AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #111 : Fundamentals Of Force And Newton's Laws

What units make up a Newton?

Possible Answers:

\(\displaystyle kg\cdot m^{2}/s^{2}\)

\(\displaystyle kg/m^{2}\cdot s\)

\(\displaystyle kg/m^{^{2}}\cdot s^{2}\)

\(\displaystyle kg/m\cdot s^{2}\)

\(\displaystyle kg\cdot m/s^{2}\)

Correct answer:

\(\displaystyle kg\cdot m/s^{2}\)

Explanation:

The answer is exemplified by the centripetal force equation: 

\(\displaystyle Force_{c} = mass\cdot (velocity_{a}^{2}/radius)\).

When we combine the units of each of the equation's components, we arrive at

\(\displaystyle kg \cdot (m/s)^{2}/m\) which, when simplified, becomes \(\displaystyle kg\cdot m/s^{2}\)

Example Question #1 : Other Force Fundamentals

Two toy cars are moving on a frictionless track. Car 1 is moving at \(\displaystyle 3 \frac{m}{s}\) when it collides into Car 2, which was at rest. After the collision, Car 2 moves at \(\displaystyle 3.4\frac{m}{s}\) and Car 1 moves at \(\displaystyle 0.6\frac{m}{s}\). If Car 1 weighs 0.05 kg, how much does Car 2 weigh?

Possible Answers:

\(\displaystyle 0.035kg\)

\(\displaystyle 0.018kg\)

\(\displaystyle 0.074kg\)

\(\displaystyle 0.214kg\)

\(\displaystyle 0.126kg\)

Correct answer:

\(\displaystyle 0.035kg\)

Explanation:

First, lets identify the given information:

\(\displaystyle v_{1i}=3m/s\)

\(\displaystyle v_{1f}=0.6m/s\)

\(\displaystyle v_{2f}=3.4m/s\)

\(\displaystyle m_{1}=0.05kg\)

Since this is a collision problem, we can apply the law of conservation of momentum:

\(\displaystyle p_{i}=p_{f}\)

\(\displaystyle p_{1i}+p_{2i}=p_{1f}+p_{2f}\)

\(\displaystyle m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f}\)

Because Car 2 is initially at rest, this equation can be simplified:

 \(\displaystyle m_{1}v_{1i}=m_{1}v_{1f}+m_{2}v_{2f}\)

The equation can then be rearranged to solve for the mass of Car 2:

\(\displaystyle m_{2}v_{2f}=m_{1}v_{1i}-m_{1}v_{1f}\)

\(\displaystyle m_{2}=\frac{m_{1}(v_{1i}-v_{1f})}{v_{2f}}\)

 By plugging in the known values, we can determine the mass of Car 2:

\(\displaystyle m_{2}=\frac{(0.05kg)(3\frac{m}{s}-0.6\frac{m}{s})}{3.4{\frac{m}{s}}}=0.035kg\)

Example Question #1 : Other Force Fundamentals

Consider a cube with a mass of \(\displaystyle 500 kg\) and sides of \(\displaystyle 50 cm\). Calculate the pressure, \(\displaystyle P\), on the floor from such a cube.

Possible Answers:

\(\displaystyle 9810 Pa\)

\(\displaystyle 1000 Pa\)

\(\displaystyle 2000 Pa\)

\(\displaystyle 1.96 Pa\)

\(\displaystyle 19620 Pa\)

Correct answer:

\(\displaystyle 19620 Pa\)

Explanation:

To calculate the pressure \(\displaystyle P\) on the floor, we need to know the force \(\displaystyle \vec{F}\) acting on the floor, as well as the area \(\displaystyle A\) over which it acts.

The force acting is simply the weight of the cube, 

\(\displaystyle F=W=mg=4905 N\)

The area is the area of one side of the cube as it rests on the floor, 

\(\displaystyle A=\left(0.5m \right )^2=0.25 m^2\) 

Therefore, the pressure \(\displaystyle P\) is given as:

\(\displaystyle P=\frac{F}{A}\)

Plug in known values and solve.

\(\displaystyle P=\frac{4905N}{0.25m^2}=19620\frac{N}{m^2}=19620Pa\)

Example Question #901 : Newtonian Mechanics

During a mission to the Moon, astronauts were traveling towards the moon at \(\displaystyle 7 \frac{miles}{second}\), however, for most of the flight to and from the Moon, they were not slammed into their seats, and could move freely about the aircraft, why is that?

Possible Answers:

Because they were not accelerating

Because there is no gravity in space

Because they were being pulled towards the Moon

None of these

The astronauts had very strong muscles and could push themselves up

Correct answer:

Because they were not accelerating

Explanation:

Once the spacecraft had reached a constant speed, so had it's occupants. Thus, the astronauts were also moving at \(\displaystyle 7\frac{miles}{second}\) and they were experiencing no forces from their seats.

\(\displaystyle F=ma\)

\(\displaystyle F=m*0\)

\(\displaystyle F=0\)

Thus, the astronauts were experiencing zero net force.

Example Question #5 : Other Force Fundamentals

Three forces act on an object that is in equilibrium and suspended in air. \(\displaystyle F_1\) is \(\displaystyle 30N\) pulling on the object \(\displaystyle 45^\circ\) from the positive horizontal axis. \(\displaystyle F_2\) is \(\displaystyle 30N\) pulling on the object \(\displaystyle 45^\circ\) from the negative horizontal axis. \(\displaystyle F_3\) is acting on the object straight down on the negative vertical axis. What is the magnitude of \(\displaystyle F_3\)

Possible Answers:

\(\displaystyle 30N\)

\(\displaystyle 42.4N\)

\(\displaystyle 0N\)

\(\displaystyle 21.2N\)

Correct answer:

\(\displaystyle 42.4N\)

Explanation:

\(\displaystyle F_1\) and \(\displaystyle F_2\) cancel each other out in the horizontal direction due to the fact that they are equal and opposite about the vertical axis. Now that the resultant force in the x-direction is zero, we must find what \(\displaystyle F_3\) must be in order to zero out the other two forces' y-components, which are:

\(\displaystyle F_{1y}=F_{2y}=30sin(45)=21.2N\)

These forces are pointing in the positive y-direction (pulling the object up). Therefore, in order to have a resultant force of zero in the vertical direction, there must be a downward force of \(\displaystyle F_3=21.2N+21.2N=42.4N\)

Since we are asked for the magnitude, we are not particularly concerned with the sign of this value.

Example Question #332 : Forces

A cowboy is lassoing a cow at the rodeo. Identify the force that is causing the object to travel in a circular path.

Possible Answers:

Normal

Electrical

Tension

Gravity

Spring

Correct answer:

Tension

Explanation:

In these conceptual problems, it is always best to draw an accurate force diagram. In string problems, there is always a tension force, and here, it is causing the circular motion. The tension force causes the lasso to swing in a circle and pulls it toward the center of the circle. 

Example Question #4 : Other Force Fundamentals

Which of the following is not one of Newton's fundamental laws?

Possible Answers:

All of these correctly represent Newton's laws

The acceleration experienced by an object is directly proportional to the force applied to that object, and inversely proportional to the mass of that object

An object in motion will remain in that state of motion unless an outside force acts on that object

In order for an object to be in motion, a force needs to act on it

For every action, there will always be an opposite reaction equal in magnitude

Correct answer:

In order for an object to be in motion, a force needs to act on it

Explanation:

To answer this question, we'll need to be familiar with Isaac Newton's fundamental laws of physics, which forms the basis of classical mechanics.

Newton's first law states that an object in motion will remain in that state of motion indefinitely, unless an outside force acts on this object.

Newton's second law can be stated as follows:

\(\displaystyle F=ma\)

This expression tells us that the acceleration experienced by an object is directly proportional to the applied force and inversely proportional to the object's mass.

Newton's third law states that for every action, there is always an equal and opposite reaction. Another way of saying this is that if one object were to exert a force on a second object, then the second object would exert the same force against the first object.

One of the answer choices states the following

"In order for an object to be in motion, a force needs to act on it."

While this may seem intuitive based on our everyday experience, and it may also show some resemblance to Newton's first law, this statement is actually incorrect. Force is not responsible for motion - force is responsible for changes in motion.

Example Question #4 : Other Force Fundamentals

You and your friend are pushing on a box with a force parallel to the ground of 40 N. The box weighs 5 kg, and the coefficient of kinetic friction between the box and the ground is 0.75. What is the net force on the box in the horizontal direction?

