AP Physics 1 - Kinetic Energy

An object has a mass, M, and a velocity, V.

What happens to the object's kinetic energy if its velocity is doubled?

The object's momentum is quadrupled.

The object's momentum is doubled.

The object's momentum is halved.

The object's momentum remains the same.

The object's momentum is one fourth of its original momentum.

Explanation

The equation for kinetic energy is:

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

Where is the object's kinetic energy, is the object's, and is the object's velocity.

We can see that the relationship between kinetic energy and velocity is quadratic, so when the velocity is doubled, the kinetic energy is quadrupled.

A bungie jumper of mass is attached to a bungie with a constant of $100 \frac{N}{m}$ . The unstretched length of the bungie is . What is the maxmimum velocity of the jumper?

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

$18.7\frac{m}{s}$

$21.3\frac{m}{s}$

$24.4\frac{m}{s}$

$11.6\frac{m}{s}$

$7.9\frac{m}{s}$

Explanation

Think about this scenario practically. After the jumper jumps, he will begin accelerating at a rate of $10\frac{m}{s^2}$ . This rate will stay constant until the bungie cord begins to stretch. At this point, the jumper has traveled a distance of . The rate of acceleration will now decrease and ultimately reach a rate of $0 \frac{m}{s^2}$ . This is the point at which the force from the bungie cord is equal and opposite to the force of gravity. This is also the point at which the jumper is traveling at his or her maxmium velocity. With all of this in mind, let's start writing expressions for the scenario.

The main expression we will use will be the one for conservation of energy:

Plugging in our expressions for these variables and removing initial kinetic energy, we get:

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

Rearranging for velocity:

$v_f = \sqrt{2gh_i - \frac{kx^2}{m}}$

We simply need to find the height distance between the jumper's initial position and the position at which the jumper is traveling at his or her greatest velocity. As previously mentioned, the point of highest velocity is the point at which the force from the bungie cord is equal and opposite to the force of gravity:

$F_{net} = 0$

Rearranging for , we get:

$x = \frac{ma}{k}$

This is the distance that the bungie is stretched. Therefore, we can say that the total height distance between the initial and final state is the length of the unstretched bungie cord plus the distance the cord has stretched:

$h_i = l + x = l + \frac{ma}{k}$

Plugging this back into the equation for final velocity, we get:

$v_f = \sqrt{2g(l+\frac{ma}{k})-\frac{kx^2}{m}}$

We have values for all of our variables, so we can simply solve for the final velocity:

$v_f = \sqrt{2(10\frac{m}{s^2})\left ( 15m+\frac{(50kg)(10\frac{m}{s^2})}{100\frac{N}{m}}\right )-\frac{(100\frac{N}{m})(5m)^2}{50kg}}$

$v_f = \sqrt{2(10\frac{m}{s^2})(20m)-50\frac{m^2}{s^2}} = 18.7\frac{m}{s}$

An object of mass moves with velocity . How fast must an object of mass move in order to have the same kinetic energy of the object of mass ?

$\frac{v^2}{2}$

$\frac{v^2}{4}$

$\frac{v^2}{8}$

Explanation

Kinetic energy is equal to $\frac{1}{2}mv^2$ . The object of mass and velocity therefore has kinetic energy equal to $\frac{1}{2}mv^2$ . Let's let the object of mass have velocity $v_{4m}$ . Therefore, its kinetic energy is $\frac{1}{2}mv_{4m}^2$ . We want to find $v_{4m}$ such that the two objects to have the same kinetic energy, so we can equate their two kinetic energies. $\frac{1}{2}mv^2=\frac{1}{2}4mv_{4m}^2$

$v^2=4v_{4m}^2$

$\frac{v^2}{4}=v_{4m}^2$

$\frac{v^2}{2}=v_{4m}$

Two objects, one of mass, , and the other of mass, , are traveling at constant velocity along a frictionless surface. The lighter object travels at , while the heavier travels at . An opposing horizontal force acts upon both objects equally, and brings them to rest. Which object takes longer to slow down, and why?

The lighter object, because the velocity is squared in the Kinetic Energy formula

The lighter object, because the mass is square rooted in the Kinetic Energy formula

The objects stop in the same amount of time, as their masses and velocities cancel out.

The heavier object, because the mass is squared in the Kinetic Energy formula

The heavier object, because the velocity is square rooted in the Kinetic Energy formula

Explanation

Examining the formula for kinetic energy will allow us to compare the two objects. The formula, $KE = .5mv^{2}$ , shows us that velocity has a higher influence in the calculation of Kinetic Energy than mass does. The velocity of the lighter object is double that of the heavier object, and because of the velocity's larger influence on the formula, the lighter one takes longer to stop, even though its mass is half that of the heavier object.

A ball with a mass of 3 kilograms is uniformly accelerated from rest and travels 110 meters in 6 seconds. What is its final kinetic energy in Joules after 6 seconds?

2017 Joules

2012 Joules

1500 Joules

1888 Joules

2300 Joules

Explanation

Find the final velocity by multiplying the average velocity by 2. Then substitute into the equation for kinetic energy.

$V_{f}=\frac{110}{6}\times2=36.67\frac{m}{s}$

$KE=.5(3kg)(36.67\frac{m}{s})^2\rightarrow \boldsymbol{2017J}$

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

Four times as great

Twice as great

The same

Eight times as great

Half as great

Explanation

Combining equations:

In an investigation into energy, a student compresses a spring a distance of 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 . She then repeats the experiment, but this time compresses the spring a distance of before releasing the spring and the cart. What would the cart's velocity be in the second experiment?

$\frac{v_0}{2}$

$\frac{v_0}{4}$

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 . 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 .

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. 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?

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

Neglect air resistance and any internal friction forces.

$2.2\frac{m}{s}$

$1.4\frac{m}{s}$

$7.6\frac{m}{s}$

$12.8\frac{m}{s}$

$4.1\frac{m}{s}$

Explanation

We will be using the equation for conservation of energy:

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:

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:

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

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.

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

A kickball of mass and radius is on top of a hill of height , 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.

$13.3\frac{m}{s}$

$16.4\frac{m}{s}$

$10.7\frac{m}{s}$

$10.1\frac{m}{s}$

$14.9\frac{m}{s}$

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:

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

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

Moment of inertia of a ring:

A bicycle wheel has a mass of and a radius of . Determine the total kinetic energy of the wheel if the bicycle is moving at $10\frac{m}{s}$ . The wheel may be approximated as a hollow ring.