MCAT Physical - MCAT Physical Sciences

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?

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

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

$16.3\frac{m}{s}$

$13.2\frac{m}{s}$

$6.8\frac{m}{s}$

$4.7\frac{m}{s}$

$8.3\frac{m}{s}$

Explanation

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:

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

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

A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS

Liquid-Solid Water Phase Change Reaction:

H2O(l) ⇌ H2O(s) + X

The scientist prepares two scenarios.

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2:

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

The same scientist in the passage measures the variables of another reaction in the lab. He knows that this reaction is spontaneous under standard conditions, with a standard free energy change of –43 kJ/mol. Using laboratory-calculated variables, he determines that the Gibbs Free Energy has a value of 0 kJ/mol. He then calculated the reaction quotient of this reaction, while knowing the equilibrium constant was 3 x 103. What is true of the reaction quotient?

It must be equal to 3 x 103

It must be less than 3 x 103

It must be more than 3 x 103

It must be zero

It must be one

Explanation

At equilibrium, reaction quotient and equilibrium constant are equal.

Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.

Over the course of the afternoon, they collide many times. Four collisions are described below.

Collision 1:

Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going $5\frac{m}{s}$ and subsequently slides into a stationary child 2. They remain linked together after the collision.

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ each.

Collision 4:

The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of $+8\frac{m}{s}$ .

In all of the above scenarios, which of the following quantities is conserved?

I. Kinetic energy

II. Potential energy

III. Momentum

III only

I and II

II and III

I and III

I, II, and III

Explanation

Kinetic energy and potential energy are interconverted. While total energy is conserved, kinetic energy is allowed to increase or decrease, provided that potential energy does the opposite. For example, as the child in collision 1 is starting at the top of a ramp, she has potential energy, but no kinetic energy. At the bottom of the ramp, all the potential energy has been turned into kinetic energy.

Momentum, on the other hand, is conserved in every collision.

What is the index of refraction for a material in which light travels at $1.85*10^{8}\frac{m}{s}$ ?

Such a material does not exist, since the speed of light is constant

Explanation

Relevant equations:

$n = \frac{c}{v}$

$c = 3 * 10^8 \frac{m}{s}$

To find index of refraction, divide the speed of light in a vacuum by the speed of light in the material:

$n = \frac{3 * 10^8}{1.85 * 10^8} \approx 1.62$

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?

50J

50kJ

20J

20kJ

Explanation

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

If the reactants and/or products in a chemical reaction are gases, the reaction rate can be determined by measuring the change of pressure as the reaction proceeds. Consider the following reaction and pressure vs. reaction rate data below.

$2XY + 2Z \rightarrow X_{2} + 2YZ$

Trial	PXY(torr)	PZ(torr)	Rate (torr/s)
1	100	200	0.16
2	200	200	0.32
3	200	100	0.04
4	200	150	0.14

What is the value of the rate constant, k, for this reaction?

k = 2 * 10-10

k = 1.6 * 10-10

k = 1.6 * 10-13

k = 2 * 10-13

Explanation

The rate law can be determined by comparing changes in reactant concentration to changes in rate. In this reaction, doubling XY doubles the reaction rate; the reaction is first-order for XY. Doubling Z increases the rate by a factor of eight, so the reaction is third-order for Z (23 = 8).

Knowing that Rate = k\[XY\] \[Z\]3, we can solve for k by using data from one of the trials. Plugging in data from trial 1, we can calculate the value of k.

0.16 = k\[100\]\[200\]3

k = 2 * 10-10

Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.

Over the course of the afternoon, they collide many times. Four collisions are described below.

Collision 1:

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ each.

Collision 4:

The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of $+8\frac{m}{s}$ .

Before collision 3 takes place, suppose that child 1 starts at the top of a ramp that is 3m long, and to the lake surface.

With what speed does child 1 hit the lake surface? Ignore friction and air resistance.

$5.4\frac{m}{s}$

$0.9\frac{m}{s}$

$55\frac{m}{s}$

$900\frac{m}{s}$

$9.8\frac{m}{s}$

Explanation

This is an application of the work-energy theorem. The amount of work done by the force of gravity to move child 1 down the ramp is equal to force * distance, as well as equal to the amount of kinetic energy picked up down the ramp.

The force down the ramp is equal to mg*sin(30o) = 600J * 0.5 = 300J

A 3kg book slides towards the right along a frictionless horizontal surface with initial velocity 5m/s, then suddenly encounters a long rough section with kinetic friction coefficient $\mu=0.25$ . How far does the book travel along the rough surface before coming to rest? (Use $\small \small g = 9.8\frac{m}{s^{2}}$ as needed)

5.1m

4.4m

3.7m

2.9m

6.5m

Explanation

We'll need to use the kinematic equation $\small v_{f}^{2}=v_{i}^{2}+2ad$ to solve for d, the distance travelled when the book has stopped ( $\small v_{f}=0$ ). Before solving for d, we need to calculate the acceleration caused by the frictional force, by using the following steps.

Find the normal force on the book, $\small F_{n}=mg$ .
Plug this normal force into $\small F_{f}=\mu_{k}F_{n}$ to solve for frictional force.
Find the acceleration caused by this frictional force, with $\small F_{f}=ma$ .

Step 1 gives $\small \small F_{n}=29.4N$ , so in step 2, $\small \small F_{f}=7.35N$ , giving an acceleration of $\small a = 2.45\frac{m}{s^{2}}$ to the left (which we will define to be the negative horizontal direction).

Returning to the original kinematic equation, $\small 0 =(5 \frac{m}{s})^{2} + 2(-2.45\frac{m}{s^{2}})d$ .

Rearranging to solve for d gives $\small d = 5.1m$

Which of the following is not accurate regarding spontaneous fission reactions?

The products have smaller binding energies per nucleon than the original atom

The two resulting nuclei are approximately equal to each other in mass and atomic number

The total kinetic energy of the products is less than the kinetic energy of the original atom

The potential energy of the system is greater after the reaction than before

The sum of the atomic weights of the products is less than the atomic weight of the original atom

Explanation

In fission a large atom, with low binding energy per nucleon, splits into two approximately equal smaller atoms, which are more stable due to greater binding energy per nucleon. All the other properties mentioned are accurate.

Suppose a ball with mass of 10kg was dropped from rest from the top of a cliff that is 80m tall. How long will it take the ball to reach the bottom of the cliff?

10s

16s

Explanation

The question tells us that the initial velocity of the ball (v0) is zero and the height (d) of the cliff is 80m. Acceleration (gravity) is 10m/s2. Using the kinematics eqation $d=v_0 t+(1/2)at^{2}$ we can solve for time.

80 = 0 + (1/2)(10)t2

80/5 = t2 = 16

t = 4s

MCAT Physical Sciences

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