Physics A Level
Chapter 6: Momentum 6.7 Understanding motion
Physics A Level
Chapter 6: Momentum 6.7 Understanding motion
In Chapter 3, we looked at Newton’s laws of motion. We can get further insight into these laws by thinking about them in terms of momentum.
Newton’s first law of motion
In everyday speech, we sometimes say that something has momentum when we mean that it keeps on moving on its own. An oil tanker is difficult to stop at sea, because of its momentum. We use the same word even when we’re not talking about an object: ‘The election campaign is gaining momentum’, for example. This idea of keeping on moving is just what we discussed in connection with Newton’s first law of motion:
An object will remain at rest or keep travelling at constant velocity unless it is acted on by a resultant force.
An object travelling at constant velocity has constant momentum. Hence, the first law is really saying that the momentum of an object remains the same unless the object experiences an external force.
Newton’s second law of motion
Newton’s second law of motion links the idea of the resultant force acting on an object and its momentum. A statement of Newton’s second law is:
The resultant force acting on an object is directly proportional to the rate of change of the linear momentum of that object. The resultant force and the change in momentum are in the same direction.
Hence:
resultant force $ \propto $ rate of change of momentum This can be written as:
This can be written as:
$F \propto \frac{{\Delta p}}{{\Delta t}}$
where F is the resultant force and $\Delta p$ is the change in momentum taking place in a time interval of Δt.
(Remember that the Greek letter delta, $\Delta $, is a shorthand for ‘change in’, so $\Delta p$ means ‘change in momentum’.) The changes in momentum and force are both vector quantities, so these two quantities must be in the same direction.
The unit of force (the newton, N) is defined to make the constant of proportionality equal to one, so we can write the second law of motion mathematically as:
$F \propto \frac{{\Delta p}}{{\Delta t}}$
Worked example 5 shows how to use this equation. This equation also shows the newton second (N s) can be used as a unit of momentum.
If the forces acting on an object are balanced, there is no resultant force and the object’s momentum will remain constant. If a resultant force acts on an object, its momentum (velocity and/or direction) will change. The equation gives us another way of stating Newton’s second law of motion:
The resultant force acting on an object is equal to the rate of change of its momentum. The resultant force and the change in momentum are in the same direction.
This statement effectively defines what we mean by a force; it is an interaction that causes an object’s momentum to change. So, if an object’s momentum is changing, there must be a force acting on it. We can find the size and direction of the force by measuring the rate of change of the object’s momentum.
A special case of Newton’s second law of motion
Imagine an object of constant mass m acted upon by a resultant force F. The force will change the momentum of the object. According to Newton’s second law of motion, we have:
$F = \frac{{\Delta p}}{{\Delta t}} = \frac{{mv - mu}}{t}$
where u is the initial velocity of the object, v is the final velocity of the object and t is the time taken for the change in velocity. The mass m of the object is a constant; hence the equation can be rewritten as:
$\begin{array}{l}
F = \frac{{m(v - u)}}{{\Delta t}}\\
= m\left( {\frac{{v - u}}{t}} \right)\\
= \frac{{27000 - 4500}}{{12}}\\
= 1875N\, \approx \,1900N
\end{array}$
The term in brackets on the right-hand side is the acceleration a of the object. Therefore, a special case of Newton’s second law is:
$F = ma$
We have already met this equation in Chapter 3. In Worked example 5, you could have determined the average force acting on the car using this simplified equation for Newton’s second law of motion.
Remember that the equation $F = ma$ is a special case of $F = \frac{{\Delta p}}{{\Delta t}}$ that only applies when the mass of the object is constant. There are situations where the mass of an object changes as it moves, for example, a rocket that burns a phenomenal amount of chemical fuel as it accelerates upwards.
Questions
13) A car of mass $1000 kg$ is travelling at a velocity of $ + 10\,m\,{s^{ - 1}}$. It accelerates for $15 s$, reaching a velocity of $ + 24\,m\,{s^{ - 1}}$. Calculate:
the change in the momentum of the car in the $15 s$ period the average resultant force acting on the car as it accelerates.
14) A ball is kicked by a footballer. The average force on the ball is $240 N$ and the impact lasts for a time interval of $0.25 s$.
Calculate the change in the ball’s momentum.
State the direction of the change in momentum.
15) Water pouring from a broken pipe lands on a flat roof. The water is moving at $5.0\,m\,{s^{ - 1}}$ when it strikes the roof. The water hits the roof at a rate of $ 10\,kg\,{s^{ - 1}}$. Calculate the force of the water hitting the roof. (Assume that the water does not bounce as it hits the roof. If it did bounce, would your answer be greater or smaller?)
