13 April 2010

Conservation laws

Problem 1.
If a 5 tons beam is raised 6 meters, what is it's potential energy? If it is raised another 8 meters in 6 seconds, what is the work done?

Solution:
The potential energy is
 
If we raise it another 8 meters then the work, which will be done, is equal to increase of the potential energy (it does not depend on the traveled time):
 

Problem 2.
(University Physics, 12th Edition, Young & Freedman 6.33) A 10-kg box moving at 5 m/s on a horizontal, frictionless surface runs into a light spring of force constant 100 N/cm. Use the work-energy theorem to find the maximum compression of the spring.

Solution:
We need to write down the energy conservation in the present problem:
The initial energy is just the kinetic energy of the box:

The final energy is the elastic energy of the spring:

Then
and
 

Problem 3.
A cat stuck up a tree and has 500 J gravitational potential energy. It then falls. Find the kinetic energy of the cat just before it is caught by the owner. Find the kinetic energy of the cat after it is caught by the owner. Find the wasted energy (after cat is caught by the owner).
Disregard air resistance.

Solution:
As the cat falls there is a transformation of potential energy into kinetic energy. Just before the cat is caught by the owner the kinetic energy is exactly equal to the initial potential energy = 500 J.

After the cat is caught by the owner, its velocity is 0. It means that the kinetic energy is equal to 0 (no motion).

The wasted energy is defined as an initial energy minus the final energy. The initial energy is 500 J, the final energy is 0. Then the wasted energy is 500 J. 

Problem 4.
A 2000 kg truck is traveling east through an intersection at 2 m/s when it is hit simultaneously from the side and the rear. One car is a 1000 kg compact traveling north at 5 m/s.  The other car is a 1500 kg midsize traveling east at 10 m/s.  The three vehicles become entangled and slide at one body.  What are their speeds and direction just after the collision?

Solution:
This is the example of perfectly inelastic collision: after the collision all parts of the system move as a whole. To find the final velocity we need to use the conservation of the net momentum.
The initial net momentum has three contributions:
- momentum of 2000 kg truck;
- momentum of 1000 kg compact;
- momentum of 1500 kg midsize.
We show these vehicles and the directions of the corresponding velocities (before the collision).
 
 
Then the (initial) net momentum is

In the final state all vehicles move with the same velocity. Then the final momentum is
Then momentum conservation takes the form
We need to rewrite this vector equation in terms of components. We introduce axis x - east direction, and axis y – north direction. Then the x- component of the above equation is
Then
 
The y- component of the above vector equation is
Then

 
Now we know x and y components of the final velocity. Then the magnitude of the final velocity is


The direction of the final velocity is characterized by the angle between the direction of velocity and axis x (east):


Problem 5.
A small body of mass is attached to a light thread which passes through a hole at the centre of a smooth table. The body is set into rotation in a circle of radius , with a speed of . The thread is then pulled down shortening the radius of the path to . What will the new linear speed and the new angular speed be?

Solution:
In the present problem we have conservation of the angular momentum. The initial angular momentum is
The final angular momentum is
Since then
From this equation we can find the final speed of the body
The angular speed can be expressed through the linear speed and the radius of circle:
Then



Problem 6.
A metal surface is illuminated one by one by photons of energy 2 eV and 4 eV respectively. The work function of the metal is 0.5 eV. What is the ratio of the maximum speeds of electrons emitted in two cases?

Solution:
In this problem we need to use the energy conservation. The work function of the metal can be considered as the potential energy of electron. Then the conservation law in the present problem takes the form: the energy of the photon is equal to the mechanical energy of electron:
The kinetic energy is
Then
and
Then the ratio of the speeds of the electrons in two cases is
 

Problem 7.
Ball A, with a mass of 2 kg, moves with a velocity 5 m/s. It collides with a stationary ball B, with a mass of 4 kg. After the collision, ball A moves in a direction 60.0 degrees to the left of its original direction, while ball B moves in a direction 50.0 degrees to the right of ball A's original direction. Calculate the velocities of each ball after the collision.

