This law shows us, for example, why a rocket moves forward as gas shoots out behind it, and why a car stops moving when it hits a tree (figure 8.10). The law says that if a body A exerts a force on another body B, then body B exerts an equal and opposite force on body A. You will remember that we met the concept of ‘paired forces’ in chapter 5. This concept is in keeping with
Newton’s Third Law, which may be briefly expressed as follows:
To every action there is an equal and opposite reaction.
A quick summary of Newton’s laws is shown in table 8.3.
Table 8.3 A summary of Newton’s laws of motion.
Law Deals with Resultant force 1st uniform motion in a straight line
or
equilibrium
no external force is present
2nd accelerated motion external force acts
3rd uniform and accelerated motion external forces may be present but internal forces cancel
You might argue that motion cannot occur if action and reaction are equal and opposite, because their resultant is zero. But consider the following case.
If I pull on a rope that is tied around a tree, there might be a state of equilibrium between my hand that pulls the rope, the rope and the tree. I pull on the rope with a certain force, P, and the rope reacts to my pull by pulling back with a tension, T, which (according to the Third Law) is equal to my pull,
P, but opposite to it. These forces, P and T, act on different bodies: the tension
of the rope acts on my hand and the muscles in my hand act on the rope. Note that these two equal forces act on different bodies. This is what the Third Law stipulates. Put another way, the law states that
To every action there is an equal and opposite reaction, these two, the action and the reaction, acting on different bodies.
To demonstrate that the rope pulls my hand, we may cut the rope. What happens then? I fall backwards. Why? Because when my hand was at rest holding on to the rope, we had a state of balanced forces: the pull of the rope on my hand and the pull of the muscles in my hand on the rope. This could be seen as a validation of Newton’s First Law (no unbalanced force, and therefore no momentum change – the momentum of my hand continues to be zero). There is also the case for Newton’s Third Law to be argued, since my pull on the rope (the action) has given rise to a tension in the rope (the reaction). These two act on different bodies, one, the tension of the rope acting on my hand, and the other, the pull of my muscles, acting on the rope.
When the rope is cut, my hand is no longer under the action of two equal and opposite forces, but under the action of only one force, that due to the muscles in my arm. This is where the Second Law comes into play. There is now no longer a tension to ‘balance’ the force in my arm. The result is that my hand moves in the direction of the now ‘unbalanced’ force, the force in my arm, which is backwards.
There are many other examples that might be cited where the principles of at least one of the laws is demonstrated. Some of the more common situations are:
Newton’s Third Law of motion ❯
ITQ8
With the help of Newton’s Second and Third Laws describe how a trampoline can slow down an athlete who lands on it in a vertical direction.
ITQ9
Which do you think is better to avoid being injured: to use a stiff trampoline to break your fall or to use a soft one? State clearly the reasons for your choice.
Worked example 8.4
A frictionless trolley, A, of mass 400 g travelling along a straight, level track at 5 m s–1 collides with another frictionless, stationary trolley, B, of mass
600 g (figure 8.11). On impact, they stick together and move along with the same speed. Calculate the speed of the two trolleys just after they collide.
( ) N TZ¶ N ),-69, ( ) (-;,9 ] Figure 8.11 Solution
Since the trolleys are frictionless, there is no external force due to friction acting on them. The track is horizontal and so the effect of gravity is zero. We use Newton’s First Law, which says that, with no external force acting on the trolleys:
momentum before collision = momentum after collision So we have
momentum before collision = (mass of A) × (initial velocity of A) = 400 g × 5 m s–1
= 2000 g m s–1
because trolley B has zero velocity and it therefore has zero momentum. This value of 2000 g m s–1 is therefore equal to the momentum after
collision. Let the final joint velocity be v. Then
momentum after collision = (mass of A + B) × (velocity of A + B) = (400 + 600) g × v
So
(400 + 600) g × v = 2000 g m s–1
giving
• The behaviour of billiard balls when struck by a cue (Laws 1 and 3). • The firing of a firearm and its recoil (Laws 1 and 3).
• The action of the jet engine of an aircraft or a spacecraft (Laws 1 and 3). • The obvious recoil of a fire hose (Laws 1 and 3).
• The action of a garden sprinkler (Laws 1 and 3).
• Alighting from a moving vehicle (such as a train) (Law 1). • Releasing air from a blown-up balloon (Laws 1 and 3).
• Putting a screw into a wall with a screwdriver while standing on a smooth floor (which law?).
• A short pendulum in a car inclines itself away from the vertical while the car accelerates or decelerates (which law(s)?).
It might well be argued that the application of the Second Law is present, but very subtly, in all of the cases mentioned above, since when the force of the Third Law comes into existence it will cause the object on which it acts to experience an acceleration. This is often manifested in a ‘jerk’ experienced by the object.
What happens to the pendulum when the car moves at constant speed in a straight line?
Worked example 8.5
A ball, A, of mass 2 kg moving along a straight line at 4 m s–1 collides
with another ball, B, of mass 4 kg moving along the same line, but in the opposite direction, at 6 m s–1. On colliding with ball B, ball A moves
backwards at 5 m s– 1. Calculate the velocity of ball B after the collision in
magnitude and direction.
Solution
Total initial momentum of the two balls before collision = 2 kg × 4 m s–1 (for A) + 4 kg × – 6 m s–1 (for B)
Here we are taking vectors acting to the right as positive and those acting to the left as negative.
So the total momentum before collision = 8 kg m s–1 + (–24 kg m s–1) = –16 kg m s–1
Since there was no external force acting on the two balls together during the impact, the total momentum after they collide must not change from the original value.
The total momentum after the collision = 2 kg × –5 m s–1 + 4 kg × v,
where v is assumed to be the unknown velocity of B after the collision. So –16 kg m s–1 = –10 kg m s–1 + 4 kg × v or –16 kg m s–1 + 10 kg m s–1 = 4 kg × v Giving 4 kg × v = – 6 kg m s–1 and v = –1.5 m s–1
The negative sign of the velocity shows that the velocity of ball B after the collision is 1.5 m s–1 in the same direction (to the left) which it had before
the collision.
Worked examples 8.4 and 8.5 illustrate the application of Newton’s First Law. We now look at examples of the application of the Second and Third Laws.
Worked example 8.6
The collision of the balls in the last example lasted 0.1 second. Calculate (i) the change in momentum of each ball; (ii) the rate of change of the momentum of each ball; and (iii) the force acting on each ball to change its momentum.
Solutions
(i) Change in momentum = final momentum – initial momentum For ball A:
change = 2 kg (–5 – 4) m s–1 = 2 × –9 kg m s–1 = – 18 kg m s–1
For ball B:
What we just did was not a proof of the laws, but a verification of them. We will now discuss investigations that can be carried out to test the laws in order to find out to what extent they are true.