physics for premedical students

Physics For Premedical Students

Sheet #2

Dr Maan A. Ibrahim

2008 - 2009

Physics For Premedical Students Sheet #2 2008-2009 By Dr Maan A. Ibrahim

Force and Newton’s Laws

Force The word force usually is a push or pull. The concept of force gives us a quantitative description of the interaction between two bodies or between a body and its environment. When a force involves direct contact between two bodies, we call it a contact force. Contact forces include the pushes or pulls you exert with your hand, the force of a rope pulling on a block to which it is tied, and the friction force that the ground exert on your feet when sliding. There are also forces, called long-range forces, that act even when the bodies are separated by empty space. You’ve experienced long range forces if you’ve played with a pair of magnets. Gravity is also a long range force; the sun exerts a gravitational pull on the earth, even over a distance of 150 million kilometers, that keeps the earth in orbit. The force of gravitational attraction that the earth exerts on your body is called your weight.

Force is a vector quantity; you can push and pull a body in different directions. Thus to describe a force, we need to describe the direction in which it acts as well as its magnitude, the quantity that describes “how much” or “how hard” the force pushes or pulls.

Newton's First Law

Newton's First Law of Motion states that a body will remain at rest or will continue to move at a constant velocity, unless an external force is applied.

This means that in order for the acceleration of a body to change, there must be a net force applied to the body. Put another way, if the forces on an object balance, there will be no acceleration (the object will continue at the same speed).

So, if we are told that a body is not accelerating (i.e. if it is moving at a constant velocity), we know that the resultant (overall) force in any one direction will be zero.

For a body in equilibrium, the net force is zero.

1


For this to be true, each component of the net force must be zero, so

Example

The following forces are acting on a body. The body moves at a constant speed of 5 m/s. Find force F.

It should be clear that F = 5. The force is therefore 5N.

Newton's Second Law

If a net external force acts on a body, the body changes momentum. The net force vector is equal to the change of momentum.

There are at least three important things of Newton’s second law:1. The above equation is a vector equation. So we can use it in component form

2. The statement of Newton’s second law refers to external forces. This means forces exerted on the body by other bodies in its environment. It’s impossible for a body to affect its own motion by exerting a force on itself; if it was possible, you could lift yourself to the ceiling by pulling up on your belt!

3. If mass is constant, we will have

2

F=20 N

N

mg


or we can write it in component form :

The acceleration has the same direction as the net force. Newton’s second law is a fundamental law of nature, the basic relation between force and motion.

Example : A worker applies a constant horizontal force with magnitude 20 N to a box with mass 40 kg resting on a level surface with negligible friction. What is the acceleration of the box?

There is no vertical acceleration, so we know that the two vertical forces must sum to zero. There is only one horizontal force, and we have

In fact, from Newton's Second Law we can derive the following equation:

Resultant Force on Body = Mass of Body × Acceleration of Body

3


This is sometimes written as F = ma, though you should make sure you understand what this means (in particular, note that F is resultant force).

Newton's Third Law

This law states that every action has an equal and opposite reaction.

For example, if a ball is placed on the table, the ball will exert a force on the table. At the same time, however, the table exerts a force on the ball (it is this force that prevents the ball from being sucked into the table!).

This "equal and opposite reaction force" is known as the normal reaction force, and the letter N or R is commonly used to represent it.

if body A exerts a force on a body B (an “action”), then body b exerts a force on body A (a “reaction”). These two force have the same magnitude but are opposite in direction. These two forces act on different bodies. The mathematical statement of Newton’s third law is

Note that the two forces described in Newton’s third law act on different bodies. This is important in problems involving Newton’s first or second law, which involve the forces that act on a body. For example, the net force acting on the football in figure above is the vector sum of the weight of the football and the force exerted by the kicker. You would not include the

force because this force acts on the kicker, not on the football.

Bon AF

Aon BF

A

B

foot

4


Some particular forces1. Weight(1). The weight W of a body is a force that pulls the body directly toward a nearby astronomical body; in everyday circumstances that astronomical body is the Earth. The force primarily due to an attraction called a gravitational attraction between the two bodies. We consider situations in which a body with mass m is located at a point where the free-fall acceleration has magnitude g, then weight can be written as (2). Since weight is a force, its SI unit is the Newton.(3). Normally we assume that weight is measured from an inertial frame. If it is, instead, measured from a non-inertial frame, the measurement gives an apparent weight instead of the actual weight.2. The normal force: When a

body is pressed against a surface, the body experiences a force that is perpendicular to the surface. The force is called the normal force N, as shown in above figure, the name coming from the mathematical term normal, meaning “perpendicular”.

3. Tension: When a cord (or a rope, cable, or other such object) is attached to a body and pulled taut, the cord is said to be under tension, as shown in the figure. It pulls on the body with a force T, whose direction is away from the body and along the cord at the point of attachment.

