TOPIC 1A: FORCES and MOTION

1.6 Momentum

Imagine you work in a sports shop. Some of the equipment is stored on very high storage platforms. Your job is to catch the equipment as it is thrown down to you.

Which would you rather catch - the netball or the large punch bag from the top platform?

**Figure 1. A sports shop - the concept of momentum**

This topic is really just common sense - the large punch bag will be much harder to catch and stop moving because it is heavy, and also it will be moving much faster when it reaches you. (It has further to fall).

This idea of 'how hard it is to stop' is called the **momentum** of an object. A heavy, fast object will have a lot of momentum.

A fast car has more momentum -is harder to stop - than a bicycle because it has more mass and a higher velocity.

The formula for momentum is pretty straightforward, and (as you would expect) it depends on the mass and the velocity of the object:

**momentum = mass x velocity**

**p = m x v**

[kg m/s] = [kg] x [m/s]

The units of momentum are simply the units of mass (kg) multiplied by the units for velocity (m/s), and are therefore **kg m/s **. (Kilograms metres per second).

For example, a 2 kg rock rolling down a hill at 3 m/s will have **a momentum of 6 kg m/s. **

One of the most important ideas in science is that whenever two objects collide, the momentum is transferred. In this video, you can see this happening as a moving metal ball transfers momentum to the ball at the other end.

**Youtube: Newton's cradle **(Jeff Regester)

When two metal balls collide later in the video, there is more momentum at the start, so therefore there is more momentum at the end after the impact with two of them moving on the other side.

This important rule is called the 'conservation of momentum':

**The total momentum before any collision ('event')
is equal to the total momentum after the collision, as long as no external forces act.**

This means that as long as nothing else pushes or pulls on the system apart from the forces involved in the collision, then momentum is conserved.

The questions below are examples of how to do these calculations - see if you can work them out before looking at the solutions.

**Questions:**

1. A 0.6 kg basketball (P) is thrown at another stationary ball (Q) at 5 m/s.

a) What is the initial momentum of the ball P?

b) Ball P stops completely after the impact (just like the 'newton's cradle' balls shown in the video above).

What is the momentum of ball Q after the impact?

a) The momentum of the ball is p = m x v,

so p = 0.6 x 5 = **3.0 kg m/s.**

b) As this momentum is conserved and ball P stops moving, all of the momentum is transferred to ball Q.

Therefore the momentum of ball Q is **3.0 kg m/s.**

2. The diagram shows a collision involving a toy trolley and a ball. The trolley of mass 2 kg is initially moving at 4 m/s. The ball is initially stationary. After the collision the trolley continues moving in the same direction at a slower speed of 3 m/s.

a) Calculate:-- i) the initial momentum of the trolley.
- ii) the final momentum of the trolley.

b) State the initial momentum of the ball.

c) Calculate the final velocity of the ball .

a) The momentum of the trolley = m x v,

so :

i) for the initial momentum, p = m x v = 2 x 4 ,

**p = 8 kg m/s **

ii) for the final momentum, p = m x v = 2 x 3,

**p = 6 kg m/s **

b) Notice the 'state' part of the question - a hint that it should be a straightforward answer not needing any maths.

The **momentum is zero** as the ball is not moving before the collision.

c) This is the tricky part and involves several stages.

From part (a) the total initial momentum for the whole system is 8 (kg m/s).
Therefore the final total momentum must also be 8 kg m/s.

As the trolley has a final momentum of 6 kg m/s (part a ii), the ball must have **2 kg m/s** of momentum at the end. This is because:

Initial momentum of trolley + ball = final momentum of trolley + ball,

So 8 + 0 = 6 + final momentum of ball,

Therefore final momentum = 2 (kg m/s).

As p = m x v, then:

v = | p |

m |

v = | 2 |

0.5 |

When we start to run, we accelerate. This means we gain momentum. The push from our legs results in a gain in momentum.

This gives us an important link between the topic of momentum and this idea of forces.

The change in momentum can be written as
Δp. Isaac Newton realised that time is also a factor. If we apply the same force over a longer time, the change in momentum will be higher. He put all these ideas together into a formula:

force (N) = |
change in momentum (kg m/s) |

time taken(s) |

F = |
Δp |

Δt |

**Example**:

A 400 g netball is thrown from rest, reaching a velocity of 5 m/s in 0.3 s. Calculate:

a) the final momentum of the ball

b) the average force applied to the ball.

**Answer:**

a) We know the momentum p = mv, and m= 400 g (0.4 kg) , v= 5 m/s^{2}.

