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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #11 : Kinetic Energy

A block with a mass of 10 kg is dropped from a height and just before it hits the ground, it has a velocity of $5 \frac{m}{s}$ . From what height was the block dropped from?

Possible Answers:

2.5m

1.25m

Correct answer:

1.25m

Explanation:

Just before impact, the block has maximum kinetic energy. Once the kinetic energy is determined, we can find the potential energy and therefore the height of the drop. We can use the following formula to determine the kinetic energy:

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

$KE = \frac{1}{2}(10kg)(25\frac{m}{s})^2=125J$

All of this energy was previously potential energy from which we can determine the height, .

125J = U = m*g*h

$125J = (10kg)(10\frac{m}{s^2})(h)$

h = 1.25m

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Example Question #61 : Work, Energy, And Power

A soccer ball of mass 0.5kg is kicked and accelerated uniformly from rest to travel 20 meters over 3 seconds into the goal. What is the final kinetic energy of the soccer ball?

Possible Answers:

KE=44.4J

KE=65J

KE=74.4J

KE=55J

KE=100J

Correct answer:

KE=44.4J

Explanation:

We need to know the final speed of the ball since we have all but this quantity in our formula for kinetic energy,

$KE=.5mv_{f}^2$

To find the final velocity we first find the acceleration:

$\Delta x=V_{0}t+.5at^2$

20m=.5a(3)^2

$a=4.444\frac{m}{s^2}$

$V_{f}=a*t=4.444\frac{m}{s^2}*3=13.333\frac{m}{s}$

$KE=(0.5)(0.5kg)(13.333\frac{m}{s})^2=44.44J$

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Example Question #14 : Kinetic Energy

A champagne cork with a mass of 10 grams accelerates off the top of the champagne bottle at $4\frac{m}{s^2}$ for 2 seconds. What is the final kinetic energy of the cork after 2 seconds?

Possible Answers:

KE=0.484J

KE=0.32J

KE=0.46J

KE=0.375J

KE=0.64J

Correct answer:

KE=0.32J

Explanation:

We can find the final velocity by multiplying the acceleration by the time. We then plug this into the formula for kinetic energy and solve. Do not forget to convert grams to kilograms.

$V=V_{0}+at$

The initial velocity of the cork was zero.

$V=(4\frac{m}{s^2})(2s)=8\frac{m}{s}$

$KE=.5mv^2=(.5)(.01kg)(8\frac{m}{s}^2)=0.32J$

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Example Question #15 : Kinetic Energy

In an investigation into energy, a student compresses a spring a distance of x_0 from its neutral length. She then puts a cart against the end of the spring and releases both the spring and the cart. She measures the kinetic energy of the cart after the spring has returned to its neutral length and finds it to be K_0 . She then repeats the experiment, but this time compresses the spring a distance of 2x_0 before releasing the spring and the cart. What would the cart's kinetic energy be in the second experiment?

Possible Answers:

$\frac{K_0}{4}$

$\frac{K_0}{2}$

2K_0

K_0

4K_0

Correct answer:

4K_0

Explanation:

The potential energy stored in a spring is given by

$U_S=\frac{1}{2}kx^2$

So the energy increases with the square of the compression. When the potential energy is converted to kinetic energy of the cart, it increases by a factor of 2^2=4 .

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Example Question #16 : Kinetic Energy

In an investigation into energy, a student compresses a spring a distance of x_0 from its neutral length. She then puts a cart against the end of the spring and releases both the spring and the cart. She measures the velocity of the cart after the spring has returned to its neutral length and finds it to be v_0 . She then repeats the experiment, but this time compresses the spring a distance of 2x_0 before releasing the spring and the cart. What would the cart's velocity be in the second experiment?

Possible Answers:

4v_0

$\frac{v_0}{2}$

v_0

$\frac{v_0}{4}$

2v_0

Correct answer:

2v_0

Explanation:

The potential energy stored in a spring is given by

$U_S=\frac{1}{2}kx^2$

So doubling the compression increases the potential energy by a factor of 2^2=4 . However, the kinetic energy:

$K=\frac{1}{2}mv^2$ , increases with the square of the velocity, so the velocity only needs to double to increase the cart's kinetic energy by a factor of .

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Example Question #17 : Kinetic Energy

In an investigation into energy, a student compresses a spring a distance of x_0 from its neutral length. She then puts a cart of mass m_0 against the end of the spring and releases both the spring and the cart. She measures the velocity of the cart after the spring has returned to its neutral length and finds it to be v_0 . She then repeats the experiment, but this time using a cart whose mass is 2m_0 before releasing the spring and the cart. What would the cart's velocity be in the second experiment?

Possible Answers:

$\frac{v_0}{4}$

2v_0

$\sqrt2v_0$

$\frac{v_0}{2}$

$\frac{v_0}{\sqrt2}$

Correct answer:

$\frac{v_0}{\sqrt2}$

Explanation:

Kinetic energy is given by:

$K=\frac{1}{2}mv^2$ , so the energy is linearly dependent on the mass, but increases with the square of the velocity. So although the mass has doubled, the velocity only needs to decrease by a factor of $\sqrt2$ to maintain the same kinetic energy. In algebraic form:
K=K_0

$\frac{1}{2}(2m)v^2=\frac{1}{2}mv_0^2$

Simplify.

