Physics - Motion and Mechanics

Your grandfather clock’s pendulum has a length of . If the clock loses half a minute per day, how should you adjust the length of the pendulum?

We should shorten the pendulum by

We should lengthen the pendulum by

We should shorten the pendulum by

We should length the pendulum by

We should lengthen the pendulum by

Explanation

We also can calculate the total number of seconds in a day.

$24h * \frac{60min}{1h} * \frac{60s}{1min} = 86400s$

There are seconds in one day.

Therefore we want our clock to swing a certain number of times with a period of to equal .

We know that our current clock has a certain number of swings with a period of to equal

So we have

We can calculate the current period of the pendulum using the equation

$T = 2\pi \sqrt{\frac{L}{g}}$

We can set up a ratio of each of these two periods to determine the missing length.

$\frac{nT_2}{nT_1} = \frac{2\pi \sqrt{\frac{L_2}{g}}}{2\pi \sqrt{\frac{L_1}{g}}$

Notice that 2, pi and g are all in both the numerator and denominator and therefore fall out of the problem.

$\frac{nT_2}{nT_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}}$

We can now solve for our missing piece.

$\frac{86400}{86360} = \frac{\sqrt{L_2}}{\sqrt{1}}$

$1.00046 = \sqrt{L_2}$

Square both sides to get rid of the square root.

We should lengthen the pendulum by

You are a witness in a court case involving a car accident. The accident involved car A of mass which crashed into stationary car B of mass . The driver of car A applied his brakes before he crashed into car B. After the collision, car A slid while car B slid 3. The coefficient of kinetic friction between the locked wheels and the road was measured to be . What was the velocity of car A before hitting the brakes?

Explanation

Knowns

$d_{Apre-collision} = 15m$

$d_{Apost-collision} = 20m$

$d_{Bpost-collision} = 30m$

$\mu = 0.8$

Unknowns

$v_{0A} = ?$

To solve the problem we must consider three different situations. The first is when the cars are skidding across the ground. While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.

The second situation is the collision itself when the first car has an initial speed and the second car is stopped. After the collision both cars are traveling with the different speeds as they skid across the ground at different distances.

The third situation is when the first car initially hits the brakes before the collision. While he is skidding to slow down, the force of friction is resisting the motion and therefore doing work to slow the car down some.

To solve this problem we must work from the end of the collision and work backwards to find out what happened before.

To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.

The work-kinetic energy theorem states that the work done is equal to the change in kinetic energy of the object.

$-W = \Delta KE$

Work is directly related to the force times the displacement of the object.

In this case, the force that is doing the work is friction.

$F_f = \mu F_N$

Since the cars are on a level surface the normal force is equal to the force of gravity.

The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.

When we put all these equations together we get

$W = \mu mgd$

We also know that kinetic energy is related to the mass and velocity squared.

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

Therefore our final equation should look like

$-\mu mgd = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_0^2$

Notice that the mass falls out of the equation since it is in every term. Also note that the final velocity is 0m/s so this will cancel out as well.

$-\mu gd = - \frac{1}{2} v_0^2$

Since both of these terms have a negative we can cancel this out as well.

$\mu gd = \frac{1}{2} v_0^2$

We can now substitute our variables to determine the velocity of each of the cars just after the crash.

$(0.8)(9.8m/s^2)(30m) = \frac{1}{2} v_{B0}^2$

$235.2 = v_{B0}^2$

$v_{B0} = 15.3m/s$

$(0.8)(9.8m/s^2)(20m) = \frac{1}{2} v_{A0}^2$

$156.8 = v_{A0}^2$

$v_{A0} = 12.5m/s$

This is the velocity of each of the cars after the collision. We must now consider our second situation of the collision itself. During this collision momentum must be conserved. The law of conservation of momentum states

$m_Av_{0A} + m_Bv_{0B} = m_Av_{fA} + m_Bv_{fB}$

We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found) and the initial value of the stopped car.

$(2000)v_{0A} + (1000g)(0m/s) = (2000kg)(12.5m/s) + (1000kg)(15.3m/s)$

Now we can solve for our missing variable.

