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

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Question

A ball is dropped off of a roof. If the ball takes four seconds to hit the ground, what is the ball's speed right before it hits the ground?

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Answer

We are given the time that the ball is in the air and the acceleration. Since the ball is dropped, we can also tell that the initial velocity is zero (starts from rest). Using these given values, we can calculate the final velocity using the appropriate motion equation.

$v_f = 0\frac{m}{s} + (-9.8\frac{m}{s^2})(4s)= -39.2\frac{m}{s}$

The final velocity is negative because the ball is traveling downward; however, the question asks for speed, allowing us to disregard the vector indicator. Speed cannot be negative.

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Question

Which of the following is not a vector quantity?

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Answer

Kinetic energy is calculated by squaring velocity (which we know is a vector). This eliminates its vector properties making kinetic energy a scalar value. All other answer choices are vector quantities.

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Question

A ball of mass 3kg is thrown into the air at an angle of 45o above the horizontal, with initial velocity of 15m/s. Instantaneously at the highest point in its motion, the ball comes to rest. Approximately what is the magnitude of acceleration at this point? Assume that air resistance is negligible.

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Answer

After being thrown, the ball is only acted on by the force of gravity. Since this force is constant throughout the motion, acceleration must also remain constant, and be equal to the gravitational acceleration of 9.8 m/s2 (approximately 10 m/s2)

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Question

A man walks two kilometers north and then two kilometers west. What is his displacement?

4 kilometers

4 kilometers northwest

2.8 kilometers northwest

8 kilometers northwest

none of these

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Answer

Choice 3 is correct. Recall that physics problems tell you where to begin and end. The only thing that matters in figuring displacement is the beginning point and end point. The man effectively walked along the hypotenuse of a right triangle whose sides measured 2 km each. Since a2 + b2 = c2, then 4 + 4 = c2, and the correct response is the square root of 8 with the direction of the displacement added for clarity.

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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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Question

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

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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

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?

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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

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).

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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?

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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.

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

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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?

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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?

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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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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

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.

If the hill is frictionless, and after both Sally and Sam reach the bottom of the hill traveling at maximum velocity, who has lost more potential energy?

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Answer

When they reach the bottom of the hill, Sam and Sally have both converted all of their potential energy into kinetic energy. We measured potential energy at 50m, so once they have lost 50m, the potential energy is 0, while the kinetic energy has reached maximum value. Because Sam has greater mass, he had more potential energy to convert to kinetic energy.

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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.

After reaching maximum velocity at the bottom of a frictionless hill identical to the one in the question, Sam deploys a parachute. How much energy must the parachute dissipate before Sam comes to a stop?

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Answer

The parachute must dissipate all of the kinetic energy that Sam has at the bottom of the hill, since there is no friction to slow him. He has 30kJ of energy, as per the equation PE = mgh.

PE = 60kg * 10m/s2 * 50m = 30,000J = KE at the bottom of the hill

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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 $5\frac{m}{s}$ 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 collision 1, imagine that child 2 was not present on the ice. How much energy would child 1 have to dissipate to the lake surface before he came to a stop? Ignore wind resistance.

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Answer

The amount of kinetic energy that child 1 has after she reaches the lake surface is the amount of energy she will dissipate to the lake before coming to a complete stop.

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Question

A 5kg ball is attached to a 10m rope. The ball is held at a horizontal angle and allowed to fall freely with a pendulum-like motion. Assume there is no air resistance.

What is the velocity of the ball when it is at its lowest point?

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Answer

This problem becomes much easier if we use the law of conservation of energy. Because there is no friction involved, all of the ball's potential energy was converted into kinetic energy when falling through the pendulum arc. As a result, the velocity at the bottom of the swing can be determined using the equation below.

$U_{1} + K_{1} = U_{2} + K_{2}$

Because the potential energy at the bottom of the swing is zero (relative to the starting point), and the velocity at the top is zero, we can simplify the equation.

$mgh_1+\frac{1}{2}(v_1)^2=mgh_2+\frac{1}{2}(v_2)^2$

$mgh_1+\frac{1}{2}(0)^2=mg(0)+\frac{1}{2}(v_2)^2$

$mgh_1 = \frac{1}{2}m(v_2)^{2}$

Since the string is 10m long, we know that the lowest point will be 10m below the starting point.

$(5kg)(10\frac{m}{s^{2}})(10 meters) = \frac{1}{2}(5kg)v^{2}$

$v = 14.1\frac{m}{s}.$

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Question

How much energy is required to a accelerate a $\small 2kg$ block from rest to a final speed of $\small 5\frac{m}{s}$ ?

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Answer

From an energy stand-point, the block starts with zero kinetic energy and zero potential energy. At the end, the block still has zero potential energy, but has a non-zero kinetic energy. Assuming there is no friction, all energy added to the block have been converted to kinetic energy.

Using the change in velocity, we can solve for the energy used to move the block.

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

We are given our mass and the change in velocity, allowing us to solve for the change in kinetic energy.

$\Delta KE=\frac{1}{2}(2kg)(5\frac{m}{s}-0\frac{m}{s})^2$

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

$\Delta KE=25J$

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Question

What is the kinetic energy of a $\small 10g$ bullet moving at $\small 750\frac{m}{s}$ ?

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Answer

Kinetic energy is given by the formula:

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

We are given the mass of the bullet and its velocity, allowing us to calculate kinetic energy from this formula.

$KE=\frac{1}{2}(0.01kg)(750\frac{m}{s})^2$

Work is often expressed in Newton-meters, but energy is usually expressed as Joules, although the units are equivalent.

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