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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #71 : Newtonian Mechanics

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

The rod is initially at rest in its horizontal position. 20J of work were applied toward rotating the rod. When the rod passes through its horizontal position now, what is the instantaneous linear velocity of mass A?

m_A=5kg, m_B = 3kg, l = 2m, h = 10m

$g=10\frac{m}{s^2}$

Neglect air resistance and any internal friction forces.

Possible Answers:

$4.1\frac{m}{s}$

$7.6\frac{m}{s}$

$1.4\frac{m}{s}$

$2.2\frac{m}{s}$

$12.8\frac{m}{s}$

Correct answer:

$2.2\frac{m}{s}$

Explanation:

We will be using the equation for conservation of energy:

E = U_i +K_i +W = U_f +K_f

We can make a few assumptions to cancel out most terms. If we assume that the reference height of this problem when the rod is horizontal is 0, we can remove both potential energies. Also, the rod is initially at rest, so we can remove initial kinetic energy. Thus, we get:

W = K_f

Breaking up kinetic energy for both masses, we get:

$W = K_{A_{f}}+K_{B_{f}}$

Substituting in expressions for linear kinetic energy (you can also do rotational, but will get to the same answer in extra steps. See the end of the solution):

$W = \frac{1}{2}m_Av_{A_{f}}^2=\frac{1}{2}m_Bv_{B_{f}}^2$

Both velocities are the same, so we can just eliminate the subscripts:

$W = \frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2$

No we just need to rearrange for velocity:

$W = \frac{1}{2}v^2(m_A+m_B)$

$v^2=\frac{2W}{(m_A+m_B)}$

$v=\sqrt{\frac{2W}{(m_A+m_B)}}$

We know each value, so time to plug and chug:

$v = \sqrt{\frac{2(20J)}{5kg + 3kg}}=2.2\frac{m}{s}$

Let's see why it didn't matter which form of kinetic energy we used. Rotational kinetic energy is the following:

$K_r=\frac{1}{2}Iw^2$

Where:

I = mr^2

$w=\frac{v}{r}$

Substituting these back in:

$K_r = \frac{1}{2}(mr^2)\left ( \frac{v}{r}\right )^2$

Canceling out radius:

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

Which is the formula for linear kinetic energy.

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Example Question #72 : Work, Energy, And Power

Consider the following system:

Spinning rod with masses at end

The rod is initially at rest and horizontal. Mass B suddenly disappears and the rod is allowed to rotate freely. Assuming the mass of the rod itself is negligible, what will be the linear velocity of mass A as it passes through its lowest point?

m_A = 20kg,l=4m,h=5m

$g = 10\frac{m}{s^2}$

Neglect air resistance and internal friction forces.

Possible Answers:

$4.2\frac{m}{s}$

$9.8\frac{m}{s}$

$1.1\frac{m}{s}$

$6.3\frac{m}{s}$

$2.9\frac{m}{s}$

Correct answer:

$6.3\frac{m}{s}$

Explanation:

We can use the law of conservation of energy to solve this:

E = U_i + K_i = U_f + K_f

Assuming a height of 0 when the mass is at its lowest point we can remove final potential energy. Also, the rod is initially at rest, so we can removed initial kinetic energy. Thus, we get:

U_i = K_f

Substituting expressions in for each, we get:

$mgh_i = \frac{1}{2}mv_f^2$

We can substitute the following expression in for our initial height:

$h_i = \frac{1}{2}l$

$\frac{1}{2}mgl=\frac{1}{2}mv_f^2$

Canceling out like terms:

gl = v_f^2

Solving for final velocity:

$v_f = \sqrt{gl} = \sqrt{10\frac{m}{s^2}(4m)} = \sqrt{40\frac{m^2}{s^2}}$

$v_f = 6.3\frac{m}{s}$

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Example Question #71 : Work, Energy, And Power

How much greater is the energy of a car crash at 60mph than 30mph .

Possible Answers:

The same

Eight times as great

Four times as great

Half as great

Twice as great

Correct answer:

Four times as great

Explanation:

KE=.5mv^2

KE_1=.5mv_1^2

KE_2=.5mv_2^2

v_2=2v_1

Combining equations:

KE_2=.5m4v_1^2=2mv^2=4KE_1

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Example Question #74 : Work, Energy, And Power

Suppose that an object accelerates from $20\: \frac{m}{s}$ to $40\: \frac{m}{s}$ in the absence of friction. How does the energy necessary to achieve this acceleration differ from when the same object accelerates from zero to $20\:\frac{m}{s}$ ?

Possible Answers:

The energy needed is two times as great

The energy needed is four times as great

The energy needed is three times as great

The mass of the object needs to be given to answer this question

The energy needed is the same in both cases

Correct answer:

The energy needed is three times as great

Explanation:

For this question, we need to consider the difference in the amount of energy needed to accelerate an object by a given amount. Since all the energy put into the system is being used to change the object's motion, we can use the work-kinetic energy theorem.

$W=\Delta KE$

First, let's consider the case where the object accelerates from at rest. We can calculate the amount of energy required by taking note of the final and initial kinetic energy. Even though we don't have the value for the mass of the object, that is okay, because we'll be comparing kinetic energy values in the end. Thus, the term for mass will cancel out.

$W_{0\rightarrow 20}=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}$

$W_{0\rightarrow 20}=\frac{1}{2}m(20\: \frac{m}{s})^{2}-0=200m\: \frac{m^{2}}{s^{2}}$

Next, we can use the same strategy in order to find the energy needed to go from $20\: \frac{m}{s}$ to $40\: \frac{m}{s}$ .