16) A golf ball has a mass of $0.046 kg$. The final velocity of the ball after being struck by a golf club is $50\,m\,{s^{ - 1}}$. The golf club is in contact with the ball for a time of $1.3 ms$. Calculate the average force exerted by the golf club on the ball.
Newton’s third law of motion
Newton’s third law of motion is about interacting objects. These could be two magnets attracting or repelling each other, two electrons repelling each other, etc. Newton’s third law states:
When two bodies interact, the forces they exert on each other are equal and opposite.
How can we relate this to the idea of momentum? Imagine holding two magnets, one in each hand. You gradually bring them towards each other (Figure 6.21) so that they start to attract each other. Each feels a force pulling it towards the other. The two forces are the same size, even if one magnet is stronger than the other. One magnet could even be replaced by an unmagnetised piece of steel and they would still attract each other equally.
If you release the magnets, they will gain momentum as they are pulled towards each other. One gains momentum to the left while the other gains equal momentum to the right.
Each is acted on by the same force, and for the same time. So, momentum is conserved. In fact, the law of conservation of momentum can be proved using Newton’s second and third laws of motion. Consider an object of mass ${m_x}$ and velocity vx colliding with a mass ${m_y}$ and velocity ${v_y}$. If the system is closed, then the force ${F_x}$ and the force ${F_y}$ on the twomasses are equal and opposite.
$\begin{array}{l}
{F_x} = - {F_y}\\
\frac{{\Delta {m_x}{v_x}}}{{\Delta t}} = - \frac{{\Delta {m_y}{v_y}}}{{\Delta t}}\\
\frac{{\Delta ({m_x}{v_x} + {m_y}{v_y})}}{{\Delta t}} = 0
\end{array}$
So, $\Delta ({m_x}{v_x} + {m_y}{v_y}) = 0$ and there has been no change in the total momentum.
EXAM-STYLE QUESTIONS
1) Which quantity has the same unit as the rate of change of momentum? [1]
A: acceleration
B: energy
C: weight
D: work
2) A railway truck of mass $8000 kg$ travels along a level track at a velocity of $2.5\,m\,{s^{ - 1}}$ and collides with a stationary truck of mass $12000 kg$. The collision takes $4.0 s$ and the two trucks move together at the same velocity after the collision.
What is the average force that acts on the $8000 kg$ truck during the collision? [1]
A: $2000 N$
B: $3000 N$
C: $5000 N$
D: $12000 N$
3) An object has mass $2.0 \pm 0.2\,kg$ and a velocity of $10 \pm 1\,m\,{s^{ - 1}}$.
What is the percentage uncertainty in the momentum of the object? [1]
A: $1\% $
B: $6\% $
C: $10\% $
D: $20\% $
4) An object is dropped and its momentum increases as it falls toward the ground.
Explain how the law of conservation of momentum and Newton’s third law of motion can be applied to this situation. [2]
5) A ball of mass $2.0 kg$, moving at $3\,m\,{s^{ - 1}}$, strikes a wall and rebounds with almost exactly the same speed. State and explain whether there is a change in:
a: the momentum of the ball [3]
b: the kinetic energy of the ball. [1]
[Total: 4]
6) a: Define linear momentum. [1]
b: Determine the base units of linear momentum in the SI system. [1]
c: A car of mass $900 kg$ starting from rest has a constant acceleration of $3.5\,m\,{s^{ - 2}}$. Calculate its momentum after it has travelled a distance of $40 m$. [2]
d: This diagram shows two identical objects about to make a head-on
collision. The objects stick together during the collision. Determine the
final speed of the objects. State the direction in which they move. [3]
[Total: 7]
7) a: Explain what is meant by an:
i- elastic collision [1]
ii- inelastic collision. [1]
b: A snooker ball of mass $0.35 kg$ hits the side of a snooker table at right angles and bounces off also at right angles. Its speed before collision is $2.8\,m\,{s^{ - 1}}$
and its speed after is $2.5\,m\,{s^{ - 1}}$. Calculate the change in the momentum of the ball. [2]
c: Explain whether or not momentum is conserved in the situation described in part b. [1]
[Total: 5]
8) A car of mass $1100 kg$ is travelling at $24\,m\,{s^{ - 1}}$. The driver applies the brakes
a: and the car decelerates uniformly and comes to rest in $20 s$.
