SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
Momentum & Impulse
K Warne
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> 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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
(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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
(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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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 >>
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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 >>
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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.
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
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.
SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
For FULL presentation click HERE >> www.warnescience.net
Hi -
This is a SAMPLE presentation only.
My FULL presentations, which contain a lot more more slides and other resources, are freely
available on my resource sharing website:
www.warnescience.net(click on link or logo)
Have a look and enjoy!
WarneScience
Top Related