Motion in Two Dimensions - MCAT Physical

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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

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 high does the ball fly?

Answer

First, solve for the initial vertical velocity of the ball.

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

Think back to the 3 main kinematics equations we know:

vf2 = vi2 + 2aΔx

vf = vi + at

Δx = vit + ½at2

We need to choose an equation that allows us to solve for the vertical height of the ball, given that we know the initial vertical velocity, final vertical velocity (zero), and the acceleration due to gravity (-9.8m/s2).

vf2 = vi2 + 2aΔy

Δy = (vf2 - vi2)/2a

$\Delta y=\frac{(0\frac{m}{s})^2-(7.1\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=2.57m$

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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 potential energy of the ball at the top of its arc?

Answer

This question asks us to consider gravitational potential energy. We are provided the mass of the ball (2kg) and know the acceleration due to gravity (9.8m/s2). Using the height of the ball at the top of its arc, we can solve for gravitational potential energy.

First, solve for the initial vertical velocity of the ball.

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

Think back to the 3 main kinematics equations we know:

vf2 = vi2 + 2aΔx

vf = vi + at

Δx = vit + ½at2

We need to choose an equation that allows us to solve for the vertical height of the ball, given that we know the initial vertical velocity, final vertical velocity (zero), and the acceleration due to gravity (-9.8m/s2).

vf2 = vi2 + 2aΔy

Δy = (vf2 - vi2)/2a

$\Delta y=\frac{(0\frac{m}{s})^2-(7.1\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=2.57m$

Now that we know the maximum height, we can find the potential energy from this position.

U = mgh

U = (2kg)(9.8m/s2)(2.57m) = 50.4J

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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 long does it take for the ball to land on the cushion?

Answer

Once again, think back to our 3 main kinematics question:

vf2 = vi2 + 2aΔx

vf = vi + at

Δx = vit + ½at2

We want an equation that allows us to determine the time the ball spends in flight. Often, it is easier to compute the time it takes for the ball to reach its maximum height, then multiply by two. However, this “shortcut” only works when the launch point and end point are at the same vertical height, as they are in this problem.

vf = vi + at

t = (vf – vi)/a

Remember to use the veritcal velocity as your initial velocity.

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

$t=\frac{(0\frac{m}{s})-(7.1\frac{m}{s})}{-9.8\frac{m}{s^2}}=0.72s$

This is one half of the flight time. We multiply by two to find the total time.

t = 2(0.72s) = 1.44s

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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.

In order to catch the ball in the center of the cushion, how far away should the cushion be placed?

Answer

First, we will need the time that the ball is in flight.

Think back to our 3 main kinematics question:

vf2 = vi2 + 2aΔx

vf = vi + at

Δx = vit + ½at2

We want an equation that allows us to determine the time the ball spends in flight. Often, it is easier to compute the time it takes for the ball to reach its maximum height, then multiply by two. However, this “shortcut” only works when the launch point and end point are at the same vertical height, as they are in this problem.

vf = vi + at

t = (vf – vi)/a

Remember to use the veritcal velocity as your initial velocity.

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

$t=\frac{(0\frac{m}{s})-(7.1\frac{m}{s})}{-9.8\frac{m}{s^2}}=0.72s$

This is one half of the flight time. We multiply by two to find the total time.

t = 2(0.72s) = 1.44s

Using this time with the initial horizontal velocity will allow us to solve for the total horizontal distance.

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

Δx = vit + ½at2

Δx = (7.1m/s)(1.44s) = 10.22m

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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 are the horizontal and vertical velocities of the ball right before it hits the ground?

Answer

The final and initial velocities are the same if air resistance can be neglected and the start and end positions are at the same height. Kinetic and potential energy must be conserved if there are no frictional forces acting against motion, so the final and initial velocities must be the same.

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

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

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Question

An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?

Answer

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use $v_y = v_o sin\theta$ .

