89 Momentum Macro

Physics Factsheet www.curriculum-press.co.uk

Number 89

Momentum - Macroscopic & Microscopic Momentum features in many sections of the A-level Physics specifications. It links with vectors and energy, and is important in the study of collisions and explosions (Conservation of Momentum) with both large objects (cars, rockets, etc) and small objects (atoms, alpha and beta particles, etc).

Example 3: A force of 10N towards the right is applied for 2s to a 3kg mass already travelling at 4ms-1 towards the right. Find the final velocity. Mass

We will begin with the general idea of momentum, and then revise the more familiar large-scale events, before finishing with a look at momentum properties at the atomic and subatomic levels.

Force

Velociy

Definition Momentum = mass × velocity

(p = mv)

Answer: F×t = mv-mu, 10×2 = 3v – (3×4), v = 10.7 ms-1 towards the right.

Momentum is a vector. In linear events, ‘+’ and ‘-‘ are often used to represent momentum towards the right and towards the left. However you must always state the direction to make the answer complete.

Example 4: A force of 10N towards the right is applied for 2s again to a 2kg mass travelling at 2ms-1 towards the left. Find the final velocity. Mass

Example1 : Find the momentum of a 500kg car travelling with a velocity of 0.005kms-1 towards the right.

Force

Answer: p = mv = 500 × 5 = 2500 kgms-1 towards the right. (Remember to use the standard units and state the direction.) Velociy

Components

Answer: Ft = mv-mu, 10×2 = 2v - 2×(-2), v = 8.0 ms-1 towards the right.

It is often necessary to separate vectors into horizontal and vertical components. Momentum is a vector. Example 2: A body has a momentum of 250 kgms-1 directed as shown in the diagram.

NB - the convention that positive stands for motion towards the right was used. (The initial velocity was towards the left.)

Conservation of momentum During a collision or explosion event, the total momentum of the system is conserved. We will confine ourselves to linear events, but this law also holds in 2 or 3 dimensions.

30o

Find the horizontal component of its momentum.

Remember that Kinetic Energy is not usually conserved in a collision. Some KE is usually converted to heat and sound.

-1

Answer: px = 250 × cos 30 = 217 kgms towards the right. Impulse and Change in Momentum The momentum of an object can be changed if a force acts on it for a period of time.

One standard collision situation involves the two objects sticking together after the collision. In this case, kinetic energy is never conserved.

We define impulse as F×t. (Ns) The impulse exerted equals the change in momentum. F×t = ∆mv = mv − mu.

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Physics Factsheet

89. Momentum - Macroscopic & Microscopic Microscopic events

Example 5 : The two balls shown stick together when they collide. 3kg

1 ms-1

5 ms-1

We will finish with a quick look at momentum considerations in atomic and subatomic Physics.

2kg

Gas Pressure (Kinetic Theory) The pressure of a gas on its container depends on the total force per second exerted by the molecules as they collide with the walls of the container.

Find the velocity after the collision, and the KE lost. Answer Momentum before: 3x5 + 2×(-1) = 13 kgms-1. Momentum after: (3+2)v.

A molecule changes momentum when it rebounds from the wall. Using the relationship Ft = ∆mv, we can work through to our normal Gas Law equations.

Momentum conserved: 5v = 13, v = 2.6 ms-1 towards the right. KE before: (½ × 3 × 52) + (½ × 2 × 12) = 38.5 J. KE after: (½ × 5 × 2.62) = 16.9 J. KE lost to heat and sound: 21.6 J. This collision is said to be inelastic. If KE is conserved in a collision, the collision is said to be elastic.

Question 6 at the end of this Factsheet helps you to follow through a set of calculations leading to the determination of the pressure of a gas sample in a container.

An “explosion” event involves an action-reaction pair pushing each other apart. The recoil of a gun is an example.

Alpha decay When an alpha particle is emitted from a nucleus, conservation of momentum insists that the remainder of the nucleus must recoil in the opposite direction.

Example 6: A 50g shell is fired from a gun of mass 12 kg. The initial velocity of the shell is 220 ms-1 in the direction shown.

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He

50g 12kg

V 220 ms-1

226

Ra

222

Rn

V

mv = MV

Example 7: For the decay reaction illustrated, v M 222 = = = 56 V m 4

Find the recoil velocity of the gun. Answers Momentum before = 0 kgms-1 Momentum after = (0.050 × 220) + 12v. 12v + 11.0 = 0, v = −0.92 ms-1. So the recoil velocity is 0.92 ms-1 towards the left.

To find the ratios of the kinetic energies of the two masses: m v2 1 56 2 ½mv2 = × = × = 56 ½MV2 M V2 56 1 So the alpha particle ends up with almost all of the kinetic energy produced in the decay.

