1.4 MOMENTUM IN TWO DIMENSIONS

Momentum

• momentum of an object to be the product of mass (m) and velocity (v).

• Momentum is a vector quantity with SI Units of kgms-1 (or Ns, since 1N = 1kgms-2).

Momentum in 2-Dimensions• Vector Form of Newton’s Second Law of Motion• Consider a particle reflecting off a surface without a change in speed.• A particle has m = 2 kg and initial velocity vi = 20 m s-1 and the collision

lasts for t = 0.2 s, then determine the average force acting on the particle.vf = -20 ms-1

Acceleration a is given by,

2200

2.0

)20()20(

msa

a

t

vva

t

va

if

vi

vf

m

Momentum in 2-Dimensions• Vector Form of Newton’s Second Law of Motion

• a = -200ms-1

• Acceleration is in opposite direction to vi.

• and since F = m a

• then F = 2 x (-200)

• = -400 N

• ie. 400 N away from the wall

• Ie, the wall applied a force of 400N on the ball away from the wall and due to Newtons 3rd Law, the ball exerted a force of 400N towards the wall.

Vector Form of Newton’s Second Law

�⃗�=𝑚�⃗�

The acceleration of an object is in the same direction as the net force acting on it.

Hence, a force changes the speed and/or the direction of motion of an object.

A ball of mass 0.50 kg moving at 4.0 m.s-1 rebounds off a wall, without any loss of speed, along the normal to the wall. The duration of the impact is 0.080 seconds. Calculate:

Calculate:

(a) The average acceleration of the ball during impact.

(b) The average force applied to the ball by the wall during impact.

(c) The average force applied to the wall by the ball during impact.

vi = 4.0 m.s-1

Example 1

vf = - 4.0 m.s-1

A ball of mass 0.50 kg moving at 4.0 m.s-1 rebounds, without any loss of speed, at an angle of 60 to the surface of a wall. The duration of the impact is 0.10 seconds. Calculate:

(a) The average acceleration of the ball during impact.

(b) The average force applied to the ball by the wall during impact.

(c) The average force applied to the wall by the ball during impact.

Example 2

Newton’s Second Law in Terms of Momentum

�⃗�=𝑚�⃗�

∴ �⃗�=𝑚∆𝑣∆𝑡

¿∆ (𝑚�⃗�)

∆𝑡

¿∆ �⃗�∆ 𝑡

but

provided that the mass of the particle is fixed.

since

¿𝑚∆ �⃗�

∆ 𝑡

A space shuttle of mass 2.20 x 104 kg is moving in space with a speed of 2.00 x 103 m.s-1. A rocket is fired that causes it to change its direction by 90 in 4.00 seconds without a change in speed.

Determine the average forceexerted on the shuttleduring firing.

Example 3

𝑝𝑖

𝑝 𝑓

A ball of mass 0.07 kg moving at 1.0 m.s-1 rebounds off a wall with a velocity of 0.60 m.s-1 as shown in the diagram. The duration of the impact is 0.010 seconds.

Calculate:

(a) The average force acting on the ball during impact.

(b) The average acceleration of the ball during impact.

(c) The average force exerted on the wall during impact.

Example 4

vi = 1.0 m.s-1 vf = - 0.60 m.s-1

A 0.30 kg steel ball strikes a floor with a speed of 10 m.s-1 at an angle of 40 with the surface. It bounces off with the same speed and angle as shown in the diagram. If the ball is in contact with the floor for 0.20 seconds, what is the average force exerted on the floor by the ball?

Example 5

Law of Conservation of Momentum

In an isolated system, the total momentum of the system remains constant.

isolated

no net external forces act upon

the system

system

a group of objects e.g. “two pucks” or“a person and the

Earth”

Law of Conservation of Momentum

Consider the collision of two particles. During impact, the first particle exerts a force on the second particle. The second particle exerts a force on the first particle.

By Newton’s Third Law =but

but (same collision)

total change in momentum is zero

Multi-Image Photographs

The collision of two objects (or the explosion of one) can be documented by taking a multi-image photograph. Using this, it is possible to determine whether momentum is conserved.

