G12 Momentum

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Momentum & Impulse

K Warne

http://www.sciencecafe.org.za/

http://www.warnescience.net/




MOMENTUM

Definition: Momentum (Symbol : p) of an object is the product of the mass and velocity of a moving body.

Momentum p = m.v= (2000)(16)`

= 32 000 kgms-1 forwards

units: kg. m.s-1

N.B. Since velocity is a vector quantity, momentum is also a vector quantity.

V = 16 m.s-1

forwards

Mass = 2000kg






1. Calculate the momentum of a cyclist who has a mass of 75 kg and rides a bike of mass

15 kg at a speed of 20 km.hr-1.

p = mv

= (15k+75)*(20/3.6)

= 500 kg.m.s-1 in the direction of the velocity

2. Calculate the momentum of a 500 tonne ship moving at 1 km.hr-1.

p = mV

= (500 x 103)*(1/3.6)

= 138 888.89 kg.m.s-1 in the direction of the velocity

3. Calculate the momentum of a car which has a mass of 1500 kg and a kinetic energy of

200 kJ.

Ek = ½ mv2 p = mV

200000= ½(1500)v2 = (1500)*(16.33)

v = √ (200000*2/1500) = 24 494.90 kg.m.s-1 in the direction of

= 16.33 m.s-1 the velocity

MOMENTUM Examples






(Newtons 2nd Law)

Newton 2: If an object experiences a resultant

force it will ......................... in the ...............

of that force.m

Fres

.. . . . . . . .a

Fres = m.a

Fres = resultant force (...)

M = mass (...)

a = acceleration (...)

F =

M = 2000kg

a: 0 16m.s-1

in 10s






(Newtons 2nd Law)Newton 2: If an object experiences a resultant force it will

accelerate in the direction of that force. Eg a 2 tonne car

accelerates from rest to 16m.s-1 in 10 s.

m Fres

.. . . . . . . .a

Fres = m.a

Fres = resultant force (N)

M = mass (kg)

a = acceleration (m.s-2)

F = m.a

= (2000).(1.6)

= 3200N in direction of motion

M = 2000kg

a = (16-0)/10

= 1.6 m.s-2

v

t






Newtons 2nd Law & Momentum Examples

• Calculate the acceleration of a 5 kg box which is pushed by

a 20 N force with a 2 N frictional force acting against it.

• What applied force would be required to accelerate a 75 kg

person vertically upward at 5 m.s-2 ?

• What would be the mass of an object that is accelerated by

a 100 N applied force from rest to 10 m.s-1 in 5s against a 3

N frictional force?

ANSWER >>






Newtons 2nd & Momentum:

The applied …………….. Force is ……. to the

………………………………….., and that this

change is in the ……………. of the …………

…………...

Fnett =

This is derived from Newton's 2nd Law the nett

force is …………………… to the change in

momentum.

mFres

.. . . . . . . .a Fres = …….

Fnet p

t






IMPULSEImpulse is the ……… of the …………………..

applied and the time over which it is

applied.

Impulse = ……….

t

Impulse changes ………………….

Fres = ma = m( ) = /

Fres .t = …………. = ………

Impulse = change in ……………….

FresFres






Increase in Momentum Example:A rocket with a mass of 2000 tones traveling at 100m.s-1 is

accelerated to a velocity of 300m.s-1 when the second stage

of the rocket engine fires, on reaching this speed the rocket

has burned up another 1500kg of fuel. Calculate:

1. the change in momentum of the rocket and

2. the magnitude of the nett force on the rocket if the

rocket took 10s to reach the new speed.

3. The impulse experienced by the rocket.

ANSWER >>






MOMENTUM CONSERVA TIONPrinciple: It states that the ……………………… of an …………. system

remains …………………. in both …………………………………..

TOTAL momentum before crash = TOTAL momentum after

…..= ……

CHANGE IN MOMENTUM

change in momentum = ……………………………………………….

p = ………………………..

A B Crash!

Initial

ma 2kg mb 3kg

Final5m/s -6m/s

va = -2m/s vb = -1.33m/s

A B






Elastic & Non-elastic Collisions

Collisions can either be:

a. Elastic (……. is …………….. and no energy lost)

b. Inelastic (…………… is …………… - some is converted to heat, sound etcJ

N.B.

Momentum is conserved in any collision between two objects, but

Kinetic energy is conserved only in a perfectly elastic collision.






Change in Momentum & Impulse

• Pi = mvi = (1)(5) = 5kg·m·s-1

• pf = mvf = (1)(-3) = -3kg·m·s-1

• p = pf – pi = -3 – 5 = -8 kg·m·s-1

• p = 8kg·m·s-1 in opposite

direction

1kg

5m.s-1

1kg3m.s-1

p 8kg.m.s-1

FORCE

t

0.05s

Impulse = F x t = p

8 = F x (0.05)

F = 8/(0.05) =

A 1 kg ball moving with a velocity of 5m.s-1 to the right collides with a

wall and bounces back with a velocity of 3m.s-1. If the collision takes

0.05s calculate the force exerted by the wall on the ball.






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G12 Momentum

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