Mechanics - SAT Subject Test in Physics

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Question

A book falls off the top of a bookshelf. What is its velocity right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Answer

The relationship between velocity and energy is:

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

We know the mass, but we need to find the total kinetic energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

$PE=2kg-9.8\frac{m}{s^2}-2.3m$

The kinetic energy will also equal , due to conservation of energy.

Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation.

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

$45.08J=\frac{1}{2}2kgv^2$

$45.08\frac{m^2}{s^2}=v^2$

$\sqrt{45.08\frac{m^2}{s^2}}=\sqrt{v^2}$

Since we are taking the square root, our answer can be either negative or positive. The final velocity of the book will be in the downward direction; thus, our final velocity should be negative.

$-6.71\frac{m}{s}=v$

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Question

A model rocket, launched vertically, travels upwards and falls to the ground. At what point during flight is the rocket's acceleration greatest?

Answer

Newton's second law tells us that force and acceleration are directly related; if there is an acceleration, then there is also a force. This principle can help conceptualize this question.

While the rocket is in the air, there is only one force acting on it: the force of gravity. We can thus conclude that the acceleration of the rocket is directly related to this force. Since the force on the rocket (the force of gravity) is constant, its acceleration is also constant.

Any object that is in projectile or free-fall motion will experience a constant acceleration due to gravity.

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Question

A crate slides across a floor for before coming to rest from its original position.

What is the force due to friction?

Answer

Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation . The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

$4\frac{m}{s}=v_i$

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

$\Delta x=v_it+\frac{1}{2}at^2$

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

Now that we have the acceleration and the mass, we can solve for the force of friction.

$F=(2.9kg)(-0.8\frac{m}{s^2})$

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Question

If the distance between two objects is reduced by two-thirds, how will the gravitational force between the objects be affected?

Answer

According to Newton's law of universal gravitation, the gravitational force between two objects is inversely proportional to the square of the distance between them.

$F_G=G\frac{m_1m_2}{d^2}$

If the distance is reduced by two-thirds, then the final distance is equal to one-third the original distance.

$d_2=\frac{1}{3}d_1$

Using this term in the equation will show that the force increases by a factor of nine.

$F_G=G\frac{m_1m_2}{(\frac{1}{3}d_1)^2}=G\frac{m_1m_2}{\frac{1}{9}d_1^2}=9(G\frac{m_1m_2}{d_1^2})$

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Question

An egg falls from a nest in a tree that is tall. A girl, away, runs to catch the egg. If she catches it right at the moment before it hits the ground, how fast does she need to run?

$\small g=-9.8\frac{m}{s^2}$

Answer

The important thing to recognize here is that the amount of time the egg is falling will be equal to the amount of time the girl is running.

Our first step will be to find the time that the egg is in the air.

We know it starts from rest above the ground, and we know the gravitational acceleration. Its total displacement will be , since it falls in the downward direction. We can use the appropriate motion equation to solve for the time:

$\Delta y =v_it+\frac{1}{2}at^2$

Use the given values in the formula to solve for the time.

$-2.8m=(0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=(-4.9\frac{m}{s^2})t^2$

$\frac{-2.8m}{-4.9\frac{m}{s^2}}=t^2$

Now that we have the time, we can use it to find the speed of the girl. Her speed will be determined by the distance she travels in this amount of time.

$v=\frac{d}{t}$

Use our values for her distance and the time to solve for her velocity.

$v=\frac{13m}{0.755s}$

$v=17.22\frac{m}{s}$

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Question

A ball is dropped from the roof a building that is tall. How long will it be before the ball hits the ground?

Answer

We can solve this problem using the kinematics equations. Note that the initial velocity will be zero, since the ball is dropped, and the acceleration will be equal to the acceleration of gravity.

$d=v_ot+\frac{1}{2}at^2$

This can be simplified, since the initial velocity is zero.

$d=\frac{1}{2}at^2$

Use the given value for the distance (the height of the building) and the acceleration of gravity to solve for the time.

$80m=\frac{1}{2}(9.8\frac{m}{s^2})t^2$

$\frac{80m}{\frac{1}{2}(9.8\frac{m}{s^2})}=16.3s^2=t^2$

$t=\sqrt{16.3s^2}=4.0s$

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Question

If the block of mass slides up a frictionless incline plane and is pulled by mass which is falling then what is the acceleration of the block on the ramp if and and the angle of incline is 30 degrees?

Answer

Begin by making a free body diagram for each block:

Use the diagram to write an equation for net force on each block:

$F_{net} =F_{t}-mg \sin (30)$

$F_{net} = Mg -F_{t}$

Since for the block on the ramp and for falling block we can substitute into these equations:

$ma =F_{t}-mg \sin (30)$

$Ma = Mg -F_{t}$

Then add the equations to get:

$Ma+ma = Mg-mg\sin (30)$

Next rearrange to isolate :

$a = \frac{M-msin(30)}{M+m}$

Substitute the given values from the question and solve:

$a = \frac{40-20sin(30)}{40+20}$

$a=\frac{1}{2}g$

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Question

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

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

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

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Question

A ball hits a brick wall with a velocity of $30\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for , what is the value of the force exerted by the wall on the ball?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Even though the ball is bouncing back at the same "speed" its velocity will now be negative, as it is moving in the opposite direction. Using this understanding we can solve for the force in our equation.

$(0.25kg* -30\frac{m}{s})-(0.25kg* 30\frac{m}{s})=F(0.1s)$

$(-7.5\frac{kg\cdot m}{s})-(7.5\frac{kg\cdot m}{s})=F(0.1s)$

$(-15\frac{kg\cdot m}{s})=F(0.1s)$

$\frac{(-15\frac{kg\cdot m}{s})}{0.1s}=F$

Our answer is negative because the force is moving the ball in the OPPOSITE direction from the way it was originally heading.