See PowerPoint (Multi image collision examples)

1. Draw diagrams showing the situation before and after impact.

2. Label the momenta of each object in these diagrams using subscripts.

3. Find equations for the total momentum before and the total momentum after impact. BE CAREFUL TO USE VECTORS.

4. Use the Law of Conservation of Momentum and solve for the unknown.

Conservation of Momentum Calculations

Conservation of Momentum

• NB. (1) No net external forces means an isolated system.

• (2) Total linear momentum refers to both size and direction (vector addition is needed to find the total)

• (3) Conserved means that it stays the same (throughout the interaction)

• (4) The objects may be deformed, broken apart, stuck together or exploded. However momentum is still conserved as long as no net external forces act.

Example

A rail wagon of mass 4.0 tones, moving 3ms-1 North collides with another wagon of mass 2.0 tones. They stick together. What is their combined velocity?

Take North to be the positive direction,

Momentum in 2-Dimensions

• The Law of Conservation of Momentum

16000

2000400012

1221

kgms v

v

vmm

fterA Momentum

wagon

wagonwagonwagon

112000

02000340002211

kgms

vmvm

Before Momentum

wagonwagonwagonwagon

Momentum in 2-Dimensions

• The Law of Conservation of Momentum

Assuming the system isolated, linear momentum is conserved.

North 2

6000

12000

600012000

1

msv

v

v

pp afterbefore

Momentum in 2-Dimensions

• The Law of Conservation of MomentumExample

A body at rest explodes breaking into 3 pieces. One piece flies off due East at 30ms-1. Another piece flies of due North at 30ms-1. What is the velocity of the 3rd piece if its mass is 3 times that of the other pieces?

1

12

11

321

3

30

30

0

kgms vmp

orthN kgms mp

East kgms mp

ppp

pp

3

fi

Cart A is travelling North with a velocity of 5.4 m.s-1 while Cart B is travelling South at 8.3 m.s-1.

The mass of Cart A is 13 kg and the mass of Cart B is 9.0 kg.

The carts collide and Cart B rebounds with a velocity of 7 m.s-1. Determine the velocity of Cart A after the collision.

Example 6

The blue car with mass 1000 kg and is travelling at 20 m.s-1 East. The red car with mass 1000 kg and is travelling at 10 m.s-1 North. What is the final velocity of the two cars stuck together?

Example 7

A spherical mass of 9.0 kg explodes into 3 pieces. One piece, of mass 3.0 kg flies off due North at 10 m.s-1 whereas another piece of mass 4.0 kg is thrown West at a speed of 13 m.s-1. Find the velocity of the third piece.

Example 8

Application: Spacecraft Propulsion

Rockets

Initially the rocket has no momentum.

When the gases leave the back of the rocket, their momentum must be balanced by the rocket gaining momentum forwards.

Application: Spacecraft Propulsion

https://www.youtube.com/watch?v=rt5dcrm8X1w

Solar Sails

Solar Sails

Consider a photon reflected from a solar sail.

The photon’s change in momentum is directed away from the sail. Hence the sail must gain momentum in the opposite direction of the photon (Conservation of Momentum).

∆ 𝑝 h𝑝 𝑜𝑡𝑜𝑛 ∆ 𝑝𝑠𝑎𝑖𝑙

sail

Consider a photon absorbed by a solar sail. Assume that the photon has initial momentum .

Solar Sails

𝑝𝑖=𝑝  sail

∆ �⃗�  =𝑝 𝑓 −𝑝𝑖

¿0 −𝑝  

¿−𝑝  

Consider a photon reflected by a solar sail. Assume that the photon has initial momentum .

Solar Sails

𝑝𝑖=𝑝  sail ∆ �⃗�  =𝑝 𝑓 −𝑝𝑖

¿−𝑝  −𝑝  

¿− 2𝑝  𝑝 𝑓=−𝑝  

Assume is the same for both absorption and reflection.

Solar Sails

Since is twice as great for a reflected photon, the force on it is twice as great compared to the absorbed photon.

Therefore, by Newton’s Third Law, the force on the sail is twice as great when the photon is reflected.

Hence the reflective sail will have twice the acceleration ().