$v_y = (75 \frac{m}{s})sin 45^o = 53 \frac{m}{s}$

Next we find how long it takes to reach the top of its trajectory using .

$0 \frac{m}{s} = 53 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 5.3s

Finally, find how high the object goes with $d = v_ot +\frac{1}{2}at^{2}$ .

$d = (53\frac{m}{s})5.3s + \frac{1}{2}(-10\frac{m}{s^2})(5.3s)^{2} = 140 m$

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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.

In Scenario 3, before they begin moving, Sam is hanging down the hill facing forward, and Sally has fixed her shoes in the snow to keep her and Sam from beginning down the hill together. If the hill is at a 30o angle to the horizontal, and Sam is stationary but tied to Sally by a rope, how much force does Sam experience down the surface of the hill?

Answer

Sam will experience force down the surface of the inclined plane as per the equation $F=mgsin(\theta)$ .

F = (60kg)(10m/s2) * sin(30o) = 300J

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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.

In Scenario 3, before they begin moving, Sam is hanging down the hill face first, and Sally has fixed her shoes in the snow to keep her and Sam from beginning down the hill together. If the hill is at a 30o angle to the horizontal, and Sam is stationary but tied to Sally by a rope, how much tension is in the rope?

Answer

Sam is pulling down the inclinced plane with a force given by the equation $F=mgsin(\theta)$ .

F = 60kg * 10m/s2 * sin(30) = 300J

Tension in the rope will be equal to this value, as these forces must cancel to result in a net force of zero.

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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.

Which of the following expressions shows the force that Sally and Sam would feel down the hill at a 30o incline if they both rode on Sam’s sled together?

Answer

This is a tricky question. The total mass of Sally, Sam, and Sam's sled is 100kg. The force down the incline is .

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Question

An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?

Answer

First, find the horizontal (x) and vertical (y) components of the velocity

$v_x = v_o cos\theta$

$v_x = (125 \frac{m}{s}) cos(30^o) = 108\frac{m}{s}$

$v_y = v_o sin\theta$

$v_y = (125\frac{m}{s})sin(30^o) = 62.5\frac{m}{s}$

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

$0 \frac{m}{s} = 62.5 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 6.25s

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

$d_x = 108 \frac{m}{s} (12.5s) = 1,350 m$

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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.

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 box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline?

Answer

To find the acceleration of the box traveling down the incline, the mass is not needed. Using the incline of the plane as the x-direction, we can see that there is no movement in the y-direction; therefore, we can use Newton's second, F = ma, in the x-direction.

There is only one force in the x-direction (gravity), however gravity is not just equal to “mg” in this case. Since the box is on an incline, the gravitational force will be equal to mgsin(30o). Substituting force into F =ma we find that mgsin(30o) = ma. We can now cancel out masses and solve for acceleration.

$F=ma\ and\ F=mgsin(\theta)$

$ma=mgsin(\theta)$

$ma=m(10\frac{m}{s^2})sin(30^o)=m(10\frac{m}{s^2})(0.5)$

$a=5\frac{m}{s^2}$

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Question

A cannon fires a cannon ball at an angle of 60o relative to the ground. The cannon ball comes out at an initial velocity of $\small 120 \frac{m}{s}$ . Assume there is no air resistance.

What is the cannon ball's maximum height off of the ground?

Answer

In order to find the maximum height of the cannonball from the ground, we can use the equation $\small v_{0}sin(\theta) = \sqrt{2gh}$ , with g being the acceleration due to gravity, and h being the height.

$\small 120sin(60^{\circ}) = \sqrt{2(10)(h)}$

$\small h = 540 m$

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Question

A cannon fires a cannon ball at an angle of 60o relative to the ground. The cannon ball comes out at an initial velocity of $\small 120 \frac{m}{s}$ . Assume there is no air resistance.

How long will the cannon ball stay in the air?

