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4.2Impulse and force–time graphs

We may rearrange F→net =Δ p→Δ t to get Δ p→=F→NET Δ t.

The quantity F→net Δ t is called the impulse the product of force and the time interval for which the force acts; it equals the change in momentum: J→=Δ p→
impulse = area under a graph of force against time of the force and is denoted by J→.

It is the product of the average net force and the time for which the force acts.

The impulse is also equal to the change in momentum. Notice that impulse is a vector; its direction is the same as that of the force (or the change in momentum).

J→=Δ p→

It can also be shown that impulse is the area under the curve in a graph of force against time.

When you jump from a height of, say, 1 m, you will land on the ground with a speed of about 4.5 m s–1. Assuming your mass is 60 kg your momentum just before landing will be 270 N s and will become zero after you land. From F net Δt = Δp, this can be achieved with a small force acting for a long time or a large force acting for a short time. You will experience the large force if you do not bend your knees upon landing—keeping your knees stiff means that you will come to rest in a short time; that is, Δt will be very small and the force large (which may damage your knees).

Figure 4.2 shows three different force–time graphs. Graph a shows a (nonconstant) force that increases from zero, reaches a maximum value and then drops to zero again. The force acts for a time interval of about 2 ms. The impulse is the area under the curve. Without calculus we can only estimate this area by tediously counting squares: each small square has area 0.1 ms × 0.2 N = 2 × 10−5 N s. There are about 160 full squares under the curve, and so the impulse is 3 × 10−3 N s. (In this case, it is not a bad approximation to consider the shape under the curve to be a triangle but with a base of 1.3 ms so that the area is then 1 2×1.3×10−3×4≈3×10−3 N s.)

Figure 4.2: Three different force–time graphs.

In graph b, the force is constant. The impulse of the force is 6.0 × (8.0 − 2.0) = 36 N s. If this force were to act on a body of mass 12 kg, initially at rest, the speed of the body after the force stops acting would be found from

Δ p=36 Ns

m v−0=36 Ns

v=36 12=3.0 m s−1

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Check yourself 2

The graph in Figure 4.3 shows the variation with time of the momentum of a body moving on a straight line.

Figure 4.3: For check yourself question 2.

Identify the part of the motion where the force acting on the body has its greatest magnitude.

Worked Example 4.3

Consider graph c in Figure 4.2. The force acts on a body of mass 3.0 kg initially at rest.

Calculate:

athe initial acceleration of the body

bthe speed at 4.0 s

cthe speed at 6.0 s.

Answer

a a=F m=12 3.0=4.0 m s−2.

bThe impulse from 0 s to 4.0 s is the area under the graph:

1 2×4.0×12=24 Ns

and is equal to the change in momentum:

m v−0

So, v=24 3.0=8.0 m s−1.

cThe impulse from 0 s to 6.0 s is 1 2×4.0×12−1 2×2.0×6.0=18 Ns. (Areas under the time axis are negative.)

So, the speed at 6.0 s is v=18 3.0=6.0 m s−1.

Worked Example 4.4

A ball of mass 0.20 kg moving at 3.6 m s–1 on a horizontal floor collides with a vertical wall. The ball rebounds with a speed of 3.2 m s–1. The ball was in contact with the wall for 12 ms. Determine the maximum force exerted on the ball, assuming that the force depends on time according to Figure 4.4.

Figure 4.4: For worked example 4.4.

Answer

The change in momentum of the ball is 0.20 × (−3.2) − 0.20 × 3.6 = −1.36 N s. The magnitude of the change is 1.36 N s. This is the area under the force–time graph.

The area is

1 2×12×10−3×F max

So:

1 2×12×10−3×F max=1.36 Ns⇒F max=0.227×10 3≈2.3×10 2 N.

Check yourself 3

A girl of mass 40 kg jumps vertically upward, lifting her centre of mass by 0.80 m after losing contact with the ground. What is the impulse she received when in contact with the ground? (Use g = 10 m s–2.)

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Links

We have seen the application of impulse in mechanics. In Unit B the concept of impulse will be applied to show that the walls of a container exert forces on molecules and so molecules exert forces on the walls. This will explain the origin of pressure in gases.

TEST YOUR UNDERSTANDING

5A student says that ‘a large force always produces a larger impulse than a smaller force’.

Explain why the student’s statement is not correct.

6A time-varying force varies with time, as shown in Figure 4.5. The force acts on a body of mass 4.0 kg.

Figure 4.5: For question 6.

aFind the impulse of the force from t = 0 to t = 15 s.

bFind the speed of the mass at 15 s, assuming the initial velocity was zero.

cState the initial velocity of the body so it is brought to rest at 15 s.

7A 0.150 kg ball moving horizontally at 3.00 m s−1 collides normally with a vertical wall and bounces back with the same speed.

aCalculate the impulse delivered to the ball.

bThe ball was in contact with the wall for 0.125 s. Find the average force exerted by the ball on the wall.

8A boy rides on a scooter pushing on the road with one foot with a horizontal force that depends on time, as shown in Figure 4.6. While the scooter rolls, a constant force of 25 N opposes the motion. The combined mass of the boy and scooter is 25 kg.

Figure 4.6: For question 8.

aFind the speed of the boy after 4.0 s, assuming he started from rest.

bDraw a graph to represent the variation of the boy’s speed with time.

9Figure 4.7 shows the variation with time of the force exerted on a ball as the ball comes into contact with a spring.

Figure 4.7: For question 9.

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aFor how long was the spring in contact with the ball?

bEstimate the magnitude of the change in momentum of the ball.

cWhat was the average force that was exerted on the ball?

10Figure 4.8 shows the variation with time of the force exerted on a body of mass 4.0 kg that is initially at rest.

Figure 4.8: For question 10.

Calculate the average power delivered to the body during the 10 s interval.

Science in context

Figure 4.9: Crumple zones.

Modern cars are designed to have ‘crumple zones’, air bags and safety belts. Discuss how each of these features helps the safety of the passengers in the case of an accident. In Figure 4.9 the front part of the car is completely destroyed, but the passenger compartment is intact.