$W_{20\rightarrow 40}=\frac{1}{2}m(40\: \frac{m}{s})^{2}-\frac{1}{2}m(20\: \frac{m}{s})^{2}$

$W_{20\rightarrow 40}=800m\: \frac{m^{2}}{s^{2}}-200m\: \frac{m^{2}}{s^{2}}=600m\: \frac{m^{2}}{s^{2}}$

Now, let's see how the two energy values differ.

$\frac{W_{20\rightarrow 40}}{W_{0\rightarrow 20}}=\frac{600m\: \frac{m^{2}}{s^{2}}}{200m\: \frac{m^{2}}{s^{2}}}=3$

Thus, the energy required to accelerate the object from $20\: \frac{m}{s}$ to $40\: \frac{m}{s}$ is triple the amount required to accelerate the same object from zero to $20\: \frac{m}{s}$ .

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Example Question #75 : Work, Energy, And Power

Deep in space Object has mass 350kg and is initially traveling with velocity $<30,50>\frac{m}{s}$ . At t=0 , it collides with Object , which has mass 500kg and is initially motionless. The two objects stick together.

Determine the final kinetic energy of the system

Possible Answers:

9.00*10^4J

2.45*10^5J

5.66*10^5J

None of these

3.95*10^5J

Correct answer:

2.45*10^5J

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v}=<12.4,20.6>\frac{m}{s}$

Determining magnitude of velocity:

$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$

Plugging in values:

$|\overrightarrow{v}|=\sqrt{12.4^2+20.6^2}$

$|\overrightarrow{v}|=24.0\frac{m}{s}$

Using definition of kinetic energy:

KE=.5mv^2

Plugging in values:

KE=.5*850*24.0^2

KE=2.45*10^5J

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Example Question #22 : Kinetic Energy

Determine the initial kinetic energy of the system.

Possible Answers:

5.95*10^5J

1.66*10^5J

8.44*10^5J

7.10*10^5J

None of these

Correct answer:

5.95*10^5J

Explanation:

Determining magnitude of velocity:

$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$

Plugging in values:

$|\overrightarrow{v}|=\sqrt{30^2+50^2}$

$|\overrightarrow{v}|=58.3\frac{m}{s}$

Using definition of kinetic energy:

KE=.5mv^2

Plugging in values:

KE=.5*350*58.3^2

KE=5.95*10^5J

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Example Question #111 : Ap Physics 1

Determine the final velocity of the two objects.

Possible Answers:

None of these

$<30,50>\frac{m}{s}$

$<6.11,84.4>\frac{m}{s}$

$<21,35>\frac{m}{s}$

$<12.4,20.6>\frac{m}{s}$

Correct answer:

$<12.4,20.6>\frac{m}{s}$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$

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Example Question #78 : Work, Energy, And Power

During time period , a rocket ship deep in space of mass $2.55*10^5\textup{ kg}$ travels from $<0,10>\textup{ km}$ to $<5,5>\textup{ km}$ . During time period , the rocket fires. During time period , the rocket travels from $<-15,0>\textup{ km}$ to $<-30,-5>\textup{ km}$ .

Time periods , , and took $10\textup{ s}$ each.

Determine the kinetic energy during time period .

Possible Answers:

$6.375*10^{10}\textup{ J}$

None of these

$9.813*10^{10}\textup{ J}$

$7.444*10^{10}\textup{ J}$

$3.235*10^{10}\textup{ J}$

Correct answer:

$6.375*10^{10}\textup{ J}$

Explanation:

KE=.5mv^2

$v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting to and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((5000-0)^2+(5000-10000)^2)}{10})^2$

$KE=6.375*10^{10}J$

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Example Question #79 : Work, Energy, And Power

Time periods , , and took $10\textup{ s}$ each.

Determine the kinetic energy during time period .

Possible Answers:

None of these

$6.789*10^{11}\textup{ J}$

$3.188*10^{11}\textup{ J}$

$3.985*10^{11}\textup{ J}$

$4.987*10^{11}\textup{ J}$

Correct answer:

$3.188*10^{11}\textup{ J}$

Explanation:

KE=.5mv^2

$v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting to and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((-15000+30000)^2+(0+5000)^2)}{10})^2$

$KE=3.188*10^{11}J$

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Example Question #81 : Work, Energy, And Power

What is the mass of a bowling ball that maintains an average kinetic energy of 75J while traveling at $7\frac{m}{s}$ ?

Possible Answers:

3.1kg

2.1kg

4.1kg

1.1kg

Correct answer:

3.1kg

Explanation:

For this question we will use the kinetic energy equation.

$K_{e}=\frac{1}{2}m(v^{2})$

Plug in and solve for the mass of the bowling ball.

$75j=\frac{1}{2}m(7\frac{m}{s})^{2}$

m=3.06Kg

The mass of the bowling ball is 3.1Kg .

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #71 : Newtonian Mechanics

Example Question #72 : Work, Energy, And Power

Example Question #71 : Work, Energy, And Power

Example Question #74 : Work, Energy, And Power

Example Question #75 : Work, Energy, And Power

Example Question #22 : Kinetic Energy

Example Question #111 : Ap Physics 1

Example Question #78 : Work, Energy, And Power

Example Question #79 : Work, Energy, And Power

Example Question #81 : Work, Energy, And Power

All AP Physics 1 Resources

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