Chapter 4: Impulse and Momentum

A. Momentum
B. Impulse, Momentum and the Second Law
C. Conservation of Momentum
D. The Coefficient of Restitution
E. Oblique Impact
F.Collisions with Spin and Friction
Problems
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We have stated Newton’s second law of motion as F = ma. In truth, Newton used a quantity other than acceleration in his original statement of the second law. Newton called this quantity momentum, which he said represented the “quantity of motion.”

A. Momentum

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We all know that a bowling ball is harder to stop than a ping pong ball traveling at the same velocity. We state this fact by saying the bowling ball has more momentum than the ping pong ball. Momentum, the ability of a body to maintain its state of motion, depends on two things mass and velocity, and is defined as the product of these two quantities.

Momentum = mass x velocity

We will use the letter “p” to represent momentum and write

p =mv (4.1)

We can see from this definition that a moving body can have a large momentum if its mass is large (a huge truck for instance), or if its velocity is large (a bullet for example). In addition, since velocity is a vector, momentum is also a vector, and the momentum of an object is in the same direction as its velocity.

B. Impulse, Momentum and the Second Law

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When you catch a baseball, a force is exerted on the baseball glove. This force comes from changing the velocity of the ball. The force if impact is proportional to the rate of change in velocity of the ball. In addition, if the object is more massive, a larger force is necessary to change its velocity.

In Chapter 3 we used F = ma to determine the size of the force necessary to cause an acceleration which resulted in a change in velocity. We will now re-express this law by replacing acceleration by its definition, the rate at which velocity changes. Begin with the second law.

F_Net = ma

and recall the definition of acceleration

Now substitute

(4.2)

There are two ways to interpret Eq 4.2. As first stated it tells us that force is the rate at which momentum changes. This is simply a restatement of Newton’s second law. Using the second statement of Eq 4.2 will define a new quantity called impulse.

Impulse = Change in momentum = p_F - p_I = F_Nett

This equation tells us that if we want to change the momentum of an object we just exert a force over some period of time. It is important to note that a large force exerted over a short period of time can give the same impulse as a small force exerted over a long period of time. If you were to jump from a chair and land with your knees locked you would stop abruptly and a large force would be exerted on your body. This is why when you land you allow your knees to flex. It extends the time over which the impulse occurs and reduces the force exerted on the body.

Question: Use the concept of impulse to explain why the “follow through” is important when throwing a football, hitting a golf ball, or kicking a soccer ball.

Examples: 1. A baseball (mass = 0.01 slug) is hit off the bat with a velocity of 120 mph (176 ft/sec) and is caught by a third baseman with quick reflexes. Assume he catches the ball in the pocket stopping it in a time of 0.02 seconds. Calculate the average force by the glove on the ball.

The negative sign means that the glove is exerting a force opposite to the motion of the ball.

2. The baseball in the previous example is now hit to a first baseman who has slower reflexes. Instead of catching the ball in the pocket it hits the part if the glove just in front of the hand is stopped in a time of 0.005 seconds. Calculate the average force on the hand.

3. A baseball thrown with a velocity of 90 mph (132 ft/sec) is hit for a home run. If the ball’s velocity when it leaves the bat is 110 ft/sec find the impulse. The mass of a baseball is 0.145 kg or 0.01 slug.

If the ball is in contact with the bat for 0.005 seconds what is the average force applied on the ball?

C. Conservation of Momentum

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We showed in the previous section that a net force is required to change the momentum of an object. Let us consider the nature of this “momentum-changing” force by breaking the net force into two parts: the net external force, and the net internal force.

where

F_NetExt = Sum of all external forces acting on the object.

F_NetInt = Sum of all internal forces acting on the object.

Consider the net internal force. From Newton’s third law of motion we know that for every force there is an equal and opposite force. As a result, when we add together all internal forces each internal “action” force is cancelled by its corresponding internal “reaction” force, and we have

This tells us that external forces are required to change the momentum of an object. A person sitting in a car cannot affect the momentum of the car by pushing on the dashboard. Similarly, the molecular forces within a baseball have no affect upon the momentum of the ball. This is because these forces are internal. An outside, or external force acting on the baseball or car is required for a change in momentum. If no external force is present, then no change in momentum is possible. This fundamental law of physics is called conservation of momentum.

If the external force on a body is zero, then the momentum of the body remains constant.

Examples:

1. A Cadillac with a mass of 2000 kg is traveling at 60 mph when it strikes a Volkswagon of mass 1000 kg initially at rest. After the collision the two cars stick together. What is their velocity after the collision?

