Newtonian Mechanics and Motion - MCAT Chemical and Physical Foundations of Biological Systems

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Question

A $\small 100g$ ball moving at $\small 2\frac{m}{s}$ strikes a stationary $\small 50g$ solid ball head on. The smaller ball moves away at $\small 1\frac{m}{s}$ . What is the velocity of the first ball after the impact?

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Answer

Momentum is conserved in all situations. Here, we have an elastic collision, meaning that the objects do not stick together after the impact. Momentum is simply the product of mass and velocity, and it has the units $\small \frac{kgm}{s}$ .

We can solve this problem by setting the initial and final momentum of the system equal to each other. The momentum of the system will be the sum of the momentum of the parts.

$m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}$

The second ball contributes no momentum initially, because its initial velocity is zero. The total initial momentum comes from the initial velocity of the larger ball.

$(0.1kg)(2\frac{m}{s})+(0.05kg)(0\frac{m}{s})=m_1v_{f1}+m_2v_{f2}$

$0.2\frac{kgm}{s}=m_1v_{f1}+m_2v_{f2}$

We know the mass of each ball and the final velocity of the smaller ball. Using these values, we can solve for the final velocity of the larger ball.

$0.2\frac{kgm}{s}=(0.1kg)v_{f1}+(0.05kg)(1\frac{m}{s})$

$0.2\frac{kgm}{s}=(0.1kg)v_{f1}+0.05\frac{kgm}{s}$

$0.15\frac{kgm}{s}=(0.1kg)v_{f1}$

$v_{f1}=\frac{0.15\frac{kg*m}{s}}{0.1kg}=1.5\frac{m}{s}$

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Question

A $\small 1000kg$ car moving east at $\small 10\frac{m}{s}$ strikes the rear of a $\small 750kg$ car, also traveling east at $\small 5\frac{m}{s}$ , and the two stay in contact after the collision. What is the final velocity of the cars?

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Answer

This is a case of inelastic collision because the objects stay in contact after the impact. The total momentum in a system is always conserved, regardless of collision type. Here, the momentum before the collision is given by the sum of the individual momentum of each car, and the momentum after the collision is given by the singular momentum of the joined vehicles.

We are given the mass of each car and their initial velocities, allowing us to completely solve the left side of this equation.

$(1000kg)(10\frac{m}{s})+(750kg)(5\frac{m}{s})=(m_1+m_2)v_3$

$13750\frac{kgm}{s}=(m_1+m_2)v_3$

We can now use the combined mass of both cars to solve for their final velocity.

$13750\frac{kgm}{s}=(1000kg+750kg)v_3$

$v_3=\frac{13750\frac{kg*m}{s}}{1750kg}=7.86\frac{m}{s}$

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Question

If a ball was thrown straight up at 10m/s and strikes the ground after two seconds, what maximum height did the ball reach?

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Answer

This question can be solved using multiple strategies. One strategy uses the equation .

vf = 0m/s

vo = 10m/s

a = -10m/s2

We use a negative acceleration because gravity is in the opposite direction of the movement of the ball. When we plug in all the values we find that d= 5m.

Another strategy uses the projectile time. In projectile motion we know that velocity is zero at the maximum height. Using only half of the projectile time, we can solve for the maximum height.

$d = \frac{1}{2}at^2=\frac{1}{2}(10\frac{m}{s^s})(1s)^2=5m$

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Question

A car is moving with a constant velocity of $10\frac{m}{s}$ when it abruptly stops, applying a constant breaking acceleration of $-5\frac{m}{s^2}$ . Over what distance does the car come to a stop?

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Answer

This problem can be easily solved using the formula, $v{_{f}}^{2} = v{_{0}}^{2} + 2ax$ , and solving for . Our initial velocity is $10\frac{m}{s}$ , acceleration is $-5\frac{m}{s^2}$ , and final velocity is $0\frac{m}{s}$ .

$(0\frac{m}{s})^2=(10\frac{m}{s})^2+2(-5\frac{m}{s^2})x$

$0=100\frac{m^2}{s^2}-10\frac{m}{s^2}x$

$-100\frac{m^2}{s^2}=-10\frac{m}{s^2}x$

$\frac{-100\frac{m^2}{s^2}}{-10\frac{m}{s^2}}=10m=x$

The car comes to a stop after applying the breaks over 10m.

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Question

If an object is dropped from a height of 450 meters above Earth, what is its velocity just before impact?

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Answer

First calculate the time it takes to hit the ground using the equation $d=v_o+\frac{1}{2}at^2$ .

We can plug in values (including acceleration due to gravity) and solve for t.

$450m=0\frac{m}{s}+\frac{1}{2}(10\frac{m}{s^2})t^2$

t = 9.49s

Next, find the final velocity with the equation .

$v_f=0\frac{m}{s}+(10\frac{m}{s})(9.49s)=95\frac{m}{s}$

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Question

An object is shot upward from the ground at a velocity of 24m/s. How long before it hits the ground?

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Answer

First find the time it takes to reach the top of its path using the equation , where the final velocity is 0m/s, initial velocity is 24m/s, and a is the acceleration due to gravity. Solve for time (t).

0 = 24 +(-10)t

t = 2.4s

Notice that this is only the time it takes the object to reach the top of its path, and the question asks for the time it takes the object to reach the ground. In order to find how long it takes to reach the ground simply double this time.

2.4s * 2 = 4.8 s

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Question

Which of the following forces is not conservative?

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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?

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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.

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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?

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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?

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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?

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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

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?

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

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

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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

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?

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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

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.

How much potential energy does Sally have at the top of the hill on her sled?

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Answer

This is a tricky question only because you have to keep your units straight. The formula is simply PE = mgh.

PE = 52kg * 10m/s2 * 50m = 26,000J = 26kJ

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

After the rope is cut and the platform falls, we want to pull the platform back up to 5m above the ground. How much energy is required to raise the platform?

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Answer

First, we need to understand what type of energy we are considering. As the resistive force to motion is due to gravity, we are talking about gravitational potential energy. We need to use the formula U = mgh.

Plugging in the values that are provided, we can solve for the potential energy (U).

U = (2kg)(9.8m/s2)(5m) = 98J

The amount of work required is equal to the change in potential energy of the platform.

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Question

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 5m/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 10m/s, slides into child 1, moving at 2m/s.

Collision 3:

The two children collide while traveling in opposite directions at 10m/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 +8m/s.

Child 1 hooks up a jet motor to a sled, and brings it to the icy lake. She sets up her sled, and launches from a velocity of +8 m/s with an acceleration of 15m/s2. After ten seconds, how fast is she traveling? Assume the lake is frictionless.

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Answer

For this question, we have to use the translational motion equations:

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Question

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

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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 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.

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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

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?

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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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