=5s\)?

Possible Answers:

They have travelled the same distance from their starting positions

They are both at rest (not moving)

They are moving at the same speed

One object is passing the other

They are at the same position

Correct answer:

They are moving at the same speed

Explanation:

Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.

Example Question #21 : Linear Motion

Two objects moving in one dimension created the following velocity vs. time graph:

Velocity time graph labeled

From the graph above, which object has traveled a greater distance from its starting position when \(\displaystyle \small t=5s\)?

Possible Answers:

None of these

It cannot be determined from the graph

Object 1

Object 2

They have travelled an equal distance

Correct answer:

Object 1

Explanation:

Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.

Example Question #2 : Interpreting Linear Motion Diagrams

An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation \(\displaystyle \small y=ax^2+bx +c\). Its motion in the horizontal direction is described by the equation \(\displaystyle \small x=v_{x}t\). What is the object's velocity is the \(\displaystyle y\) direction in terms of its horizontal position \(\displaystyle \small x\)?

Possible Answers:

\(\displaystyle \small v_{y}=(2ax+b)v_{x}\)

\(\displaystyle \small a\)

\(\displaystyle \small b\)

\(\displaystyle \small (2av_{x}t+b)v_{x}\)

\(\displaystyle \small 2ax+b\)

 

 

 

Correct answer:

\(\displaystyle \small v_{y}=(2ax+b)v_{x}\)

Explanation:

The y velocity is the time derivative of the \(\displaystyle \small y\) position, and not the \(\displaystyle \small x\) derivative. In order to find it, use the chain rule:

\(\displaystyle \frac{dy}{dt}=\left(\frac{dy}{dx}\right) \frac{dx}{dt}\) 

\(\displaystyle \frac{dy}{dx}= 2ax + b\)

Of course, \(\displaystyle \frac{dx}{dt} =v_{x}\)

Example Question #1 : Interpreting Linear Motion Diagrams

Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are \(\displaystyle 3\: kg\) and \(\displaystyle 5\: kg\).

Assume the pulley has negligible mass and friction.

Img1

Possible Answers:

\(\displaystyle 1.96\:m/s^2\)

\(\displaystyle 2.45\:m/s^2\)

\(\displaystyle 9.8\:m/s^2\)

\(\displaystyle 2.88\:m/s^2\)

\(\displaystyle 1.4\:m/s^2\)

Correct answer:

\(\displaystyle 2.45\:m/s^2\)

Explanation:

Img2

From the force diagram above, we can see that tension \(\displaystyle (T)\) is pulling up on both sides of the string and gravity is pulling down on both blocks. With this information we can write 2 force equations:

\(\displaystyle 5kg: 5g-T=5a\)

\(\displaystyle 3kg: -3g+T =3a\)

If we add the two equations together, we get:

\(\displaystyle 2g = 8a\)

where \(\displaystyle g = 9.8\:m/s^2\)

Solving for \(\displaystyle a\), we get 

\(\displaystyle a = 2.45\:m/s^2\)

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