To solve this we use conservation of momentum.

2. A rifle of mass 5 kg fires a 20 gram (0.02 kg) bullet with a velocity of 500 m/sec. What is the recoil velocity of the gun?

Once again we can use conservation of momentum to solve this. Initially the gun and the bullet are at rest so the momentum is zero.

After being fired the bullet has momentum mv to the right and the gun has momentum MV to the left. Conservation of momentum requires the final momentum of the system to be zero.

Momentum of gun to left = momentum of bullet to right = 0

D. The Coefficient of Restitution

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The tendency of a body to return to it normal shape once it has been deformed (i.e., its elasticity) differs from one body to another. Some return very quickly to their original shape, while others do so less quickly. Because there is no way of directly calculating the elasticity if a body, it is necessary to rely in the results if experiments to predict the outcome of any given impact.

Newton investigated the properties of elastic bodies and the results of impacts between then and formulated the following empirical law (Newton’s law of impact):

If two bodies moving toward one another collide the difference between their velocities before impact is proportional to the difference between their velocities after impact.

With the help of a picurte, it is simple to write thies in algebraic terms.

Where the u’s represent the velocities before impact and the v’s represent velocity after impact.

To write this as an equation we introduce a proportionality constant, e, know as the coefficient of restitution.

A special and important case of this problem is when one of the bodies is very massive and has no velocity before or after the collision. (For example, a ball thrown against the wall or floor.) The problem is simplified since v₂ = 0 and u₂ = 0. We then have

(4.4)

This special case shows us that the maximum value of the coefficient of restitution is e = 1. This corresponds to a perfectly elastic collision in which the speed of the ball is unchanged after the collision. If the collision is not perfectly elastic, then Eq. (4.4) tells us what the fractional decrease of the velocity is as the result of a collision with a fixed surface.

If any changes are made in the conditions, such as using a different ball or a different landing surface, the value of e changes. For example, the coefficient of restitution for a basketball and a hardwood floor is 0.75, but if the basketball is replaced by a softball,e drops to a value of 0.33. When a volleyball is dropped on a hardwood floor the coefficient of restitution is 0.75, but on grass it is 0.40. For these examples it is quite evident that the nature of both impacting bodies determines what value e takes. It is clear that it would be incorrect to refer to the “coefficient of restitution of a body,” for this coefficient depends not just on one of the impacting bodies but on both of them.

Example: Measuring the coefficient of restitution

Because velocities are somewhat more difficult to measure than distances it is desirable to convert the form of Eq (4.4) into yet another form. This is done by considering the ball’s motion before and after impact with the floor and using appropriate equations of uniformly accelerated motion to arrive at expressions for v₁ and u₁.

We want to find v_F = u₁ =velocity the ball strikes.

We want to find v_F = u₁ =velocity the ball strikes.

Now substitute these values of u₁ and v₁ into Eq. 4.4.

(4.5)

This gives us an expression of e in terms of h_d the height from which the ball is dropped, and h_b the height to which is subsequently bounced.

E. Oblique Impact

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There are few examples in sports where two bodies collide with each other directly. In most cases the collision is not “head-on,” and we call these oblique impacts. To analyze oblique impacts in detail it is convenient to consider several special cases.

1. Nonspinning ball, e = 1 and no friction.

To examine this case it is necessary to look at the vertical and horizontal components of velocity.

Because the ball and the floor are imagined to be perfectly smooth, and there are no horizontal forces acting on either body, so there are no forces that can alter the horizontal velocity. The horizontal velocity must therefore be the same after impact as before, v_H = u_H. In the vertical direction there is a force that changes the velocity. But, since e=1 this just reverses the direction and leaves the speed unchanged.

The result is that the Angle of Incidence = Angle of Reflection

2. Nonspinning ball, e < 1, and no friction.

Once again there are no horizontal forces so the horizontal velocity is unchanged. But e < 1, so the ball rebounds with reduced vertical speed.

It is easy to see that in this case the angle of incidence is less than the angle of reflection.

3. Nonspinning ball, e<1, with friction.

When friction is present a horizontal force is possible. This horizontal force will reduce the horizontal component of velocity. The vertical component of velocity will also be reduced since e<1. In this case no blanket statement can be made relating the angles of incidence and reflection. However, it should also be noted that the reflected ball will be spinning because of the horizontal force of friction.