Calculate the change in momentum of the car. [2]
Calculate the braking force on the car. [2]
Determine the braking distance of the car. [2]
[Total: 6]
9) A marble of mass $100 g$ is moving at a speed of $0.4\,m\,{s^{ - 1}}$ in the x-direction.
a: Calculate the marble’s momentum. [2]
The marble strikes a second, identical marble. Each moves off at an angle of ${45^ \circ }$ to the x-direction.
b: Use the principle of conservation of momentum to determine the speed of each marble after the collision. [3]
c: Show that kinetic energy is conserved in this collision. [2]
[Total: 7]
10) A cricket bat strikes a ball of mass $0.16 kg$ travelling towards it. The ball initially hits the bat at a speed of $25\,m\,{s^{ - 1}}$ and returns along the same path with the same speed. The time of impact is $0.0030 s$. You may assume no force is exerted on the bat by the cricketer during the actual collision.
a: Determine the change in momentum of the cricket ball. [2]
b: Determine the force exerted by the bat on the ball. [2]
c: Describe how the laws of conservation of energy and momentum apply to this impact and state whether the impact is elastic or inelastic. [4]
[Total: 8]
11) a: State the principle of conservation of momentum and state the condition under which it is valid. [2]
b: An arrow of mass $0.25 kg$ is fired horizontally towards an apple of mass $0.10 kg$ that is hanging on a string, as shown in Figure 6.23.
The horizontal velocity of the arrow as it enters the apple is $30\,m\,{s^{ - 1}}$. The apple was initially at rest and the arrow sticks in the apple.
i- Calculate the horizontal velocity of the apple and arrow immediately after the impact. [2]
ii- Calculate the change in momentum of the arrow during the impact. [2]
iii- Calculate the change in total kinetic energy of the arrow and apple during the impact. [2]
iv- A rubber-tipped arrow of mass $0.25 kg$ is fired at the centre of a stationary ball of mass $0.25 kg$. The collision is perfectly elastic.
Describe what happens and state the relative speed of separation of the arrow and the ball. [2]
[Total: 10]
12) a: State what is meant by:
i- a perfectly elastic collision [1]
ii- a completely inelastic collision. [1]
b: A stationary uranium nucleus disintegrates, emitting an alpha-particle of mass $6.65 \times {10^{ - 27}}kg$ and another nucleus X of mass $3.89 \times {10^{ - 25}}kg$.
i- Explain why the alpha-particle and nucleus X must be emitted in exactly opposite directions. [2]
ii- Using the symbols ${v_\alpha }$ and ${v_x}$ for velocities, write an equation for the conservation of momentum in this disintegration. [1]
iii- Using your answer to part b ii, calculate the ratio ${v_\alpha }:{v_x}$ after the disintegration. [1]
[Total: 6]
13) a: State two quantities that are conserved in an elastic collision. [1]
b: A machine gun fires bullets of mass $0.014 kg$ at a speed of $640\,m\,{s^{ - 1}}$.
i- Calculate the momentum of each bullet as it leaves the gun. [1]
ii- Explain why a soldier holding the machine gun experiences a force when the gun is firing. [2]
iii- The maximum steady horizontal force that a soldier can exert on the gun is $140 N$. Calculate the maximum number of bullets that the gun can fire in one second. [2]
[Total: 6]
14) Two railway trucks are travelling in the same direction and collide. The mass of truck X is $2.0 \times {10^4}kg$ and the mass of truck Y is $3.0 \times {10^4}kg$. This graph shows how the velocity of each truck varies with time.
a: Copy and complete the table. [6]
| Final kinetic energy / J |
Initial kinetic energy / J |
Change in momentum / kg $m\,{s^{ - 1}}$ |
|
| truck X | |||
| truck Y |
b: State and explain whether the collision of the two trucks is an example of an elastic collision. [2]
c: Determine the force that acts on each truck during the collision. [2]
[Total: 10]
SELF-EVALUATION CHECKLIST
After studying the chapter, complete a table like this:
| I can | See topic… | Needs more work | Almost there | Ready to move on |
| define and use linear momentum | 6.2 | |||
| state and apply the principle of conservation of momentum to collisions in one and two dimensions | 6.3, 6.5 | |||
| relate force to the rate of change of momentum | 6.2 | |||
| state all three of Newton’s laws of motion | 6.7 | |||
| recall that, for a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation | 6.3 | |||
| discuss energy changes in perfectly elastic and inelastic collisions. | 6.3 |