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Question

Near the surface of the earth, a projectile is fired from a canon at an angle of $\theta$ degrees above the horizon and an initial velocity of meters per second. Which of the following expressions gives the time it takes the projectile to reach its maximum height?

Answer

At the maximum height of projectile motion, $v_{y}=0$ and because this takes place near the surface of the earth we know which we can plug into the equation:

$v_{y} = v_{0} + at$

$0 = v_{0} + gt$

We can then rearrange for

$t = \frac{v_{0}}{g}$

Next substitute this value into the equation $v_{y} = v \sin \theta$ to get the correct answer

$t = v \sin \frac{\theta }{g}$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

How long is the rock in the air?

Answer

We first need to find the vertical component of the velocity.

$v_{i}\sin(\Theta)=v_{iy}$

We can plug in the given values for the angle and the initial velocity to find the vertical component.

$30\frac{m}{s} \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the horizontal distance that the rock travels?

Answer

We first need to find the horizontal component of the initial velocity.

$v_{i}* \cos(\Theta)=v_{x}$

We can plug in the given values for the angle and initial velocity and solve.

$30\frac{m}{s}* \cos(35^{\circ})=v_{x}$

$30\frac{m}{s}* (0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity.

$\Delta x =v_xt$

We know , but now we need to find the time the rock is in the air.

We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance.

$\Delta x =v_xt$

$\Delta x =24.57\frac{m}{s}* 4.02s$

$\Delta x=98.77m$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what angle to the horizontal will the rock impact the ground?

Answer

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i} \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

$\tan^{-1}(\frac{v_y}{v_x})=\Theta$

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

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Question

If air resistance is negligible, 8 seconds after it is released, what would be the velocity of a stone dropped from a helicopter that has a horizontal velocity of 60 meters per second?

Answer

We are looking for total velocity, which in this case has both a horizontal and vertical component.

Because the helicopter is flying horizontally we know $v_{y0}=0$

We can assume that this takes place near the surface of the earth so

We can plug this into the equation:

$v_{yf}=v_{y0} +at$

$v_{yf}=0 +(10)(8)$

$v_{yf}=80$

Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so:

$v_{x}=60$

Next we must use vector addition to add the horizontal and vertical components of velocity. Because this horizontal and vertical velocities are perpendicular the sum will be the hypotenuse of a right triangle:

$60^{2}+80^{2}=v^{2}$

$v=100\frac{m}{s}$

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Question

Suppose a planet has a mean distance from the sun four times that of Earth's. How many Earth years would it take this planet to orbit the sun?

Answer

Kepler's third law gives the relationship between the period of a planet's orbit (year) and the radius of its orbit (distance form the sun):

$T^2\propto R^3$

The square of the period is proportional to the cube of the radius. We can set the product of these values equal to a constant, since they are proportional.

$T^2R^3=\text{constant}$

In our question, we know that the distance is increased by a factor of four and the period for Earth is 1 year.

Solve for the period of the new planet.

$\frac{(R_E)^3}{64(R_E)^3}=(T_2)^2$

$\frac{1}{64}=(T_2)^2$

$\frac{1}{8}=T_2$

The period for the new planet is one-eighth the period of Earth, meaning that in one Earth year this planet will complete only one-eighth its orbit. It will take eight Earth years to equal one complete orbit for this planet.

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Question

Which of these is not an example of Newtonian mechanics?

Answer

Newtonian mechanics apply to all objects of substantial mass travelling at significantly slower than the speed of light.

Newton's law of universal gravitation, Newton's second law, momentum, and the equation for mechanical energy all fall under Newtonian mechanics.

The mass-energy equivalence suggests that mass can change as the speed of an object (such as an electron) approaches the speed of light. Newtonian mechanics assume that mass is constant, and do not apply to objects approaching the speed of light.

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Question

A wrench is being used to tighten a bolt. The center of the bolt is the pivot point, and the wrench is long. If a force of is applied at the midpoint of the wrench and that force is parallel to the wrench, what is the torque?

Answer

The equation for torque is

$\tau=r\perp F$

The force is applied meters from the pivot. and in this case because the force is parallel, the perpendicular component of force is zero, so:

$\tau= (.5)(0)$

$\tau=0N\cdot m$

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Question

A girl bounces on a massless pogo stick. If the spring constant for the stick is $15,000\frac{N}{m}$ , what is the maximum compression of the spring?

$g=-9.8\frac{m}{s^2}$

Answer

There are two forces at work here: the force due to gravity and the restoring force of the spring. We can set these two forces equal to one another because the forces must be in equilibrium when the spring is compressed at its maximum point.

Expand this equation by using the formulas for gravitational and spring force, respectively.

Plug in our given values for the girl's mass, gravitational acceleration, and the spring constant. Using these values, we can solve for the displacement of the spring.

$(36kg)(-9.8\frac{m}{s^2})=-(15,000\frac{N}{m})(x)$

$-352.8N=-15,000\frac{N}{m}*x$

$\frac{-352.8N}{-15,000\frac{N}{m}}=x$

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Question

A certain spring has an equilibrium length of . When of force are applied, the spring length becomes . What is the spring constant?

Answer

The equation for spring force is:

The force applied to the spring will be equal and opposite the force of the spring itself. In this case, the force applied is , so the force of the spring will be .

The change in distance will be , or . Using this value, we can solve for the spring constant.

$k=\frac{20N}{0.10m}=200\frac{N}{m}$

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Question

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Answer

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

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