Electron diffraction

Exam Hint: Conservation of momentum calculations are fairly standard. Use a simple system such as stating momentum before, momentum after, and then equating the two. Take care with units and direction.

When an electron is accelerated through a voltage, V, it gains kinetic energy which can be found from ½mv2 = eV. As momentum p = mv, we can replace ½mv2 with p2/2m. √(2meV), where m is the mass of Setting p2/2m = eV, we find that p =√ the electron (9.1 × 10-31 kg).

Changing mass

Then λ = h/p lets us calculate the de Broglie wavelength for the electron.

Remember that momentum depends on both mass and velocity. Momentum changes can be caused by a change in mass or a change in velocity. Sand pouring onto a moving conveyor belt would slow the belt down (if the motor were turned off) as mass times velocity must stay the same for the sand/conveyor belt system.

For low energies (small voltages) this all works. But for large energies, calculations show that the electron speed is greater than the speed of light. As the electron speed approaches the speed of light, the electron mass increases from the rest mass value up to relativistic values, and the Newtonian equations no longer work.

These systems are less common than changes in velocity, but you should be aware that changes in mass cause changes in momentum.

Example 8: find the speed of an electron accelerated through (a) 1000V, (b) 1MV.

sand

Answer: (a) v2 = (2eV) / m, v = 1.9 × 107 ms-1. (b) v = 5.9 × 108 ms-1. (greater than the speed of light in a vacuum)

conveyor belt

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Physics Factsheet

89. Momentum - Macroscopic & Microscopic For very high voltages, we must use the equation for the momentum of photons (p = E/c) to work out an approximate value for the de Broglie wavelength of the electron.

(f) p = F / A = (1.25 × 10-14) / 0.25 = 5.0 × 10-14 Nm-2. This is a very small pressure, because this is a very small number of molecules.

Beta decay and neutrinos

(e) F = ∆mv / t = 2.5 × 10-23 × 5.0 × 108 = 1.25 × 10-14 N

Finally, cloud chamber photographs can show a decayed neutron (a proton) recoiling in a non-linear situation from the beta particle it has just emitted.

(d) Before: mv = 2.5 × 10-26 × 500 = 1.25 × 10-23 kgms-1 After: mv = -1.25 × 10-23 kgms-1

beta

(c) 1 / 0.002 = 500 s-1 by each molecule. Total = 500 × 1.0 × 106 = 5.0 × 108 collisions s-1.

neutron

Conservation of momentum predicts that a third particle must be involved. This is the neutrino, which must be heading in an upwards direction to satisfy our conservation law.

(b) t = s / v = (2 × 0.5)/500 = 0.002 s. 6. (a) One-third of them, or 1.0× 106 molecules.

Questions 5. (30 × 0.10) + 1.5v = 0, v = 2.0 ms-1 backwards.

1. Find the momentum of a ball of mass 82g moving towards the left with a speed of 30ms-1.

4. Change in momentum as the egg comes to rest is the same for each surface, but the time taken to decelerate is greater with the carpet. Ft = ∆p predicts that for ∆p fixed, a greater time of contact will mean a smaller force applied to the egg.

2. A 20 kg projectile is launched with a speed of 15 ms-1 at an angle of 25o above the horizontal.

25o

3. ∆p = Ft = 3500 × 10 = 35 000 kgms-1 Final p = 55 000 kgms-1 towards the right.

Find its initial vertical momentum.

2. py = 20 × 15 × sin25 = 127 kgms-1 upwards.

3. A car has a momentum of 20 000 kgms-1 towards the right. Its engine provides a thrust of 5000N for 10 seconds, also towards the right. All resistive forces add to 1500N. Fine the momentum of the car at the end of the 10s period.

1. p = mv = 0.082 × 30 = 2.46 kgms-1 towards the left.

Answers

1500N

5000N

Po = 20 000 kg ms-1

4. Explain why an egg is more likely to break when dropped onto a concrete floor rather than onto a carpet. 5. A child of mass 30kg jumps forward off a stationary skateboard of mass 1.5 kg. The child’s forward speed is 0.10 ms-1. what is the recoil velocity of the skateboard? 6. There are 3.0 × 106 molecules of a gas moving randomly in a cubic box of sides 0.5 m at a speed of 500 ms-1. (a) How many molecules are effectively moving in the x-plane? z

x

y

(b) How often will each of these collide with the shaded wall? (c) How many collisions per second with this wall will there be in total? (d) What is the momentum of each molecule before and after each collision with the shaded wall? (Assume elastic collisions, and assume that the mass of each molecule is 2.5 × 10-26 kg.) (e) Use Ft = ∆mv to find the force of all the collisions with this wall in one second. (f) Find the pressure the gas exerts on this wall, and comment on the result.

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Acknowledgements: This Physics Factsheet was researched and written by Paul Freeman The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136