Newtonian Mechanics and Motion - MCAT Physical

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Question

A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?

Answer

First solve for the potential energy of the pendulum at the height of 2.4m.

PE = mgh

PE = (405kg)(10m/s2)(2.4m) = 9720J

This must be equal to the maximum kinetic energy of the object.

KE = ½mv2

9720J = ½mv2

Plug in the mass of the object (405 kg) and solve for v.

9720J = ½(405kg)v2

v = 6.9m/s

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Question

An empty mining cart has a mass of and is traveling down a track that has a slope of $40^{\circ}$ to the horizontal. The cart is traveling at a rate of $10\frac{m}{s}$ when an operator notices a disturbance on the track ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled with the wheels locked?

$\mu_s = 0.6,\ \mu_k=0.3$

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

Answer

We need the equation for conservation of energy for this problem:

We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.

Substituting our equations for each variable, we get:

$\frac{1}{2}mv_f^2 = mgh_i + \frac{1}{2}mv_i^2 - \mu_kNd$

Rearranging for final velocity we get:

$v_f = \sqrt{2gh_i+v_i^2-\frac{2\mu_kNd}{m}}$

If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of $\frac{m^2}{s^2}$ , which will ultimately give us units of $\frac{m}{s}$ , which is what we want.

We have values for all variables except two: height and normal force.

Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:

$sin(40^{\circ})=\frac{h_i}{20 m}$

Now we just need to find the normal force. The following diagram will help visualize this calculation.

If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.

Therefore, we can say that:

$N = mgcos(40^{\circ}) = (15kg)(10\frac{m}{s^2})cos(40^{\circ})= 114.9 N$

Now that we have all of our variables, it's time to plug and chug:

$v_f=\sqrt{(2\cdot10\cdot12.856)+(10)^2-\frac{2\cdot0.3\cdot114.9\cdot20}{15}}$

$v_f=\sqrt{257.1+100-91.9} = 16.3 \frac{m}{s}$

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Question

Which of the following forces is not conservative?

Answer

Conservative forces are forces that do not lose energy to heat, sound, or light. Of these answers, energy is completely conserved and transferred from kinetic energy to potential energy, or vice versa. Gravitational forces, electrostatic forces, and elastic forces all work by providing a potential that will work in the same direction as the motion of an object or particle, allowing kinetic and potential energy to interconvert. Frictional forces lose energy as heat when sliding across a surface, and the more force (the more rough the surface), the more energy that is lost.

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Question

Which of the following is not a conservative force?

Answer

Friction is a non-conservative force, meaning that the work it does depends on the path taken by the object. For example, moving a brick in a long zig-zag across the table will generate more heat from friction than moving it in a straight line across the table.

Electric and gravitational forces are conservative. This can be tested by knowing a constant equation to calculate the energy associated with these forces; such equations are applicable regardless of path. No such equation exists for frictional energy.

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Question

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

Find the horizontal component of velocity once the ball has left the cannon.

Answer

This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.

vx = (10m/s)(cos(45o)) = 7.1m/s

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Question

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

What is the initial vertical component of velocity of the ball?

Answer

This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.

vy = (10m/s)(sin(45o)) = 7.1m/s

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Question

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

What is the horizontal acceleration of the ball during its flight?

Answer

Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.

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Question

Which factors increase the maximum velocity of a pendulum?

Answer

Both the length of the pendulum's string and the angle of displacement affect the maximum velocity of the pendulum. Increasing the length of the pendulum's string and increasing the angle of displacement both increase the distance the pendulum must travel in a single period, increasing its potential energy at its maximum height, and therefore the maximum velocity at its lowest point.

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Question

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

How much energy is stored in the spring before the ball is launched?

Answer

In order to determine how much energy is stored, we first need to understand what type of energy we want to consider. A spring stores potential energy; the potential energy of the spring is maximized at maximal displacement from its resting state. In order to compute the potential energy stored, we need both the spring constant (100N/m) and the displacement from resting (1m).

PEs = ½k(Δx)2 = ½(100N/m)(1m)2 = 50J

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Question

What is the period of a pendulum that has a string length of 9.8m?

Answer

The key to answering this question is to recall the following important formula for a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$ .

$T=2\pi\sqrt{\frac{9.8}{9.8}}=2\pi$

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Question

A 200-gram object moves in the +x direction at 4m/s and collides with an identical object moving in the –y direction at 3m/s. If the two objects stick together after the collision, what is the magnitude of their resulting velocity?

Answer

Use conservation of momentum in each direction: $\small p_{xi}=p_{xf}$ and $\small p_{yi}=p_{yf}$

The initial momentum in the x direction is provided only by the first object.

$\small \small \small p_{xi }= mv_{xi}=(0.2kg)(4 \frac{m}{s})= 0.8 \frac{kgm}{s}$

This also equals the final x-momentum, when the two objects move together at the same velocity.

$\small p_{xf}=(m+m)v_{xf}$

$\small \small 0.8\frac{kgm}{s}=(0.4kg)v_{xf}$

$\small \small v_{xf}=2 \frac{m}{s}$

The initial momentum in the y direction is provided only by the second object.

$\small p_{yi }= mv_{yi}=(0.2kg)(-3 \frac{m}{s})= -0.6 \frac{kgm}{s}$

Again, this equals the final y-momentum.

$\small p_{yf}=(m+m)v_{yf}$

$\small -0.6\frac{kgm}{s}=(0.4kg)v_{xf}$

$\small v_{yf}=-1.5 \frac{m}{s}$

Combine the two velocities using the Pythagorean theorem.

$\small a^{2}+b^{2}=c^{2}$

$\small (2\frac{m}{s})^{2} +(-1.5\frac{m}{s})^{2}= v_{f}^{2}$

$\small v_{f}=2.5 \frac{m}{s}$

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Question

Which of the following is a correct description of an inelastic collision?