Answer

In order to solve for the amount of time, we need to know the vertical initial velocity of the cannonball. This is given by the equation $\small v_{0}sin(60^{\circ})$ . Using the initial velocity of $\small \small 120\frac{m}{s},$ we determine that the vertical initial velocity is $\small 104\frac{m}{s}.$

Knowing this, we can solve the amount of time it will take for the cannon ball to reach its maximum peak, with a velocity of $\small 0\frac{m}{s}$ using the equation $\small v = v_{0} + at$

$\small 0 = 104\frac{m}{s} + (-10\frac{m}{s^{2}})(t)$

$\small t = 10.4 s$

This is the amount of time it takes for the ball to reach its peak, not the total time it is in the air. Since the ball must drop from the maximum height, we double the time and arrive at the correct answer of 20.8s.

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Question

A cannon fires a cannon ball at an angle of relative to the ground. The cannon ball comes out at an initial velocity of $\small 120 \frac{m}{s}$ . Assume there is no air resistance.

How far will the ball be from the cannon once it hits the ground?

Answer

We must first determine the amount of time that the ball is in the air, then use the horizontal velocity to solve for the distance travelled.

In order to solve for the amount of time, we need to know the vertical initial velocity of the cannonball. This is given by the equation $\small v_{0}sin(60^{\circ})$ . Using the initial velocity of $\small \small 120\frac{m}{s},$ we determine that the vertical initial velocity is $\small 104\frac{m}{s}.$

Knowing this, we can solve the amount of time it will take for the cannon ball to reach its maximum peak, with a velocity of $\small 0\frac{m}{s}$ using the equation $\small v = v_{0} + at$

$\small 0 = 104\frac{m}{s} + (-10\frac{m}{s^{2}})(t)$

$\small t = 10.4 s$

This is the amount of time it takes for the ball to reach its peak, not the total time it is in the air. Since the ball must drop from the maximum height, we double the time and find that total flight time is 20.8s.

The next step is determining the horizontal initial velocity from the cannon. This is determined using the equation below.

$\small v_{0}cos(60^{\circ}) = 120cos(60^{\circ}) = 60\frac{m}{s.}$

Since we assume that there is no air resistance, we can conclude that the horizontal velocity of the cannonball will remain constant throughout its flight. As a result, the total displacement equation is simplified to the following: $\small x = x_{0} + v_{0}t$ .

$\small x = 0 + (60\frac{m}{s})(20.8 s)$

$\small x = 1250 m$

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Question

A cannon fires a cannon ball at an angle of 60o relative to the ground. The cannon ball comes out at an initial velocity of $\small 120 \frac{m}{s}$ . Assume there is no air resistance.

Which of the following statements is true?

Answer

The only acceleration on the cannon ball is done by gravity, which is constant throughout the ball's flight. Since there is no air resistance, the horizontal velocity of the ball is also constant, however, the vertical velocity of the ball is not constant, because the velocity of the ball changes throughout its flight.

It is $\small 103.9\frac{m}{s}$ at the beginning ( $\small v_{0}sin(60^{\circ})$ )and is $\small 0\frac{m}{s}$ at the maximum height of the cannon ball.

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Question

A 2kg box is at the top of a frictionless ramp at an angle of . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

When the box is released, how long will it take the box to reach the ground?

Answer

We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation $\small cos(60^{\circ}) = \frac{30}{x}$ .

$\frac{1}{2}=\frac{30m}{x}\rightarrow x=60m$

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of $\small gsin\Theta$ . Since the angle is 60o, the acceleration on the box is $\small 8.66 \frac{m}{s^{2}}.$

$gsin\Theta=(10)sin(60)=8.66\frac{m}{s^2}$

Finally, we can plug these values into the following distance equation and solve for time.

$\small x = x_{0} - \frac{1}{2}(gsin\Theta )t^{2}$

$\small 0 = 60 - \frac{1}{2}(8.66)t^{2}$

$\small \small t = 3.72 s$

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