Solution:
In the present problem we need to use the momentum conservation law: the net momentum of two balls before the collision is equal to the net momentum of two balls after the collision. It is important that momentum is a vector. Then the net momentum of two balls is a vector sum of the momentum (vector) of ball A and the momentum (vector) of ball B.
Therefore, the initial net momentum of two balls is
A momentum of a ball is a product of a mass of a ball and its velocity. Then
 
The final momentum is
 
Then the momentum conservation law takes the form
or
.......................(1)

 

 
To solve this vector equation we need to introduce coordinate system (axis x and axis y) as shown in the figure.
We follow the standard procedure of finding the components of vectors and rewrite the vector equation (1) in terms of x and y-components:
x-component of equation (1):
y-component of equation (1):
 
Finally, we have system of two equations with two unknown variables
 
From the second equation we obtain
We substitute this expression in the first equation:
Then
Therefore the velocities of the balls after collision are 22.9 m/s and 12.8 m/s (the initial kinetic energy is less than the final energy, which means that during the collision there is an additional source of energy). 

Problem 8.
A toy car “1” of mass 0.30 kg moves along a frictionless surface with a velocity of 0.20 m/s. It collides with another toy car “2”, with a mass of 0.40 kg and a speed of 0.10 m/s in the same direction. After the collision, toy car “1” continues to move in the same direction with a velocity of 0.15 m/s. Calculate the speed of toy car “2” after the collision.

Solution:
This problem is similar to problem 7.
We need to use the momentum conservation law: the net momentum of two toy cars before the collision is equal to the net momentum of two cars after the collision.
Then the net momentum of two cars is a vector sum of the momentum of car “1” and the momentum of car “2”.
Therefore, the initial net momentum of two cars is
A momentum of a car is a product of a mass of a car and its velocity. Then
 
The final momentum is
 
Then the momentum conservation law takes the form
or
............................(1)
 
 
To solve this vector equation we need to introduce coordinate system. Since we have motion along a line then it is enough to introduce only one axis: axis x (as shown in the figure).

We follow the standard procedure of finding the components of vectors and rewrite the vector equation (1) in terms of x-components:
x-component of equation (1):
Then

Problem 9.
A skater of mass 80 kg initially at rest speeds up to a final speed of 10.0 m/s along a straight line and towards the East direction.
(a) Find the momentum of the skater while at rest.
(b) Find the momentum of the skater while traveling with its final speed.
(c) Find the change in momentum of the skater.
(d) Find the impulse acted on the skater.
(e) If that impulse exerted on the skater acts for 4 s, find the average force acting on the skater.

Solution:
In this problem we need to use the definition of momentum and impulse.
Momentum of an object is a vector, which is defined as the product of a mass of the object and its velocity:
Therefore the direction of momentum is the same as the direction of velocity and the magnitude of momentum is equal to the product of the mass of the object and its speed (the magnitude of velocity).

(a) If the skater is at rest then its speed is zero. Then the momentum of the skater is 0.

(b) The final speed of the skater is10 m/s, its mass is 80 kg, then the final momentum is
 
(c) The change in the momentum is equal to the difference between the final momentum and initial momentum:

(d) The impulse acted on the skater is equal to the change of its momentum:

(e) At the same time he impulse acted on the skater is equal to the product of the (average) force acted on the skater and the time
We know time: . Then we can find the force:
 

Problem 10.
A constant force of 80 N acts for 8 s on a box of mass 10 kg horizontally that initially rests on a horizontal frictionless surface.
(a) Find the change in the box's momentum.
(b) Calculate the final speed of the box after the 8 s have passed.

Solution:
In this problem we need to use the definition of an impulse of a force and the relation between the impulse and the chance of the momentum of the box.
If a constant force acts on a box for a time then the impulse of the force is
In the present problem a constant force of 80 N acts for 8 s. Then and . Then the impulse is

(a) The change in the box's momentum is equal to the impulse of the force acting on the box:

(b) The initial speed of the box is zero. Then the initial momentum of the box is zero. The change of the momentum is equal to the difference between the final momentum and the initial momentum (which is zero). Then

Now we know the final momentum of the box. To find the final speed we just need to divide the final momentum by the mass of the box:
 

Problem 11.
At an amusement park there is a roller coaster ride ("ride of Steel"). After the first drop, riders are moving at the speed of 120 km/h, entering an underground tunnel. Given the fact that the roller coaster was moving at a speed of 4 km/h at the top of the hill, determine the vertical drop that these participants fell through. Neglect friction.