1. Friction(1). If we slide or attempt to slide a body over a surface, the motion

5


will be resisted by a bonding between the body and the surface. The resistance is regarded as a single force f, called the frictional force, or simply friction. This force is directed along the surface, opposite the direction of the intended motion, as shown in the figure. If in some situation, the friction can be negligible, the surface is then said to be frictionless.(2). If the body does not move, then the static frictional force fs and the component of F that is parallel to the surface are equal in magnitude, and fs

is directed opposite that component of F.(3). The magnitude of fs has a maximum value fs,max that is given by

where is the coefficient of static friction and N is the magnitude of the normal force. If the magnitude of the component of F that is parallel to the surface exceeds , then the body begins to slide along the surface.(4). If the body begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value fk given by . Where is the coefficient of kinetic friction.

Example 1:

A net force of 10 Newton’s acts on a box which has a mass of 2 kg. What will be the acceleration of the box?

Solution:

This is about as straightforward as it can get - Newton's Second Law says that the acceleration of an object equals the net force on it divided by its mass:

Note that if the net force is in Newton’s and the mass is in kilograms, the acceleration is in m/s2. Why? Here's a way to think about it. The force "1 Newton" is defined to be the amount of force that will accelerate an object of mass 1 kg at 1 m/s2.

6


Example 2:

How much horizontal net force is required to accelerate a 1000 kg car at 2 m/s2?

Solution:

Newton's 2nd Law relates an object's mass, the net force on it, and its acceleration:

NEWTON'S SECOND LAW(with friction)

OBJECTIVE: To examine our model of friction by investigating the sliding of a block up a ramp.

THEORY: The application of Newton's 2nd Law to this block-ramp system yields similar results as in last week's set-up except that in this lab we are including friction.

Consider Figure 1 where the block is moving up the ramp. A hanging weight (mass m2) is attached to the string. The string exerts a tension (T) on both objects. Application of Newton's 2nd Law gives for the block:

7


Fx = T - fk - Mg sin = Ma(1)

Fy = N - Mg cos = 0(2)

and for the hanging weight:

Fx = 0(3)

Fy = -T + m2 g = m2 a(4)

(Note that we have taken down to be the positive direction for the hanging weight so that the accelerations of the block and hanging weight have the same sign.)

Eq. (2) yields the normal force as:

N = Mg cos (5)

Solving for the kinetic friction force by combining Eqs. (1)& (4) gives:

fk = (m2 - M sin ) g - (m2 + M) a(6)

For each ramp angle, then, there is a different sized normal force and kinetic friction force as described by Eqs. (5)&(6). However, our model of friction states that the ratio of the kinetic friction force to the normal force (which is the coefficient of kinetic friction) is constant,

= fk / N .

The coefficient depends only on the kind of surfaces that are in contact. It is independent of contact area. There is a slight dependence of on speed, with decreasing as the object travels faster, but we will ignore this effect since the block will not be moving at high speeds.

Example

8


1) A 90 kg woman stands in an elevator. Find the force which the floor of the elevator exerts on the woman a) when the elevator has an upward acceleration of 2 m/sec2; b) when the elevator is rising at constant speed; c) when the elevator has a downward acceleration of 2 m/sec2.

Solution: We have a simple application of Newton's 2nd Law of motion. We need only draw the vector force diagram for all forces acting on the woman. There are only two forces present, these are:

Fearth on woman = W = mg .

Ffloor on woman = N .

We select a CS with 'y' chosen upwards. The basic equation involved with the problem is the 2nd law:

Fnet = m a .

a). ay 0; Fy = N - mg = m ay N = m(g + a) = (90)(9.8 + 2) = 1062 N.b). a 0; Fy = N - mg = m ay N = m(g + a) = (90)(9.8 - 2) = 702 N.

c) a = 0 (equilibrium) Fy = N - mg = 0 N = mg = (90)(9.8) = 882 N.

Equilibrium

If the sum of the forces on an object is zero, then it is in equilibrium (even if moving with constant velocity). In 2-D,

Fx = 0, Fy = 0

Example: Mass resting on a table.

The two forces acting on the mass are the upward force exerted by the table, n, (n for ‘normal’ or perpendicular direction) and the downward pull of gravity, mg. Thus,

9


Fy = n – mg = 0, or n = mg

Example: Mass suspended by ropes. Find the tensions, T1, T2, and T3, in the ropes.

The tensions are the forces with which the ropes pull on the knots which tie the ropes together. The mass is in equilibrium, so

Fy = T3 – mg,

so T3 = mg = (20 kg)(9.8 m/s2) = 196 N

The knot is in equilibrium. We show the forces acting on the knot on a x-y coordinate system. Then

Fx = -T1cos(30o) + T2cos(50o) = 0

Fy = T1sin(30o) + T2sin(50o) – T3 = 0

Or, 0.866T2 = 0.643T1

0.5T1 + 0.766T2 = 196 N

Combining these two eqs to solve for T1 and T2

–

0.5T1 + 0.766(0.643T1/0.866) = 196

1.069T1 = 196, or T1 = 183 N

Then, T2 = 0.643T1/0.866 = 0.643(183)/0.866, or T2 = 136 N

Example: A 4-kg block is pulled along a level frictionless surface by a 20-N horizontal force. What is the acceleration of the block?

10


A ‘freebody’ diagram is given to the right, showing all the forces acting on the mass. Applying the 2nd law,

Fx = max, or 25 N = (5 kg)ax, or

ax = 20/4 = 5 m/s 2

What is the normal force exerted on the block by the table?

Fy = may = 0 (it can’t accelerate up or down)

Or, n – mg = 0, or n = mg = (4kg)(9.8 m/s2) = 39.2 N

Example:

The block in the example above is pulled with the 20-N force at an angle of 30o above the surface.

The free body diagram is given to the right, and the forces are also shown on an x-y coordinate system. Now,

Fx = Fcos = max

20N cos(30o) = 17.3 N = (4 kg)ax

ax = 17.3/4 = 4.3 m/s 2

Fy = Fsin + n – mg = 0

n = mg - Fsin

= (4kg)(9.8 m/s2) – (20N)sin(30o)

n = 29.2 N

Example: Inclined plane without friction.

11


Find the acceleration down the plane and the normal force exerted on the block by the plane.For simplicity, we choose a coordinate system with the y-axis perpendicular to the plane and resolve the forces into x- and y-components. This way, we know that ay = 0 and we only have to find ax. Then,

Fy = n – mg cos = ay = 0, n = mg cos

Fx = mg sin = max, ax = g sin

So the inclination of the plane reduces the normal force below that for a level surface (mg). If = 30o, for example, then the normal force is 0.866 mg and the acceleration down the incline is 4.9 m/s2.

Example:

A 5-kg block is pulled along a horizontal surface with a 35-N horizontal force. The coefficient of kinetic friction between the block and the surface is 0.4. Find the acceleration of the block.

Fy = n – mg = 0, n = mg = (5 kg)(9.8 m/s2) = 49 N

fk = kn = kmg = 0.4(5 kg)(9.8 m/s2) = 19.6 N

Fx = F – fk = max

35 N – 19.6 N = (5 kg)ax, ax = 3.1 m/s 2

Example:

12


The block in the above example is pulled with a force of 35 N at an angle of 15o above the horizon. Find the acceleration.

We will refer to the diagram to the right, where we have resolved the applied force into x- and y- components.

Fy = n + Fsin – mg = 0, n = mg - Fsin

= (5 kg)(9.8 m/s2) –(35 N)sin(15o) = 39.9 N

fk = kn = 0.4(39.9 N) = 16.0 N

Fx = Fcos – fk = max (35 N)cos(15o) – 16.0 N = (5 kg)ax

ax = 3.56 m/s 2

Note that the acceleration is actually larger than when pulled horizontally with the same force. Does this make sense? Would this be the case if the angle were much larger than 15o?

Example: Mass and pulley system.

Two masses are connected by a string draped over a light, frictionless pulley, as shown to the right. The weight of the hanging mass pulls the other mass along the surface. Friction exists between mass 1 and the surface. We want to find the tension in the cord and the acceleration of the masses.

We apply Newton’s 2nd law to each mass separately.

Mass 1: Fy = n – m1g = 0, n = m1g

13


Fx = T – fk = m1a, or

T - km1g = m1a

Mass 2:

Fy = m2g – T = m2a (down is positive)

There are two unknowns (T and a) and two equations (boxed). If we add the two equations (left 1 + left 2 = right 1 + right 2) we will get

m2g - km1g = (m1 + m2)a

or,

Knowing a, we can find T using one of the boxed equations.

Rotational Motion and the Law of Gravity

This chapter deals with rotational kinematics – the relationships between rotational position, velocity, acceleration and time. Rotational dynamics will be discussed in the next chapter. Dynamics deals with energy, momentum, and forces. Universal gravitation is also discussed in this chapter in part since satellite and planetary motion involves rotation.

Angular displacement, velocity, and acceleration

Angle () can be defined in terms of radius (r) and arc length (s) on a circle as

Unless otherwise stated, is usually measured in the counterclockwise direction from the positive x-axis. If s and r are measured in the same units

14


(e.g., m), then the above equation gives is in radians. Angle can also be measured in degrees or revolutions.

1 rev = 2 rad = 360o

Average angular velocity is given by

The instantaneous angular velocity is just the average angular velocity in the limit of very short time interval. Depending on units for and t, units for can be rad/s, deg/s, rev/s, rev/min (rpm), etc.

Average angular acceleration is given by

Units can be rad/s2, deg/s2, rev/s2, rev/min2, etc.

The relationships above are mathematically similar to those for motion in 1-D. Thus, we can get the equations for constant angular acceleration from those for constant linear acceleration by appropriately changing the variables.

1-D motion with constant a Rot. motion with constant

It is important that the units in the above equations be compatible. For example, if is in rad/s2, then should be in rad/s, t in s, and you will get in rad. Also, don’t mix the left and right equations above. For example, the equation would not make sense. Also, don’t confuse and a, although they look similar.

15


Example:

A wheel increases its rotational velocity from 200 rpm to 300 rpm in 10 sec.

What is its angular acceleration in rad/s2?

How many turns did the wheel make?

Relationship between angular and linear motionSince s = r, then it follows that

, or

, or

In the above, vt is the tangential speed of a point going around a circle and at is the component of acceleration tangent to the circle.

Example:

A merry-go-round rotates at a constant angular speed. It takes 20 sec to make a complete revolution. What is the speed of a rider who is 4 m from the center?

16


The speed of a rider increases as he moves further from the center.

Centripetal acceleration

An object moving in a circle can have a component of acceleration tangent to its path if it has an angular acceleration ( ). This is related to the change in the speed. In addition, the object can have a radial component (towards or away from the center) because it is moving in a curved path, even is its speed is constant. This is also called the centripetal acceleration. These two components of acceleration are

.

The diagram to the right shows how the centripetal acceleration is obtained for uniform circular motion (constant speed). Because of the change in direction, the velocity changes by an amount v which is directed toward the center of the circle. From the similarity of the triangles formed by sides s and r and by v and v, we have

So, for uniform circular motion, the acceleration is always directed toward the center of the circle even though the speed is constant.

Example:

A car goes around a circular track of circumference 2000 m with constant speed of 1 minute. What is the magnitude of the acceleration of the car?

17


Combined tangential and centripetal acceleration

As mentioned previously, an object can have both tangential and radial (centripetal) components of acceleration if its speed is changing and it is changing direction. Since these are mutually perpendicular components of the acceleration, then the magnitude of the total acceleration is

, where

Centripetal force

By Newton’s 2nd law, an object moving in a circle with constant speed must have a net force directed toward the center of the circle given by

Example:

Suppose that the car going around the track in the above example has a mass of 1500 kg. The net force acting on the car is then

What is the origin of the centripetal force on the car going around the track?

If the track is flat with no banking, then the centripetal force is entirely due to friction

18


between the tires and the track. This would require a coefficient of friction given by

(Note: This might be considered static friction even though the car is moving if the tires are not sliding.)

To avoid the possibility of the car skidding because of not enough friction, the track can be banked so that the normal force has a radial component. Then at just the right bank angle no friction is required and

which gives for the optimum angle of bank

.

Example:

What is the effective weight of a person when at the top and when at the bottom of a Ferris wheel?

Bottom: Top:

19


The effective weight is the normal force exerted by the Ferris wheel seat on the person. At the bottom,

At the top,

So, the person ‘weighs’ more at the bottom and less at the top. The

fractional change in his effective weight is

Example:

A Ferris wheel has radius r = 8 m and it takes 30 sec to make one revolution. What is the effective change in the weight of a 160 lb person at the bottom and top of the Ferris wheel?

Work, Energy & Power

Work Done

Suppose a force F acts on a body, causing it to move in a particular direction. Then the work done by the force is the component of F in the direction of motion × the distance the body moves as a result. Work done is measured in joules (which has symbol J).

20


So if we have a constant force of magnitude F newtons, which moves a body a distance s (meters) along a flat surface, the work done is F × s joules.

Now suppose that this force is at an angle of a to the horizontal. If the body moves a distance of s meters along the ground, then the work done is F cos a × s (since F cos a is the component of the force in the direction of motion).

Work Done Against Gravity

Now suppose that the force we are considering is one which causes a body to be lifted off of the ground. We call the work done by the force the "work done against gravity". This is equal to mgs joules, where s is the vertical distance moved by the body, m is the mass of the body and g is the acceleration due to gravity. [Compare with "gravitational potential energy" below].

Energy

The energy of a body is a measure its ability to do work.

Kinetic Energy

The kinetic energy (K.E.) of a body is the energy a body has as a result of its motion. A body which isn't moving will have zero kinetic energy, therefore.

K.E. = ½ mv2

where m is the mass and v is the velocity of the body.

Gravitational Potential Energy

21

http://www.mathsrevision.net/alevel/pages.php?page=77


Gravitational potential energy (G.P.E.) is the energy a body has because of its height above the ground.

G.P.E. = mgh

where h is the height of the body above the ground.

There are also other types of potential energy (such as elastic potential energy). Basically, the total potential energy measures the energy of the body due to its position.

Conservation of Energy

If gravity is the only external force which does work on a body, then the total energy of the body will remain the same, a property known as the conservation of energy.

Therefore, providing no work is done:

Initial (PE + KE) = final (PE + KE)

Connection Between Energy and Work Done

If energy is not conserved, then it is used to do work.

In other words, the work done is equal to the change in energy. For example, the work done against gravity is equal to the change in the potential energy of the body and the work done against all resistive forces is equal to the change in the total energy.

Power

Power is the rate at which work is done (measured in watts (W)), in other words the work done per second.

It turns out that:

Power = Force × Velocity

22

http://www.mathsrevision.net/alevel/pages.php?page=80#energy

http://www.mathsrevision.net/alevel/pages.php?page=80#energy


For example, if the engine of a car is working at a constant rate of 10kW, the forward force generated is power/velocity = 10 000 / v, where v is the velocity of the car (the 10 was changed to 10 000 so that we are using the standard unit of W rather than kW).

Example

A car of mass 500kg is traveling along a horizontal road. The engine of the car is working at a constant rate of 5kW. The total resistance to motion is constant and is 250N. What is the acceleration of the car when its speed is 5m/s?

The equation of motion horizontally (from Newton's Second Law):

5000 - 250 = 500a v

when v = 5: 500a = 750a = 1.5

so the acceleration is 1.5ms -2

Linear Momentum and Collisions

This chapter is about the concept of linear momentum.

Linear momentum is defined as

p = mv [kgm/s]

23

http://www.mathsrevision.net/alevel/pages.php?page=87


Momentum is a vector whose direction is the same as the velocity. So, in 2-D

px = mvx, py = mvy

Momentum is not the same as kinetic energy. K = ½ mv2 is a scalar and has no direction. Also, K depends on the square of the speed. Because of this difference, two different objects can have the same momentum but have different kinetic energies.

Example:

Car A has a mass of 2000 kg and is traveling at 10 m/s north. Car B has a mass of 1000 kg and is traveling at 20 m/s north. Both cars have the same momentum –

pA = (2000 kg)(10 m/s) = 20,000 kg-m/s, direction = north

pB = (1000 kg)(20 m/s) = 20,000 kg-m/s, direction = north

However, they have different kinetic energies -KA = ½ (2000 kg)(10 m/s)2 = 100,000 J

KB = ½ (1000 kg)(20 m/s)2 = 200,000 J

Newton’s 2nd law can be written in terms of change in momentum:

This is the same as F = ma, since

.

Impulse

Impulse is defined as

24


Impulse = Fdt = Favet [Ns = kgm/s]

The impulse exerted by a force depends on the time of application and the average force during that time. From Newton’s 2nd law, the impulse on an object is the same as its change in momentum:

Favet = p = mvf - mvi

Example:

A pitcher throws a baseball toward a batter with a speed of 40 m/s toward a batter. The batter hits the ball straight back towards the pitcher with a speed of 30 m/s. A baseball has a mass of about 145 g.

What is the impulse exerted on the ball by the bat? Assume right = + , left = -

Impulse = p = mvf – mvi = m(vf – vi) = (0.145 kg)(30 m/s – (-40 m/s))

= (0.145 kg)(70 m/s) = 10.15 kg-m/s

If the contact time was 5 milliseconds, what was the average force exerted by the bat on the ball?

Impulse from a variable force

If the force varies during a contact, then the impulse depends on the average force during the contact. This is equivalent to finding the impulse from the area under the force versus time graph.

25


(A point of possible confusion - The above is similar to, but not the same as, the work energy theorem, which states that Favex ( = area under F vs x curve) = KE.)

Reducing the impact force during a collision

The damage that results from a collision is a consequence of the size of the impact force. So, to reduce the damage we need to spread out the time of impact.

Example:

An egg dropped from a height of 1 m onto a concrete floor will break, but it will not break if dropped from the same height onto a pillow. Why? The impulse is the same in both cases (same p). The pillow increases time of contact (t) and thus reduces the average force (F).

Other examples would include a safety airbag in a car, a padded dashboard, a flexile bumper on a car, …

Conservation of Linear MomentumConsider a collision between two balls. During the collision the balls exert equal and oppositely directed forces on each other (Newton’s 3rd law). That is,

F12 = -F21

Since the contact times are the same for both masses,

F12 t = -F21 t

26


If all other forces can be neglected, then from Newton’s 2nd law

p1 = - p2

(p1 + p2) = ptotal = 0

Or, ptotal = constant

The above argument applies to an arbitrary number of particles. If Fext = 0, then the total momentum of a system remains constant in time.

Example:

A 10-kg gun fires a 25-g bullet with a speed of 300 m/s. What would be the recoil speed of the gun? Assume the gun is held lightly so that its recoil is not restricted.

pf = pi

mbvb + mgvg = 0

vg = -mbvb/mg = -(0.025)(300)/10 = - 0.75 m/s

Note: If you hold the gun tightly, its recoil speed will be much less since your mass adds to the gun in the above calculation.

Collisions

Collisions can be classified as elastic or inelastic. In an elastic collision kinetic energy is conserved. In an inelastic collision kinetic energy is not conserved. When two colliding objects stick together the collision is referred to as completely inelastic. In a completely inelastic collision you have the maximum loss of kinetic energy; however, not all the kinetic energy is necessarily lost.

27


Two colliding billiard balls may be nearly elastic (but not completely). If you throw a piece of putty and it sticks to a wall, then the collision is completely inelastic.

Note: Whether the collision is elastic or inelastic, momentums is always conserved.

Example:

A 5000-kg truck traveling at 10 m/s makes a head-on collision with a 1000-kg car traveling at 30 m/s. The car and truck become entangled and stick together on impact. What is their common velocity immediately after the collision?

Choose the initial direction of motion of the truck as the positive direction. Then

Note that we used -30 m/s for the velocity of the car. The fact that we got +3.3 m/s means that the entangled vehicles move in the direction of the truck’s initial velocity.

1) A freight car weighing 25 tons runs into another freight car of the same weight. The first was moving at 6 mi/hr (8.8 ft/sec) and the second was at rest. If the cars are coupled together after the collision, what is their final speed?

Solution: We have a collision problem in 1-dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown. We have conservation of linear momentum: pAi pB i = pA f pB f .

28


We have a perfectly inelastic collision problem. Hence: m1 v1i m2

v2i = (m1 + m2) vf

For x-components this becomes : m1 v1 + 0 = (m1 + m2) vf .

Since m1 = m2 , we have: vf = (1/2) v1 = 3 mi/hr (4.4 ft/sec).

2) Two blocks are travelling toward each other. The first has a speed of 10 cm/sec and the second a speed of 60 cm/sec. After the collision the second is observed to be travelling with a speed of 20 cm/sec in a direction opposite to its initial velocity. If the weight of the first block is twice that of the second, determine: (a) the velocity of the first block after collision; (b) whether the collision was elastic or inelastic.

Solution: We have a collision problem in 1-dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown.

Since the surface is frictionless, and since no work is performed by either mg or the normal, then the net force acting on the system is 0, and we have conservation of linear momentum:

p1i p2i = p1i p1f

Thus adding the x-components we have: m1v1i - m2v2i = m1 v1f + m2

v2f

Since m1 = 2 m2 we find: 2 v1i - v2i = 2 v1f + v2f (2)(10) - (60) = 2 v1f + 20

29


Thus 2 v1f = - 60 and v1f = - 30 cm/sec ('-' means to left).08-2

The initial KE is given by: KEI = (1/2) m1 (v1I)2 + (1/2) m2 (v2I)2 . This gives:

= (1/2)(2 m2)(10)2 + (1/2) m2 (60)2 = (1/2)(200 + 3600) m2 = 1900 m2

The final KE is: KEf = (1/2) m1 (v1f)2 + (1/2) m2 (v2f)2 . This gives:

= (1/2)(2 m2)(30)2 + (1/2) m2 (20)2 = (1/2)(1800 + 400) m2 = 1100 m2

Since KEf is not equal to KEI, the collision is inelastic.

3) A block of mass 200 g, sliding with a speed of 12 cm/sec on a smooth level surface, makes a head-on, elastic collision with a block of unknown mass, initially at rest. After the collision the velocity of the 200 g block is 4 cm/sec in the same direction as its initial velocity. Determine the mass of the 2nd block and its speed after the collision.

Solution: We have a collision problem in 1-dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown.

Since the surface is frictionless, and since no work is performed by either mg or the normal, then the net force acting on the system is 0, and we have conservation of linear momentum:

p1I p2I = p1I p1f

Thus adding the x-components we have: m1v1I + 0 = m1 v1f + m2 v2f .

This gives: (200)(12) = (200)(4) + m2 v2f .

30


Since we have 2 unknowns we look for an 'energy condition'. We are told that the collision is elastic. Hence, we also have conservation of KE and write:

(1/2)m1(v1I)2 + (1/2)m2(v2I)2 = (1/2)m1(v1f)2 + (1/2)m2(v2f)2

We may cancel the (1/2) factors, and we obtain: (200)(12) 2 + 0 = (200)(4)2 + m2(v2f)2 .

Thus (m2 v2f) v2f = (200)(8) v2f = (200)(144) - (200)(16) or 8 v2f = 128.

Hence: v2f = 16 cm/sec, and m2 = (200)(8)/(16) = 100 gm .

4) A 0.005 kg bullet going 300 m/sec strikes and is imbedded in a 1.995 kg block which is the bob of a ballistic pendulum. Find the speed at which the block and bullet leave the equilibrium position, and the height which the center of gravity of the bullet-block system reaches above the initial position of the center of gravity.

Solution: A 'ballistic pendulum' is a device which can be used to measure the muzzle velocity of a gun. The bullet is fired horizontally into the block of wood and becomes imbedded in the block. The block is attached to a light rod and can swing like a pendulum. After the collision the 'bob' swings upward and the maximum height it reaches is determined. From this information plus the masses of the bullet and block, one can determine the velocity of the bullet. The critical point to note in this problem is that we have two distinct problems:

Problem 1: A collision problem. Apply Conservation of Linear Momentum and energy relation. Problem 2: A work-energy type problem. Apply conservation of total mechanical energy.

In component form (for x-components) this gives:

31


m vAix + M vBix = (m + M) v fx. Since vBi = 0, we have:

vf = (m vAi) /(m + M) = (.005)(300)/(2.00) = 0.75 m/sec.

(Note: We were able to 'solve' the collision problem without an 'energy relation' since the collision was a perfectly inelastic collision. That is the two objects had the same final velocity. This condition is equivalent to an 'energy relation', since for such a collision the loss of KE is a maximum possible amount).

Part 2: In the work-energy part of the problem we note that the only force which performs work is gravity. Hence, we have only conservative forces present, and we have conservation of total mechanical energy.

We draw the figure indicating 'initial' and 'final' situations. We may choose the 0 level for gravitational potential energy anywhere we like. Hence, select UI = 0.

Then: KEI + UI = KEf + Uf .

(1/2)(m + M) vf2 = 0 + (m + M)g h

Here vf is the 'initial' velocity in this part of the problem ( 0.75 m/sec) and (m+M) is the combined mass of bullet & block.

Thus: h = v f2/2g = (.75)2/(2)(9.8) = .0287 m or 2.87 cm.

Note the reversal of this problem. If we know the masses and measure 'h', then from part 2 we can calculate 'v f' (the initial velocity of bullet & block in the 2nd part of the problem). This is the same as vf , the final velocity in the collision problem. Thus using this we can calculate vAi the 'muzzle' velocity of the bullet. Examples : Answer and Explanation:

1. An object is raised above the ground gaining a certain amount of potential energy. If the same object is raised twice as high it gains __________.

32


a. four times as much potential energy b. twice as much potential energy c. neither of these

Answer and Explanation:

Answer: B

The potential energy of an object is directly proportional to the height of the object above the ground (or some other arbitrary "zero-level") in accordance with the equation

PE = m*g*h

If the h in the equation is doubled (the object is raised twice as high), then the PE is doubled as well.

2. When an object is lifted 10 meters, it gains a certain amount of potential energy. If the same object is lifted 20 meters, its potential energy is _____.

a. less b. the same c. twice as much

d. four times as much e. more than 4 time as much

Answer: C

The potential energy of an object is directly proportional to the height of the object above the ground (or some other arbitrary "zero-level") in accordance with the equation

PE = m*g*h

If the h in the equation is doubled (say from 10 m to 20 m), then the PE is doubled as well.

3. A 1000 kg car and a 2000 kg car are hoisted the same distance at constant speed in a gas station. Raising the more massive car requires ____________.

33


a. less work b. as much workc. twice as much work.

d. four times as much work

e. more than 4 times as much work

Answer: C

The amount of work done by a force to displace an object is found from the equation

W = F*d*cos(Theta)

The force required to raise the car at constant speed is equivalent to the weight (m*g) of the car. Since the 2000-kg car weight 2X as much as the 1000-kg car, it would require twice as much work to lift it the same distance.

4. An object that has kinetic energy must be ____________.

a. moving b. falling c. at an elevated position

d. at rest e. none of these

Answer: A

Pay attention to the keywords "must be." Kinetic energy is the energy of motion and an object must be moving if it has kinetic energy. The object could be falling and could be at an elevated position; the object must not be at rest if it has kinetic energy.

5. An object that has potential energy has this energy because of its _____________.

a. speed b. acceleration c. momentum d. position e. none of these

34


Answer: D

Potential energy is the energy of position and any object which has potential energy owes this PE to the fact that it has some given position other than the so-called "zero-level position."

6. An arrow is drawn so that it has 40 J of potential energy. When fired horizontally, the arrow will have a kinetic energy of ________.

a. less than 40 J b. more than 40 J c. 40 J

Answer: C

A drawn arrow has 40 J of stored energy due to the stretch of the bow and string. When released, this energy is converted into kinetic energy such that the arrow will have 40 J of kinetic energy upon being fired. Of course, this assumes no energy is lost to air resistance, friction or any other external forces and that the arrow is shot horizontally.

7. A 2 kg mass is held 4 m above the ground. What is the approximate potential energy of the mass with respect to the ground?

a. 20 J b. 40 J c. 60 J d. 80 J e. none of these

Answer: D

The potential energy of an object is found from the equation

PE = m*g*h

where m is mass, h is height, and g=10 m/s/s (approx.). Plugging and chugging into this equation yields PE=(2 kg)*(10 m/s/s)*(4 m) = 80 J.

35


8. A 2 kg mass has 40 J of potential energy with respect to the ground. Approximately how far is it located above the ground?

a. 1 m b. 2 m c. 3 m d. 4 m e. none of these

Answer: B

The potential energy of an object is related to its height by the equation

PE = m*g*h

where m is mass, h is height, and g=10 m/s/s (approx.). Plugging and chugging into this equation yields 40 J = (2 kg)*(10 m/s/s)*(h); rearranging and solving for h yields an answer of 2 m.

9. A ball is projected into the air with 100 J of kinetic energy which is transformed to gravitational potential energy at the top of its trajectory. When it returns to its original level after encountering air resistance, its kinetic energy is ____________.

a. less than 100 J b. 100 J c. more than 100 J d. not enough information given

Answer: A

During any given motion, if external forces do work upon the object, then the total mechanical energy will be changed. If external forces do negative work (i.e., F*d*cos(Theta) is a negative number), then the final TME is less than the initial TME. In this case, air resistance does negative work to remove energy from the system. Thus, when the ball returns to its original height, their is less TME than immediately after it was thrown. At this same starting height, the PE is the same as before. The reduction in TME is made up for by the fact that the kinetic energy has been reduced; the final KE is less than the initial KE.