F. Collisions with Spin and Friction

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The previous discussions of direct oblique impact assumed that the incident ball was not spinning. If the ball is spinning at the moment it makes contact with a surface, the magnitude of friction is changed according to the rate and direction of that spin. As a direct consequence, the speed and direction the ball after impact are also changed. We will consider the two important cases of spinning balls – back spin and top spin.

1. Back spin. The ball shown below has back spin. It is traveling with velocity u that is broken down into its vertical and horizontal components. Furthermore the ball is spinning with velocity v_spin to the right on the bottom.

Back spin:

Because of the back spin on the ball the bottom of that ball is traveling with an increased velocity to the right. When the ball strikes this results in an increased force that tends to slow down the horizontal velocity of the ball and decreases the angle of reflection. The force will also slow the spin of the ball, and of it is large enough, may even reverse the spin.

2. Top spin. The ball shown below has top spin. It is spinning with velocity v_spin to the right on the top, and v_spin to the left on the bottom.

Top spin:

Because of the topspin , the bottom of the ball is traveling with either decreased velocity to the right, no net velocity, or velocity to the left. Three results are possible depending on the magnitude of v_spin:

a. If v_spin = u_H the bottom of the ball has no net horizontal velocity and the horizontal force on impact is zero. As a result the horizontal velocity is unchanged.

b. If v_spin < u_H the horizontal velocity of the ball is reduces at the bottom. This subsequently reduces the horizontal force. The horizontal velocity will be reduced after impact but to a lesser degree that is experienced in the nonspinning case.

c. If v_spin > u_H the bottom of the ball will have a velocity which is opposite to the motion of the ball. Upon impact this will result in a force which increases the horizontal velocity!

Summary

Momentum Elastic Collision

Impulse Inelastic Collision

Net Force Coefficient of Restitution

Internal Forces Newton’s Law of Impact

External Forces Oblique Impact

Conservation of Momentum Collisions with Spin

Problems

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1. Which has more momentum, a 10kg object moving at 3 m/sec or a 20 kg object moving at 2 m/sec?

2. A baseball has a mass of 0.15 kg. A pitcher throws a 90 mph (40 m/sec) fastball which is caught by the catcher.

a. What is the change in momentum of the ball?

b. What is the impulse on the catcher’s mitt?

c. If this impulse is exerted over a time of 0.05 seconds what is the force on the catcher’s mitt over this time?

3. A baseball (mass = 0.15 kg is thrown towards a batter with a velocity of 40 m/sec (90 mph). The batter hits the ball and returns it directly back with a velocity of 60 m/sec (135 mph).

a. What is the change in momentum of the ball?

b. What is the impulse on the ball?

c. If the ball is in contact with the bat for 0.0025 seconds what is the force on the ball during this time?

4. A Volkswagon with a mass of 1000 kg is traveling at 60 mph when is strikes a Cadillac of mass 2000 kg initially at rest. After the collision the two cars stick together. What is their velocity after the collision?

5. In the following problem we will try to estimate the compress ional force on the body of a 160 lb student if he falls a height of 1 ft and lands with his legs locked.

a. What is the mass of the student in slugs?

b. How fast is he traveling when he strikes the ground?

c. Use the results of parts a and b to find the impulse exerted on the student by the floor.

d. The answer you obtained in part b should be 8 ft/sec. We can use this answer to find how long the impulse of the floor is exerted. Assuming with legs locked that the feet and legs compress 0.05 ft. Find the time over which the impulse is exerted.

e. Use the results of parts c and d to find the force exerted on the student’s legs. (Note: A compress ional force of 4,500 lb is required to break the tibia.)

6. A 192 lb fullback is running with a velocity of 8 ft/sec. A 160 lb defensive back tries to stop this fullback by colliding with him head-on. How fast must the defensive back run?

7. Try problem 4 with a 288 lb defensive lineman trying to stop the fullback.

8. In the original Olympics the Greeks had an event similar to the modern day long jump. The difference being that the Greek jumpers would run with weights in their hands. While they were in the air they would throw the weights backward. How would this technique increase the distance of their jump?

9. The coefficient of restitution of a basketball on a hardwood floor is 0.76. If a basketball is dropped from a height of one meter (waist level for normal people), to what height will is rebound?

10. A nonspinning ball strikes a particular surface at an oblique angle. The coefficient of restitution is 0.5 and the surface is frictionless. How are the horizontal and vertical components of velocity changes as a result of this collision?

11. What is the effect of topspin or backspin on a ping pong ball as it hits the table? What about sidespin?

12. Why does a superball pick up speed after its second bounce?