Answer

Momentum is conserved in any collision, but kinetic energy is only conserved in elastic collisions. So, an inelastic collision has conservation of momentum, but not conservation of kinetic energy.

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Question

A ball with mass of 2kg is dropped from the top of a building this is 30m high. What is the approximate velocity of the ball when it is 10m above the ground?

Answer

Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic energy gained.

Change in gravitational potential energy can be found using the difference in mgh. $mgh_{f} - mgh_{i} = 2kg10\frac{m}{s^{2}}10m - 2kg10\frac{m}{s^{2}}30m = -400 Joules$

So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.

$400 Joules = \frac{1}{2}mv^{2}$

$400 = v^{2}$

$v = 20 m/s$

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Question

A stone of mass m sits atop a hill of height h. As it rolls downhill, which of the following is true?

Half way down the hill, $\dpi{100} \small PE=\frac{1}{2}mgh$

Half way down the hill, $\dpi{100} \small PE=\frac{1}{4}mgh$

Half way down the hill, $\dpi{100} \small PE=mg\sqrt{h}$

Half way down the hill, PE still equals mgh.

None of these is true.

Answer

4. Choice 1 is correct because initially, all of the mechanical energy in the stone was potential energy and none was kinetic energy: ME = KE + PE. PE is “stored” in the stone-hill system by rolling the stone up hill. It is obvious that it takes half as much energy to roll the stone half way up the hill, compared with rolling it to the top. At the bottom of the hill, all of the PE will have been converted into KE, given by the formula $\dpi{100} \small KE=\frac{1}{2}mv^{2}$

Since the PE was ½ mgh when rolling the stone half way up hill, it is the same as it rolls down hill.

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Question

Consider a spring undergoing simple harmonic motion. When the spring is at its maximum velocity __________.

Answer

Kinetic energy is highest when the spring is moving the fastest. Conversely, potential energy is highest when the spring is most compressed, and momentarily stationary. When the force resulting from the compression causes the spring to extend, potential energy decreases as velocity increases.

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Question

Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.

They then decide to sled down the hill, but disagree about who will go first.

Scenario 1:

Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.

Scenario 2:

Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.

Scenario 3:

Unable to agree, Sam and Sally tether themselves with a rope and go down together.

At the bottom of a neighboring hill, a neighbor watches Sally and Sam come down the hill. Sally is traveling 15m/s and Sam is traveling 10m/s. From the moment the neighbor begins watching, to just after they both come to a stop, who has dissipated more heat in the form of friction? (Assume all friction is lost as heat).

Answer

Sally has greater kinetic energy in this example than does Sam. From the moment when the neighbor begins watching we can calculate the kinetic energy. Once stopped, all of the kinetic energy will have been dissipated.

Sally's KE = 1/2 (52kg) (15m/s)2 = 5850J

Sam's KE = 1/2 (60kg) (10m/s)2 = 3000J

All of this energy will be dissipated as friction before Sam and Sally come to a stop.

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Question

A rock is dropped from a given height and allowed to hit the ground. The velocity of the rock is measured upon impact with the ground. Assume there is no air resistance.

In order for the velocity of the rock to be doubled before impact, which of the following is necessary?

Answer

We can compare height and velocity by comparing the equations for potential and kinetic energy. This is possible because the rock initially has no kinetic energy (velocity is zero) and has no potential energy upon impact (height is zero). Using conservation of energy will yield the comparison below.

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

Because velocity is squared in the equation for kinetic energy, it requires a quadrupling of height in order to double the velocity.

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

$mg(xh)=\frac{1}{2}m4(v)^2$

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Question

A boulder drops from rest off of a cliff. Find its velocity at before impact.

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

Answer

Conservation of energy dictates that the initial energy and final energy will be equal.

$PE_{i} + KE_{i} = PE_{f} + KE_{f}$

In this case, the boulder starts with zero kinetic energy and ends with both kinetic and potential energy.

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

$mgh_i+\frac{1}{2}m(0\frac{m}{2})^2=mgh_f+\frac{1}{2}mv_f^2$

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

We can cancel the mass from each term and plug in the given values to solve for the velocity at a height of $\small 7m$ .

$(10\frac{m}{s^2})(100m)=(10\frac{m}{s^2})(7m)+\frac{1}{2}v_f^2$

$930\frac{m^2}{s^2}=\frac{1}{2}v_f^2$

$1860\frac{m^s}{s^2}=v_f^2$

$43\frac{m}{s}=v_f$

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Question

Ignoring air resistance, which of the following is true regarding the motion of a pendulum?

Answer

Energy must be conserved through the motion of a pendulum. Let point 1 represent the bottom of the oscillation and point 2 represent the top. At point 1, there is no potential energy, using point 1 as our "ground/reference," thus all of the system energy is kinetic energy. At point 2, the velocity is zero; thus, the kinetic energy is zero and all of the system energy is potential energy. At the highest point in the swing, potential energy is at a maximum, and at the lowest point in the swing, kinetic energy is at a maximum.

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Question

A block of wood is floating in space. A bullet is fired from a gun and hits the block, embedding itself in the wood and generating heat. Which of the following is conserved?

Answer

Momentum is always conserved in a system, when not experiencing external forces.

An inelastic collision can be identified if the two objects stick together after the collision occurs, such as the bullet becoming embedded in the wood. In an inelastic collision, kinetic energy is not conserved. Since mechanical energy is the sum of kinetic and potential energy, mechanical energy is also not conserved. This lack of conservation is due to the conversion of some of the kinetic energy to heat and sound. The kinetic energy decreases as the heat energy increases, resulting in a non-constant temperature.

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