Solution:
In the problem we need to use energy conservation law: since the friction is zero then the mechanical energy, which is the sum of kinetic energy and gravitational potential energy, is constant.
We need to use the correct SI units, which means that the speed should be measured in m/s. Then

Now we need to write the energy conservation law. The initial mechanical energy is the sum of initial potential energy and the initial kinetic energy:
The final energy is

The initial energy is equal to the final energy. Then
From the equation we obtain
We know the initial velocity, the final velocity, the initial height (it is zero). Then we can find the final height (vertical drop):
 
 

Problem 12.
A man drops a 10 kg rock from the top of a ladder of height 5 m.
What is its speed just before it hits the ground?
What is its Kinetic Energy when it reaches the ground?

Solution:
In this problem we have free fall motion. We can solve this problem either by using the kinematics equations, which describe the motion with constant acceleration (free fall motion) or by using the energy conservation law.
Below we solve this problem by using the energy conservation law.
We assume that there is no friction, then we have the conservation of the net mechanical energy of the rock. The mechanical energy of the rock is a sum of the kinetic energy and gravitational potential energy:
The kinetic energy is determined by the speed of the rock:
The gravitational potential energy depends on the height of the rock:
Then the mechanical energy is
The conservation of the mechanical energy means that the mechanical energy in the initial state is equal to the mechanical energy in the final state:
or
.......................................(1)
 
The initial velocity is zero:
the initial height is 5 m:
the final height is zero (ground):
We do not know the final velocity:

We substitute these values in the equation (1) and obtain

Now we can find the final kinetic energy:
 

Problem 13.
A cyclist intends to cycle up a 8 degrees hill whose vertical height is 150 m. If each complete revolution of the pedals moves the bike 6 m   along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses. The pedals turn in a circle of diameter 30 cm. The total mass of the cyclist and his bike is 100 kg.

Solution:
In this problem we need to use generalized work-energy theorem: work done by an external force is equal to the change of the net mechanical energy of the system:

We assume that the cyclist moves with constant speed. Then the initial and the final kinetic energies are the same. Therefore in the above expression we need to take into account only the gravitational potential energy:
where is the total mass of the cyclist and bike.

Then
Now we need to define the initial and the final states of the system. We introduce the final state as the state after one complete revolution (relative to the initial state).
 

We know that after one complete revolution the cyclist moves 6 meters (as show in the figure). Then the change in the height of the cyclist is
Therefore the change in the gravitation potential energy of the cyclist+bike is
 
 
The work done by the force is
During one revolution the pedal travels a distance of
(circumference of a circle with diameter ). Where is the diameter of the circle. Then the work done by the force is
 
This work is equal to the change in gravitational potential energy:
From this expression we can find the force:

Problem 14.
A body of mass 5 kg slides a distance of 6 m down a rough Inclined plane 30 degree. Then it moves on frictionless horizontal surface and compresses a spring. The coefficient of kinetic friction is 0.1 and the spring constant is 300 N/m. Find the maximum compression of the spring.

Solution:
In this problem we need to use the work-energy theorem, which determines the relation between the work done by the friction force and the change of the net mechanical energy.
According the work-energy theorem the work done by the friction force (which is always negative) is equal to the difference between the final mechanical energy and the initial mechanical energy.
................................................. (1)

The mechanical energy is the sum of kinetic energy, gravitational potential energy, and elastic energy of the spring.

The initial velocity of the body is zero, the initial compression of the spring is zero, the initial height of the body is . Then the initial mechanical energy of the body is
 
In the final state we have a maximum compression of the spring. It means that in the final state the velocity of the body is zero. The final height of the body is 0 (ground level). Then the final mechanical energy of the body is
 
Then equation (1) takes the form
Now we need to find the work done by the friction force. The work done by the friction force (see figure) is
where is the displacement of the body over incline plane. The friction force is determined by the normal force and the coefficient of kinetic friction:
To find the normal force we need to write down the second Newton's law for the motion along the incline. There are three forces acting on the body: gravitational force, normal force, and the friction force. Then the second Newton's law takes the form:
...........................................................(2)
 
 

The direction of the acceleration is along the incline. It means that the acceleration has only x component (see figure). The y-component of acceleration is zero. Then to find the normal force we need to write down only the y-component of equation (2):
From this equation we can find the normal force:
Now we can find the friction force
and the work done by the friction force
 
Then we can find the maximum (final) compression of the spring: