AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #2 : Centripetal Force And Acceleration

We can determine the centripetal force exerted by the nucleus on an electron. The diameter of an atom is , and an electron is moving at in a circular motion around the nucleus. The mass of an electron is roughly 

What is the centripetal force exerted by the nucleus on an electron? 

Possible Answers:

Correct answer:

Explanation:

The formula for centripetal force is:

Where  is the mass of the object,  is its velocity, and  is the radius of the circle made by the motion of the object around the center. We can model an atom like a circle, with its nucleus being its center. Since we are given the diameter of an atom, its radius will be

 

Example Question #1 : Centripetal Force And Acceleration

Imagine a car driving over a hill at a constant speed. Once the car has reached the apex of the hill, what is the direction of the acceleration?

Possible Answers:

Upward

Opposite direction of motion

Downwards

In direction of motion

No acceleration

Correct answer:

Downwards

Explanation:

If we imagine the hill as a semi-circle, it appears that the car is moving along a circle. At the apex of the hill, the car's acceleration points downwards as this points towards the center of the circle. If an object travels in a circular fashion, at a constant speed, the direction of acceleration is always towards the center of the circle. This type of acceleration arises do to the change in velocity. Although the speed is constant, the direction changes. 

Example Question #1041 : Ap Physics 1

The passengers in a car driving over a hill at a speed of   feel weightless.

 

Calculate the radius of curvature of the hill.

Possible Answers:

Correct answer:

Explanation:

We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.

The weight is positive, because it points towards the center of the circle, while the normal force  is negative because it points away from the circle.

Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.

 

Example Question #1041 : Ap Physics 1

Imagine a table that has a hole drilled through the center. Through this hole is a string that connects one mass that hangs under the table, to a mass that is on top of the table and spinning. mass M weight 10 kg, mass m weighs 5 kg and the radius, R, of the circle mass M traces out is 50 cm. At what speed, v, does the mass M need to rotate at in order to keep mass m suspended?

 

Possible Answers:

Correct answer:

Explanation:

For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration. 

I have stated previously that this force must be equal to the gravititational force acting on mass m. So,

Solving for v,

Plugging in values yields a speed of 1.6 m/s

Example Question #11 : Centripetal Force And Acceleration

 object moves in a circular motion with a diameter of . What is the magnitude of this object's velocity?

Possible Answers:

Correct answer:

Explanation:

Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.

In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:

The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is , we can find the radius by taking half of this value, which is . Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.

Solving for velocity, we can plug in values to obtain:

Example Question #1041 : Ap Physics 1

At the 2015 NCAA Track Championships, a competitor in the hammer is in the midst of his throw. During a period of his pre-throw spinning, his hammer is moving at a constant angular velocity of  in its circular orbit. Being a regulation hammer, it weighs , and the length of its rope is . What is the magnitude of the force exerted towards the competitor along the hammer's rope?

Possible Answers:

Correct answer:

Explanation:

Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as . By plugging in the numbers given by the problem we get  which, when simplified gives us approximately  or 

Example Question #51 : Circular And Rotational Motion

A  ball is being swung faster and faster in a horizontal circle. When the speed reaches , the tension () is  and the string breaks. What is the radius  of the circle?

Possible Answers:

Correct answer:

Explanation:

In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:

 

Example Question #11 : Centripetal Force And Acceleration

You are riding a rollercoaster going around a vertical loop, on the inside of the loop. If the loop has a radius of , how fast must the cart be moving in order for you to feel three times as heavy at the top of the loop?

Possible Answers:

Correct answer:

Explanation:

To solve this problem, it is important to remember the signs of each force acting on the coaster.  Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.

Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster.  We want the speed at which a rider would feel three times as heavy, or three times their weight.

Example Question #51 : Circular And Rotational Motion

A car's tire can rotate at a frequency of (revolutions per minute). Given that a typical tire radius is , what is the centripetal acceleration of the tire? 

Possible Answers:

Correct answer:

Explanation:

Centripetal acceleration  can be given as:

, where  is the linear velocity,  is the radius of the circle. 

To determine linear velocity, we need to determine how quickly the wheel spins.

Firstly, let's convert from rpm to Hertz. 

The wheel makes 50 full revolutions in one second. To determine linear velocity,

, where  is frequency, which we just determined and  is circumference of the tire. 

By the definition of the circumference of a circle,

Therefore, 

Example Question #51 : Circular And Rotational Motion

A ferris wheel ride of diameter 10m has a cart moving at . What is the centripetal acceleration of the cart?

Possible Answers:

Correct answer:

Explanation:

The formula for centripetal acceleration is given by:

 is the centripetal acceleration,  is the velocity and  is the radius of the circular path. We substitute in our known values and then solve for alpha which will give us the centripetal acceleration of the ferris wheel.

Be aware that the radius is half of the diameter, so  will be 5 meters.

This will give us a centripetal acceleration of 

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