$(2000kg)v_{0A} = 40300 kgm/s$

$v_{0A} = 20.15m/s$

Now we must consider our third situation which is when car A is braking. We can return to our work-kinetic energy theorem equation again as the force of friction is what is slowing the object down.

$-\mu mgd = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_0^2$

This time however, the final velocity is not 0m/s as the car A is still moving when he crashes into car B. The final velocity is the velocity that we determined the car was moving with before the collision in the second part of the problem. The masses do still cancel out in this case.

$-\mu gd = \frac{1}{2} v_f^2 - \frac{1}{2} v_0^2$

$-(0.8)(9.8m/s^2)(15m) = \frac{1}{2} (20.15m/s)^2 - \frac{1}{2} (v_0)^2$

$-117.6 = 203.01 - \frac{1}{2} (v_0)^2$

$-320.6 = -\frac{1}{2} (v_0)^2$

While traveling along a highway, a specific automobile is capable of an acceleration of about $1.6m/s^{2}$ . At this rate, how long does it take to accelerate from to ?

Explanation

Knowns:

$a = 1.6m/s^2{}$

$v_{0}= 65km/h$

$v_{f}= 120km/h$

Unknowns:

$\Delta t = ?$

Equation:

$a=\frac{{}v_{f} - v_{0}}{\Delta t}$

The most important thing to recognize here is that the initial and final velocities are not in the correct units. The first step is to convert both of these values to .

$\frac{65km}{h} * \frac{1h}{60 min} * \frac{1 min}{60 s} *\frac{100m}{km} = 18.06m/s$

$\frac{120km}{h} * \frac{1h}{60 min} * \frac{1 min}{60 s} *\frac{100m}{km} = 33.33m/s$

Then rearrange your equation to solve for (the missing variable).

${\Delta t}=\frac{{}v_{f} - v_{0}}{a}$

Now plug in the variables and solve.

${\Delta t}=\frac {33.33m/s - 18.06m/s}{1.6m/s^2}$

$\Delta t = 9.55s$

A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v. Is the normal force that the seat exerts on the rider at the top of the wheel?

Less than the force the seat exerts at the bottom of the wheel

More than the force the seat exerts at the bottom of the wheel

The same as the force the seat exerts at the bottom of the wheel

Explanation

The centripetal force is what is acting on the rider. At the top of the Ferris wheel, the normal force is pointing up, and the gravitational force is pointing down. The sum of these two forces must equal the centripetal force pointing downward toward the center of the circle. Therefore the normal force must be smaller than the gravitational force. At the bottom of the Ferris wheel, the same forces are present. However, the sum of these forces must equal the centripetal force point upward toward the center of the circle. Therefore the normal force must be greater than the gravitational force. Since the normal force must be greater than the gravitational force at the bottom and less than the gravitational force at the top, the force at the bottom must be greater than the force on the top.

A person jumps from the roof of a house high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of . If the mass of his torso (excluding legs) is , find the average force exerted on his torso by his legs during deceleration.

Explanation

To solve this problem we need to divide up the situation into two parts. During the first part the person is jumping from the roof of a house and is therefore undergoing freefall and accelerated motion due to the force of gravity. Therefore we will need to use kinematic equations to solve for the final velocity as the person lands. In the second part of the problem, the person decelerates their torso through a specific distance and comes to a stop. We will then need to calculate the acceleration of the torso during this second part to determine the average force applied.

Let us start with kinematics to determine the speed of the torso as it hits the ground.

Knowns

We can use the kinematic equation

$v_f^2 = v_i^2 +2a\delta x$

$v_f = \sqrt{ 0 + (2)(-9.8m/s^2)(-3.2m)}$

This is the velocity of the torso as it hits the ground. We will know use the same equation with new variables to determine the acceleration of the torso as it comes to a stop.

Knowns

$v_f^2 = v_i^2 +2a\delta x$

Rearrange to get the acceleration by itself

$\frac{v_f^2-v_i^2}{2\delta x} = a$

$a = \frac{0-(7.9m/s)^2}{2(0.65)$

We can now plug this into Newton’s 2nd Law to find the average force acting on the object.