10. A woman lifts a box from the floor. She then carries with constant speed to the other side of the room, where she puts the box down. How much work does she do on the box while walking across the floor at constant speed?

a. zero J b. more than zero J c. more information needed to determine

36


Answer: A

For any given situation, the work done by a force can be calculated using the equation

W = F*d*cos(Theta)

where F is the force doing the work, d is the displacement of the object, and Theta is the angle between the force and the displacement. In this specific situation, the woman is applying an upward force on the box (she is carrying it) and the displacement of the box is horizontal. The angle between the force (vertical) and the displacement (upward) vectors is 90 degrees. Since the cosine of 90-degrees is 0, the woman does not do any work upon the box. A detailed discussion of a similar situation

11. A car moving at 50 km/hr skids 20 m with locked brakes. How far will the car skid with locked brakes if it is traveling at 150 km/hr?

a. 20 m b. 60 m. c. 90 m d. 120 m e. 180 m

Answer: E

When a car skids to a stop, the work done by friction upon the car is equal to the change in kinetic energy of the car. Work is directly proportional to the displacement of the car (skidding distance) and the kinetic energy is directly related to the square of the speed (KE=0.5*m*v^2). For this reason, the skidding distance is directly proportional to the square of the speed. So if the speeds is tripled from 50 km/hr to 150 km/hr, then the stopping distance is increased by a factor of 9 (from 20 m to 9*20 m; or 180 m).

12. Which has greater kinetic energy, a car traveling at 30 km/hr or a half-as-massive car traveling at 60 km/hr?

a. the 30 km/hr car b. the 60 km/hr car c. both have the same kinetic energy

Answer: B

This problem is complicated by the fact that no mass is given for the two cars; only the ratio of mass is known. The complication can be resolved in one of two ways: 1) make up a mass for each car - such as 10 kg and 5 kg,

37


or 2) assign m as the mass of one of the cars and (1/2)m as the mass of the second car. Then use the kinetic equation

KE = 0.5*m*v2

and plug and chug. The mass can then be determined for each car and compared. Using the second method yields the following results for the two cars:

KE for the 30 km/hr car:

KE = 0.5*m*(30 km/hr)2

KE = 0.5*m*(900 km2/hr2)

KE = 450*m km2/hr2

KE for the 60 km/hr car:

KE = 0.5*(0.5m)*(60 km/hr)2

KE = 0.5*(0.5m)*(3600 km2/hr2)

KE = 900*m km2/hr2

13. A diver who weighs 500 N steps off a diving board that is 10 m above the water. The diver hits the water with kinetic energy of ___________.

a. 10 J b. 500 J c. 510 J d. 5000 J e. more than 5000 J.

Answer: D

The use of the work-energy theorem and a simple analysis will yield the solution to this problem. Initially, there is only PE; finally, there is only KE. Assuming negligible air resistance, the kinetic energy of the diver upon hitting the water is equal to the potential energy of the diver on top of the board.

PEi = KEf

m*g*hi = KEf

Substituting 500 N for m*g (500 N is the weight of the diver, not the mass) and 10 m for h will yield the answer of 5000 J.

38


14. A 2500 N pile driver ram falls 10 m and drives a post 0.1 m into the ground. The average impact force on the ram is _________.

a. 2500 N b. 25000 N c. 250,000 N d. 2,500,000 N

Answer: C

The use of the work-energy theorem and a simple analysis will yield the solution to this problem. Initially, there is only PE; finally, there is neither PE nor KE; external work has been done by an applied force upon the pile driver. Assuming negligible air resistance, the kinetic energy of the diver upon hitting the water is equal to the potential energy of the diver on top of the board.

PEi + Wext = 0

PEi = - Wext

m*g*hi = - F*d*cos(Theta)

Substituting 2500 N for m*g (2500 N is the weight of the driver, not the mass); 10 m for h; 0.1 m for the displacement of the driver as caused by the upward applied force exerted by the ram; and 90 degrees for Theta (the angle between the applied force and the displacement of the ram) will yield the answer of 250000 N for F.

15. A person on the edge of a roof throws a ball downward. It strikes the ground with 100 J of kinetic energy. The person throws another identical ball upward with the same initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the ground with a kinetic energy of ______________.

a. 100 J b. 200 J c. less than 100 J d. more than 200 J e. none of these

Answer: A

Quite surprisingly to many, each ball would hit the ground with the same speed. In each case, the PE+KE of the balls immediately after being thrown is the same (they are thrown with the same speed from the same height).

39


Upon hitting the ground, they must also have the same PE+KE. Since the PE is zero (on the ground) for each ball, it stands to reason that their KE is also the same. That's a little physics and a lot of logic - and try not to avoid the logic part by trying to memorize the answer.

16. A 10 N object moves at 1 m/s. Its kinetic energy is approximately ____________.

a. 0.5 J b. 1 J c. 10 J d. more than 10 J

Answer: A

The KE of any object can be computed if the mass (m) and speed (v) are known. Simply use the equation

KE=0.5*m*v2

In this case, the 10-N object has a mass of 1 kg (use Fgrav = m*g). The speed is 1 m/s. Now plug and chug to yield KE=0.5 J.

40

physics for premedical students

Documents