\(\displaystyle g=9.81\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle +13.2N\)

\(\displaystyle +3.2N\)

\(\displaystyle +4.7\)

\(\displaystyle +8.3N\)

\(\displaystyle +1.6N\)

Correct answer:

\(\displaystyle +3.2N\)

Explanation:

The force due to kinetic friction is calculated by the equation 

\(\displaystyle F_{k}=\mu _{k}F_{N}\)

where \(\displaystyle F_{N}\) is the normal force on the box from the ground. The frictional force is in the opposite direction of the force applied by you and your friend. Therefore, the net force can be calculated:

\(\displaystyle F_{net}= F_{app}-F_{k}\)

\(\displaystyle F_{net}=F_{app}-\mu _{k}F_{N}\)

\(\displaystyle F_{net}=F_{app}-\mu _{k}mg\)

\(\displaystyle F_{net}=40N-(0.75)(5kg)(9.81\frac{m}{s^2})=+3.2N\)

Example Question #1 : Springs

A block with a mass of 1kg is hanging vertically from a spring that has a constant of \(\displaystyle 40\frac{N}{m}\). If the block is pulled downward 1.5m past equilibrium and released, what is the velocity of the block as it passes back through the equilibrium position?

\(\displaystyle g=10\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 8.4\frac{m}{s}\)

\(\displaystyle 12.1\frac{m}{s}\)

\(\displaystyle 5.8\frac{m}{s}\)

\(\displaystyle 7.7 \frac{m}{s}\)

\(\displaystyle 6.3\frac{m}{s}\)

Correct answer:

\(\displaystyle 7.7 \frac{m}{s}\)

Explanation:

This problem deals with the conservation of energy in the form of a spring. The formula for the conservation of energy is:

\(\displaystyle E = U_i +K_i = U_f +K_f\)

We can eliminate initial kinetic energy to get:

\(\displaystyle U_i = U_f +K_f\)

We have two forms of potential energy in this problem: gravitational and spring. Therefore, we can rewrite the expression to be:

\(\displaystyle U_{is} = K_f + U_{fg}\)

There is no initial gravitational potential energy and no final spring potential energy. Subsituting our expressions for potential and kinetic energy, we get:

\(\displaystyle \frac{1}{2}kx^2 = \frac{1}{2}mv_f^2 + mgh\)

Rearranging for final velocity, we get:

\(\displaystyle v_f = \sqrt{\frac{kx^2}{m}-2gh}\)

\(\displaystyle x\) is the original displacement (the same as \(\displaystyle h\)).

We have all of the values, so we can solve:

\(\displaystyle v_f = \sqrt{\frac{40\cdot(1.5)^2}{1} - (2\cdot 10\cdot 1.5)}\)

\(\displaystyle v_f = \sqrt{60} = 7.7 \frac{m}{s}\) 

Example Question #902 : Newtonian Mechanics

A spring with a constant of \(\displaystyle 3\frac{N}{m}\) is hanging from the ceiling of an elevator. A block of mass 1kg is attached to the end of the spring. If the elevator begins to accelerate at a rate of \(\displaystyle 2\frac{m}{s^2}\), how far will the equilibrium position of the spring shift?

Possible Answers:

\(\displaystyle 1.5m\)

\(\displaystyle 2m\)

\(\displaystyle 0m\)

\(\displaystyle 0.37m\)

\(\displaystyle 0.67m\)

Correct answer:

\(\displaystyle 0.67m\)

Explanation:

Before the elevator begins to accelerate, the block has a weight of:

\(\displaystyle F = mg = (1kg)(10\frac{m}{s^2}) = 10N\)

While the elevator is accelerating, the block has a weight of:

\(\displaystyle F = m(g+a) = (1kg)(10\frac{m}{s^2}+2\frac{m}{s^2}) = 12 N\)

Now we can say that the weight of the block has effectively increased by 2N.

We can use this increase in force to find out how much the equilibrium position will shift:

\(\displaystyle F= kx\)

Rearranging for x (displacement), we get:

\(\displaystyle \Delta x = \frac{\Delta F}{k} = \frac{2N}{3\frac{N}{m}} = 0.67 m\)

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