8.2 Impulse - College Physics for AP® Courses

Learning Objectives

By the end of this section, you will be able to:

Define impulse.
Describe effects of impulses in everyday life.
Determine the average effective force using graphical representation.
Calculate average force and impulse given mass, velocity, and time.

The information presented in this section supports the following AP^® learning objectives and science practices:

3.D.2.1 The student is able to justify the selection of routines for the calculation of the relationships between changes in momentum of an object, average force, impulse, and time of interaction. (S.P. 2.1)
3.D.2.2 The student is able to predict the change in momentum of an object from the average force exerted on the object and the interval of time during which the force is exerted. (S.P. 6.4)
3.D.2.3 The student is able to analyze data to characterize the change in momentum of an object from the average force exerted on the object and the interval of time during which the force is exerted. (S.P. 5.1)
3.D.2.4 The student is able to design a plan for collecting data to investigate the relationship between changes in momentum and the average force exerted on an object over time. (S.P. 4.1)
4.B.2.1 The student is able to apply mathematical routines to calculate the change in momentum of a system by analyzing the average force exerted over a certain time on the system. (S.P. 2.2)
4.B.2.2 The student is able to perform analysis on data presented as a force-time graph and predict the change in momentum of a system. (S.P. 5.1)

The effect of a force on an object depends on how long it acts, as well as how great the force is. In Example 8.1, a very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller than the tennis racquet’s force) would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum $Δ p$ .

By rearranging the equation $F_{net} = \frac{Δ p}{Δ t}$ to be

Δ p = F_{net} Δ t,

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we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity $F_{net} Δ t$ is given the name impulse. Impulse is the same as the change in momentum.

Impulse: Change in Momentum

Change in momentum equals the average net external force multiplied by the time this force acts.

Δ p = F_{net} Δ t

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The quantity $F_{net} Δ t$ is given the name impulse.

There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a car, and certainly the airbags, allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant, whether an air bag is deployed or not, but the force (to bring the occupant to a stop) will be much less if it acts over a larger time. Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident.

Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs can be immense if you land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with a parachute, extends the time over which the force (on you from the ground) acts.

Making Connections: Illustrations of Force Exerted

The force exerted by the barrier on the block is a constant 15 N from t=0 until t=0.24 s, at which time the force drops to zero.

Figure 8.3 This is a graph showing the force exerted by a fixed barrier on a block versus time.

A 1.2-kg block slides across a horizontal, frictionless surface with a constant speed of 3.0 m/s before striking a fixed barrier and coming to a stop. In Figure 8.3, the force exerted by the barrier is assumed to be a constant 15 N during the 0.24-s collision. The impulse can be calculated using the area under the curve.

Δ p = F Δ t = (15 N) (0.24 s) = 3.6 kg \cdot m/s

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Note that the initial momentum of the block is:

p_{i n i t i a l} = m v_{i n i t i a l} = (1.2 kg) (- 3.0 m/s) = - 3.6 kg \cdot m/s

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We are assuming that the initial velocity is −3.0 m/s. We have established that the force exerted by the barrier is in the positive direction, so the initial velocity of the block must be in the negative direction. Since the final momentum of the block is zero, the impulse is equal to the change in momentum of the block.

Suppose that, instead of striking a fixed barrier, the block is instead stopped by a spring.Consider the force exerted by the spring over the time interval from the beginning of the collision until the block comes to rest.

The force exerted by the spring on the block steadily increases from 0 at t=0 up to 30 N at t=0.24 s.

Figure 8.4 This is a graph showing the force exerted by a spring on a block versus time.

In this case, the impulse can be calculated again using the area under the curve (the area of a triangle):

Δ p = \frac{1}{2} (base) (height) = \frac{1}{2} (0.24 s) (30 N) = 3.6 kg \cdot m/s

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Again, this is equal to the difference between the initial and final momentum of the block, so the impulse is equal to the change in momentum.

Example 8.3

Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall

Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of $30º$ from the perpendicular, and bounces off at an angle of $30º$ from perpendicular to the wall.

(a) Determine the direction of the force on the wall due to each ball.

(b) Calculate the ratio of the magnitudes of impulses on the two balls by the wall.

Strategy for (a)

In order to determine the force on the wall, consider the force on the ball due to the wall using Newton’s second law and then apply Newton’s third law to determine the direction. Assume the $x$ -axis to be normal to the wall and to be positive in the initial direction of motion. Choose the $y$ -axis to be along the wall in the plane of the second ball’s motion. The momentum direction and the velocity direction are the same.

Solution for (a)

The first ball bounces directly into the wall and exerts a force on it in the $+ x$ direction. Therefore the wall exerts a force on the ball in the $- x$ direction. The second ball continues with the same momentum component in the $y$ direction, but reverses its $x$ -component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum.

