AP Physics C

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AP Physics C: Electricity and Magnetism › AP Physics C

Questions 1 - 10

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

I and III

I and II

I, II, and III

I only

II only

Explanation

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:

$R_x = v_{ox}t$

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 $8:\uptext{s}$ ?

$2.75:\uptext{m}$

$1.10:\uptext{m}$

$0.85:\uptext{m}$

$2.20:\uptext{m}$

$1.65:\uptext{m}$

Explanation

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 $2.75:\uptext{m}$

Two solid cylinderical disks have equal radii. The first disk is spinning clockwise at $18\frac{rad}{s}$ and the second disk is spinning counterclockwise at $4\frac{rad}s{}$ . The second disk has a mass three times larger than the first. If both spinning disks are combined to form one disk, they end up rotating at the same angular velocity and same direction. Find this angular velocity after combination.

$1.5\frac{rad}{s}$

$4.5\frac{rad}{s}$

$2.0\frac{rad}{s}$

$8.3\frac{rad}{s}$

$10.5\frac{rad}{s}$

Explanation

For the first disk, we have the information below.

$\omega_1=18\ \frac{rad}{s}$

$m_1=\text{mass of 1st disk}$

$R_1=\text{radius of 1st disk}$

$I_1=\frac{1}{2}m_1R_1^2$

For the second disk, we have the information below.

$\omega_2=-4\ \frac{rad}{s}$

(The minus sign indicates counterclockwise while the positive indicates clockwise.)

$I_2=\frac{1}{2}m_2R_2^2=\frac{1}{2}(3m_1)R_1^2=\frac{3}{2}m_1R_1^2$

To do this problem, we use conservation of angular momentum. Before the disks are put in contact, the initial total angular momentum is given by the equation below.

$L_i=I_1\omega_1+I_2\omega_2=\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2$

It is just the sum of the angular momentums of each disk. When the disks are combined together, then final angular momentum can be found by the following equation.

$L_f=I_1\omega+I_2\omega=\frac{1}{2}m_1R_1^2\omega+\frac{3}{2}m_1R_1^2\omega=2m_1R_1^2\omega$

Set this equal to the initial angular momentum.

$\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2=2m_1R_1^2\omega$

Simplify and solve for $\omega$ .

$\omega=\frac{1}{4}\omega_1+\frac{3}{4}\omega_2$

$\omega=\frac{1}{4}(18\ \frac{rad}{s})+\frac{3}{4}(-4\ \frac{rad}{s})$

$\omega=1.5\ \frac{rad}{s}$

Starting from rest, a skateboarder travels down a 25o incline that's 22m long. Using conservation of energy, calculate the skateboarder's speed when he reaches the bottom. Ignore friction.

$13.5 \frac{m}{s}$

$21.8\frac{m}{s}$

$6.3\frac{m}{s}$

$2.9\frac{m}{s}$

$32.8\frac{m}{s}$

Explanation

Conservation of energy states that .

The skateboarder starts from rest; thus, and . At the bottom of the incline, $K_2=\frac{1}{2}mv^2$ and .

$mgh=\frac{1}{2}mv^2$

Solve for v.

$v=\sqrt{2gh}$

Using trigonometry, $h=22\sin(25)$ .

$v=\sqrt{2(9.8\ \frac{m}{s^2})(22)(\sin(25))}$

$v=13.5\ \frac{m}{s}$

An object is pushed across a rough surface with a force of 53N. The rough surface exerts a frictional force of 3.47N on the object. If the object is pushed 7.9m, how much work is done on the object?

Explanation

The defintion of work is:

$W=F_{net}\cdot d$

The net force on this object is:

$F_{net}=F_{app}-F_{friction}$

We can calculate this term using the given values:

$F_{net}=53N-3.47N=49.53N$

The distance is given. Substituting these values:

A 25kg child climbs up a tree. How much work is required for him to climb up this tree to a height of three meters?

Explanation

The forces acting on the child are the force of gravity and the upward force provided by the child himself. Choosing the upward direction as positive, Newton's second law applied to the child gives the following equation.

$\Sigma F_y=F_c-mg=0$

$F_c=mg=(25\ \text{kg})(9.8\ \frac{m}{s^2})=245\ \text{N}$

To calculate the work done by the child to bring himself three meters up the tree, we use the work equation below.

$W=F_c\cdot d=(245\ \text{N})(3\ \text{m})=735\ \text{J}$

Which of the following could be used as units of impulse?

$\frac{kg*m}{s}$

$\frac{kg*m}{s^2}$

$\frac{kg*m^2}{s}$

$\frac{kg*m^2}{s^2}$

$\frac{kg*m}{s^3}$

Explanation

Relevant equations:

$J = \Delta p = m\Delta v$

$J = F \Delta t$

Impulse is defined as change in momentum, so has the same units as momentum. These units can easily be found using the given equations.

$J=m\Delta v=(kg)(\frac{m}{s})=\frac{kg*m}{s}$

$J=F\Delta t=(N)(s)=N*s$

$2.75:\uptext{m}$

$1.10:\uptext{m}$

$0.85:\uptext{m}$

$2.20:\uptext{m}$

$1.65:\uptext{m}$

Explanation

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

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

, 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 $2.75:\uptext{m}$

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

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

I and III

I and II

I, II, and III

I only

II only

Explanation

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.

$R_x = v_{ox}t$

A 25kg child climbs up a tree. How much work is required for him to climb up this tree to a height of three meters?