$v^2=\frac{1}{2}v_0^2$

$v=\frac{v_0}{\sqrt2}$

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Example Question #18 : Kinetic Energy

There is a truck and a car on a highway going the same speed in the same direction. The car has mass while the truck has mass .

What is the correct statement about the two vehicle's kinetic energies?

Possible Answers:

The car has exactly $\frac{1}{2}$ the kinetic energy of the truck

The car and the truck have the same kinetic energy

The car has exactly $\frac{1}{2}$ the kinetic energy of the truck.

The car has double the kinetic energy of the truck

The truck has four times the kinetic energy of the car

Correct answer:

The car has exactly $\frac{1}{2}$ the kinetic energy of the truck

Explanation:

The equation for kinetic energy is:

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

Where is the mass of an object and is the velocity of the object.

We see from the equation that mass has a linear relationship with kinetic energy and the problem statement gives that the vehicles have the same velocity. Because the car has half the mass of the truck and the same speed, the car has half as much kinetic energy as the truck.

Without specific numbers given, we can always substitute easy numbers into the equation to help understand the concept. If we say the velocity in both cases is $v=1\frac{m}{s}$ , and the car's mass is 1 kg then we have:

$\frac{1}{2}*(1kg)*(1\frac{m}{s})^{2}=\frac{1}{2}J$

We know the truck's mass is double the car's, which yields:

$\frac{1}{2}*(2kg)*(1\frac{m}{s})^{2}=1J$

This makes it clear that the car has half the kinetic energy of the truck.

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Example Question #61 : Work, Energy, And Power

Moment of inertia of hollow sphere: $I=\frac{2mr^2}{3}$

A kickball of mass .450kg and radius 12.5 cm is on top of a hill of height 15m , at the edge of a straight incline to the bottom. Suppose the ball was just barely pushed over the edge. Calculate the ball's velocity at the bottom of the hill. It may be assumed that the ball is essentially hollow. Ignore any losses due to friction, as well as any velocity from the initial push.

Possible Answers:

$13.3\frac{m}{s}$

$10.1\frac{m}{s}$

$10.7\frac{m}{s}$

$14.9\frac{m}{s}$

$16.4\frac{m}{s}$

Correct answer:

$13.3\frac{m}{s}$

Explanation:

There will be two types on kinetic energy, rotational and translational.

Using conservation of energy:

$E_{initial}=E_{final}$

$KE_{initial}+PE_{initial}=KE_{final}+PE_{final}$

Initially, kinetic energy will be zero, and in the final state, potential energy will be zero.

$mgh=.5mv^2+.5I\omega^2$

Recall:

$I=\frac{2mr^2}{3}$

and

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

Combine equations:

$mgh=.5mv^2+.5*\frac{2mr^2}{3}(\frac{v}{r})^2$

Solve for

$v=\sqrt{\frac{6}{5}gh}$

Plug in values:

$v=\sqrt{\frac{6}{5}*9.8*15}$

$v=13.3\frac{m}{s}$

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Example Question #20 : Kinetic Energy

Determine the kinetic energy of a mole of oxygen gas at 300K .

The formula for kinetic energy given temperature is:

$KE=\frac{3}{2}n*r*T$ , where is the number of moles, $r=8.314\frac{J}{Kmol}$ , and is temperature.

Possible Answers:

3741J

$2.25*10^{-27}J$

$2.25*10^{27}J$

7482J

Correct answer:

3741J

Explanation:

To solve this question, we simply plug in given values into the equation for kinetic energy:

$KE=\frac{3}{2}n*r*T=\frac{3}{2}*1mol*8.314\frac{J}{Kmol}*300K=3741J$

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Example Question #71 : Work, Energy, And Power

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 spinning and mass B has an instantaneous linear velocity of $5\frac{m}{s}$ . What is the instantaneous linear kinetic energy of mass A?

$l = 5m, h = 20 m, m_{A} = 5kg, m_{B}=2kg$

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

Possible Answers:

25.0J

22.2J

12.5J

31.3J

5.0kg

Correct answer:

12.5J

Explanation:

We only need to know the equation for linear kinetic energy to solve this problem:

$KE_{linear} = \frac{1}{2}mv^2$

In the problem statement, we were told that the instantaneous velocity of mass B was $5\frac{m}{s}$ . Since both masses are attached to a rigid rod and spinning in a uniform circle, we know that:

$v_B = 5\frac{m}{s}$

$v_{A}=v_{B}$

$v_{A} = 5\frac{m}{s}$

Plugging this into our original equation, we get:

$KE_{linear} = \frac{1}{2}(5kg)(5\frac{m}{s^2}) = 12.5J$

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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 #11 : Kinetic Energy

Example Question #61 : Work, Energy, And Power

Example Question #14 : Kinetic Energy

Example Question #15 : Kinetic Energy

Example Question #16 : Kinetic Energy

Example Question #17 : Kinetic Energy

Example Question #18 : Kinetic Energy

Example Question #61 : Work, Energy, And Power

Example Question #20 : Kinetic Energy

Example Question #71 : Work, Energy, And Power

All AP Physics 1 Resources

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