Two skaters push off of each other in the middle of an ice rink. If one skater has a mass of and an acceleration of , what is the acceleration of the other skater if her mass is ?

Explanation

For this problem, we'll use Newton's third law, which states that for every force there will be another force equal in magnitude, but opposite in direction.

This means that the force of the first skater on the second will be equal in magnitude, but opposite in direction:

Use Newton's second law to expand this equation.

We are given the mass of each skater and the acceleration of the first. Using these values, we can solve for the acceleration of the second.

From here, we need to isolate the acceleration of the second skater.

Notice that the acceleration of the second skater is negative. Since she is moving in the opposite direction of the first skater, one acceleration will be positive while the other will be negative as acceleration is a vector.

A satellite orbits above the Earth. The satellite runs into another stationary satellite of equal mass and the two stick together. What is their resulting velocity?

$m_E=5.97x10^{24}kg$

$G=6.67x10{-11}m^3/kgs$

$1.33x10^{11}m/s$

$2.7x10^{-3}m/s$

Explanation

We can use the conservation of momentum to solve. Since the satellites stick together, there is only one final velocity term.

We know the masses for both satellites are equal, and the second satellite is initially stationary.

Now we need to find the velocity of the first satellite. Since the satellite is in orbit (circular motion), we need to find the tangential velocity. We can do this by finding the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means .

Solve for for the satellite. To do this, use the law of universal gravitation.

$F=\frac{Gm_1m_2}{r^2}$

Remember that r is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=\frac{Gm_sm_E}{(r_E+h)^2}$

$F_G=\frac{(6.67x10^{-11}m^3/kgs)(2500kg)(5.97x10^{24}kg)}{(6.37x10^5m+3.8x10^8m)^2}$

$F_G=\frac{(6.67x10^{-11}m^3/kgs^2)(1.49)(1028kg)}{21.45x10^{17}m^2}$

$F_G={(6.67x10^{-11}m^3/kgs^2)(1.03x10^{11}kg^2m^2)$

Now that we know the force, we can find the acceleration. Remember that centripetal force is Fc=m∗ac. Set our two forces equal and solve for the centripetal acceleration.

$2.7x10^{-3}m/s^2=a_c$

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

$a_c=\frac{v^2}{r}$

$2.7x10^{-3}m/s^2=\frac{v^2}{1.45x10^{17}m}$

$(2.7x10^{-3}m/s^2)(1.45x10^{17}m)=v^2$

$2.85x10^{15}m^2/s^2=v^2$

$\sqrt{2.85x10^{15}m^2/s^2}$

This value is the tangential velocity, or the initial velocity of the first satellite. We can plug this into the equation for conversation of momentum to solve for the final velocity of the two satellites.

$1.3325x10^{11}kgm/s=(5000kg)v_3$

$\frac{1.3325x10^{11}kgm/s}{5000kg}=v_3$

The area under the curve on a Force versus time (F versus t) graph represents

Impulse

Momentum

Work

Kinetic energy

Explanation

If we were to examine the area under the curve of a constant force applied over a certain amount of time we would have a graph with a straight horizontal line. To find the area of that rectangle we would multiply the base times the height. The base would be the time (number of seconds the force was applied). The height would be the amount of force applied during this time. Force*time is equal to the impulse acting on the object which is equal to the change in momentum of the object.

Two equal mass balls (one green and the other yellow) are dropped from the same height and rebound off of the floor. The yellow ball rebounds to a higher position. Which ball is subjected to the greater magnitude of impulse during its collision with the floor?

It’s impossible to tell since the time interval and the force have not been provided in the problem

Both balls were subjected to the same magnitude of impulse

The yellow ball

The green ball

Explanation

The impulse is equal to the change in momentum. The change in momentum is equal to the mass times the change in velocity.

The yellow ball rebounds higher and therefore has a higher velocity after the rebound. Since it has a higher velocity after the collision, the overall change in momentum is greater. Therefore since the change in momentum is greater, the impulse is higher.

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $3\frac{m}{s}$ , what is the force of friction?