These changes mean the change in momentum for both balls is in the $- x$ direction, so the force of the wall on each ball is along the $- x$ direction.

Strategy for (b)

Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball.

Solution for (b)

Let $u$ be the speed of each ball before and after collision with the wall, and $m$ the mass of each ball. Choose the $x$ -axis and $y$ -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.

p_{xi} = mu; p_{yi} = 0

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p_{xf} = - mu; p_{yf} = 0

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Impulse is the change in momentum vector. Therefore the $x$ -component of impulse is equal to $- 2 mu$ and the $y$ -component of impulse is equal to zero.

Now consider the change in momentum of the second ball.

p_{xi} = mu cos 30º; p_{yi} = –mu sin 30º

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p_{xf} = - mu cos 30º; p_{yf} = - mu sin 30º

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It should be noted here that while $p_{x}$ changes sign after the collision, $p_{y}$ does not. Therefore the $x$ -component of impulse is equal to $- 2 mu cos 30º$ and the $y$ -component of impulse is equal to zero.

The ratio of the magnitudes of the impulse imparted to the balls is

\frac{2 mu}{2 mu cos 30º} = \frac{2}{\sqrt{3}} = 1 . 155 .

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Discussion

The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative $x$ -direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive $x$ -direction.

Our definition of impulse includes an assumption that the force is constant over the time interval $Δ t$ . Forces are usually not constant. Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force $F_{eff}$ that produces the same result as the corresponding time-varying force. Figure 8.5 shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times $t_{1}$ and $t_{2}$ . That area is equal to the area inside the rectangle bounded by $F_{eff}$ , $t_{1}$ , and $t_{2}$ . Thus the impulses and their effects are the same for both the actual and effective forces.

Figure is a graph of force, F, versus time, t. Two curves, F actual and F effective, are drawn. F actual is drawn between t sub1 and t sub 2 and it resembles a bell-shaped curve that peaks mid-way between t sub 1 and t sub 2. F effective is a line parallel to the x axis drawn at about fifty five percent of the maximum value of F actual and it extends up to t sub 2.

Figure 8.5 A graph of force versus time with time along the

x

-axis and force along the $y$ -axis for an actual force and an equivalent effective force. The areas under the two curves are equal.

Making Connections: Baseball

In most real-life collisions, the forces acting on an object are not constant. For example, when a bat strikes a baseball, the force is very small at the beginning of the collision since only a small portion of the ball is initially in contact with the bat. As the collision continues, the ball deforms so that a greater fraction of the ball is in contact with the bat, resulting in a greater force. As the ball begins to leave the bat, the force drops to zero, much like the force curve in Figure 8.5. Although the changing force is difficult to precisely calculate at each instant, the average force can be estimated very well in most cases.

Suppose that a 150-g baseball experiences an average force of 480 N in a direction opposite the initial 32 m/s speed of the baseball over a time interval of 0.017 s. What is the final velocity of the baseball after the collision?

Δ p = F Δ t = (480) (0.017) = 8.16 kg \cdot m/s

8.28

m v_{f} - m v_{i} = 8.16 kg \cdot m/s

8.29

(0.150 kg) v_{f} - (0.150 kg) (- 32 m/s) = 8.16 kg \cdot m/s

8.30

v_{f} = 22 m/s

8.31

Note in the above example that the initial velocity of the baseball prior to the collision is negative, consistent with the assumption we initially made that the force exerted by the bat is positive and in the direction opposite the initial velocity of the baseball. In this case, even though the force acting on the baseball varies with time, the average force is a good approximation of the effective force acting on the ball for the purposes of calculating the impulse and the change in momentum.

Making Connections: Take-Home Investigation—Hand Movement and Impulse

Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why?

Making Connections: Constant Force and Constant Acceleration

The assumption of a constant force in the definition of impulse is analogous to the assumption of a constant acceleration in kinematics. In both cases, nature is adequately described without the use of calculus.

Applying the Science Practices: Verifying the Relationship between Force and Change in Linear Momentum

Design an experiment in order to experimentally verify the relationship between the impulse of a force and change in linear momentum. For simplicity, it would be best to ensure that frictional forces are very small or zero in your experiment so that the effect of friction can be neglected. As you design your experiment, consider the following:

Would it be easier to analyze a one-dimensional collision or a two-dimensional collision?
How will you measure the force?
Should you have two objects in motion or one object bouncing off a rigid surface?
How will you measure the duration of the collision?
How will you measure the initial and final velocities of the object(s)?
Would it be easier to analyze an elastic or inelastic collision?
Should you verify the relationship mathematically or graphically?