Therefore F = 0.4 x 5 = **2 kg m/s.**

b) The force applied is F = Δp/Δt,

So therfore F = 2/0.3 **= 6.67 N** (to 3 sig, figs.)

When a ball is hit or kicked, the momentum changes rapidly. We say that there has been an** impulse** on the ball. In fact, any **change in momentum Δp** is called an impulse.

We can rearrange the formula above to find the impulse:

**Impulse (change in momentum) = force x time taken**

**Δp or Δ(mv) = FΔt**

Impulse is just another term used when describing how momentum can chage when a force is acting.

**Questions.**

3. A 100 g toy car accelerates from rest to 2 m/s in 5 seconds.

What force is needed to make the car accelerate?

We know that:

**F = 0.04 N**

F = | Δp | = | (mv - mu) |

t | t |

We know from the question that:

- m = 100 g = 0.1 kg (remember to convert units to kg!)
- u = 0 m/s
- v = 2 m/s
- t = 5 s

Putting all of these into the formula gives:

F = | (0.1 x 2) - (0.1 x 0) |

5 |

F = | (0.2) - (0) | = | 0.2 |

5 | 5 |

4. On a roller-coaster ride, passengers are held safely in their seats by seat belts. The belt provides a force that will push backwards on the passengers and slow them down in an emergency.

The **maximum** force that the seat belts can apply is 2 kN.

If a 100 kg passenger slows down from 15 m/s to rest, calculate:

a) The change in momentum of the passenger.

b) The shortest time in which the passenger can be stopped safely.

a) The initial velocity here is 'u' and equals 15 m/s. The final velocity v = 0, so:

Change in momentum Δp = mv - mu,

so Δp = (100 x 0) - (100 x15),

**Δp = -1500 kg m/s. **(or just 1500 kg m/s - the negative sign tells us that the momentum is reducing instead of increasing. however the **change** is momentum is still 1500 kg m/s, so both answers will be marked as correct).

Change in momentum Δp = mv - mu,

so Δp = (100 x 0) - (100 x15),

b) We know that:

F = | Δp | or F = | (mv - mu) |

t | t |

so the maximum force 2000 = | 1500 |

t |

So the time will be:

t = | 1500 |

2000 |

**t = 0.75 seconds **at the shortest.**
**If the time is shorter than this, the force will exceed 2 kN and the seatbelt may break.

5. A 300 g ball is kicked, changing the velocity from 2 m/s to 6 m/s in 0.15 s. Calculate:

a) the impulse acting on the ball.

b) the force acting on the ball.

a) We know that impulse = Δ(mv). The mass is constant, so we can skip a step and just find the change in veloicty and muliply it by the mass:

impulse = m Δv
= 0.3 (kg) x 4 (m/s) **= 1.2 kg m/s**

b) The formula to find the force from the momentum change is F = Δp/t. The impulse of 1.2 kg m/s from part (a) is equal to the change in momentum Δp.

Therefore F = Δp/t = 1.2/0.15 **= 13.3 N** (to 3 sig. figs.)

**Extension.**

Where does the formula for the force come from? If you want to learn more about this, click the extension button here.

We have learned about 2 different formulas for acceleration:

a) F = m x a

b) a = | Δv | or a = | v - u |

t | t |

Can you combine these 2 formulas together to give the formula for force and momentum given above?

Question 4 above shows us that when we need to use seatbelt, they provide a force in an emergency that can slow us down safely and prevent injuries. However the force on our bodies depends on the **change in momentum** and also the** time taken** to do this. If the time is longer, the force needed will be smaller.

Consider this car used in 'stock car' races. If it hits another car, it will slow down rapidly.

**Figure 2. A car used in 'stock car racing' where collisions are part of the entertainment.**

If the car slows down from top speed to rest several times, the change in momentum will always be the same. However, as:

F = | Δp |

t |

And if Δp is constant, then the smaller the value of 't' is , the higher the force needed to cause the deceleration.

This could be dangerous to the drivers.

You will notice on this car that the front of the car has many dents - the front end crumples in. This is called a 'crumple zone'. It means that the front end **slowly** absorbs the impact . This makes the value of 't' a bit longer, and so the force a bit smaller. The impact for the driver is not a severe - safer for the driver.

**Top Tip:**

When asked these questions on impact safety in exams, remember to answer in **terms of physics**.

- Show the formula and then...
- talk about how time increases...
- making less force on the driver for the same change in momentum:

**Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:**