Motion - AP Physics C Electricity & Magnetism

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

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Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

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Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

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Question

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?

Answer

The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.

$F_c=ma=\frac{mv^2}{r}$

Remember that centripetal acceleration is given by the following equation.

$a=\frac{v^2}{r}$

Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.

$T=\frac{mv^2}{r}$

Since we're trying to find the speed of the ball, we solve for v.

$v=\sqrt{\frac{Tr}{m}}$

We know the following information from the question.

We can use this information in our equation to solve for the speed of the ball.

$v=\sqrt{\frac{(87N)(1.8m)}{1.6kg}}}$

$v=9.9\frac{m}{s}$

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Question

In uniform circular motion, the net force is always directed ___________.

Answer

The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.

In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.

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Question

A car moves around a circular path of radius 100m at a velocity of $20\frac{m}{s}$ . What is the coefficient of friction between the car and the road?

Answer

The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.

$F_c=m\frac{v^2}{r}$

$F_f=\mu F_N=\mu mg$

$F_c=F_c\rightarrow m\frac{v^2}{r}=mg\mu$

We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:

$\frac{v^2}{r}=g\mu$

$\mu=\frac{v^2}{rg}$

We can use our given values to solve:

$\mu=\frac{(20\frac{m}{s})^2}{(100m)(9.8\frac{m}{s^2})}=0.4$

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Question

An object of mass 10kg undergoes uniform circular motion with a constant velocity of $10\frac{m}{s}$ at a radius of 10m. How long does it take for the object to make one full revolution?

Answer

The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:

$t=\frac{d}{v}=\frac{2\pi r}{v}$

Use the given radius and velocity to solve for the time per revolution:

$t=\frac{2\pi(10m)}{10\frac{m}{s}}=2\pi\approx6.3s$

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Question

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?

Answer

The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.

$F_c=ma=\frac{mv^2}{r}$

Remember that centripetal acceleration is given by the following equation.

$a=\frac{v^2}{r}$

Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.

$T=\frac{mv^2}{r}$

Since we're trying to find the speed of the ball, we solve for v.

$v=\sqrt{\frac{Tr}{m}}$

We know the following information from the question.

We can use this information in our equation to solve for the speed of the ball.

$v=\sqrt{\frac{(87N)(1.8m)}{1.6kg}}}$

$v=9.9\frac{m}{s}$

Compare your answer with the correct one above

Question

In uniform circular motion, the net force is always directed ___________.

Answer

The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.

In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.

Compare your answer with the correct one above

Question

A car moves around a circular path of radius 100m at a velocity of $20\frac{m}{s}$ . What is the coefficient of friction between the car and the road?

Answer

The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.

$F_c=m\frac{v^2}{r}$

$F_f=\mu F_N=\mu mg$

$F_c=F_c\rightarrow m\frac{v^2}{r}=mg\mu$

We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:

$\frac{v^2}{r}=g\mu$

$\mu=\frac{v^2}{rg}$

We can use our given values to solve:

$\mu=\frac{(20\frac{m}{s})^2}{(100m)(9.8\frac{m}{s^2})}=0.4$

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Question

An object of mass 10kg undergoes uniform circular motion with a constant velocity of $10\frac{m}{s}$ at a radius of 10m. How long does it take for the object to make one full revolution?

Answer

The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:

$t=\frac{d}{v}=\frac{2\pi r}{v}$

Use the given radius and velocity to solve for the time per revolution:

$t=\frac{2\pi(10m)}{10\frac{m}{s}}=2\pi\approx6.3s$

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Question

A ball of mass is tied to a rope and moves along a horizontal circular path of radius as shown in the diagram (view from above). The maximum tension the rope can stand before breaking is given by $T_{max}$ . Which of the following represents the ball's linear velocity given that the rope does not break?

Answer

This is a centripetal force problem. In this case the tension on the rope is the centripetal force that keeps the ball moving on a circle.

$T=F_C=\frac{mv^2}{r}$

If we want for the rope not to break, then the tension should never exceed $T_{max}$ .

$T_{max}\geq\frac{mv^2}{r}$

Now we just solve for velocity:

$v\leq\sqrt{\frac{rT_{max}}{m}}}$

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Question

A ball of mass is tied to a rope and moves along a horizontal circular path of radius as shown in the diagram (view from above). The maximum tension the rope can stand before breaking is given by $T_{max}$ . Which of the following represents the ball's linear velocity given that the rope does not break?

Answer

This is a centripetal force problem. In this case the tension on the rope is the centripetal force that keeps the ball moving on a circle.

$T=F_C=\frac{mv^2}{r}$

If we want for the rope not to break, then the tension should never exceed $T_{max}$ .

$T_{max}\geq\frac{mv^2}{r}$

Now we just solve for velocity:

$v\leq\sqrt{\frac{rT_{max}}{m}}}$

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