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KineticEk = ½ mv2 Potential Ep = mgh
Heat EH = cmT Heat EH = ml(changing temperature) (changing state)
Electrical EE = Pt = IVt light sound nuclear
Work done Ew = Fd energy E = Pt
Energy is conserved, this means the total energy remains the same
Energy Revision
h
At top of slope
Ek = 0
Ep = mgh
At bottom of slope
Ek = ½mv2
Ep = 0
Total energy is conserved
Assume no friction
Ep at top = Ek at bottom
mgh = ½mv2
gh = ½ v2 ie. mass is not importantv = 2gh
Problems using conservation of energy.
With friction
Total energy is conserved
Ep at top = Ek at bottom + Ew
mgh = ½mv2 + Fd
Tutorial questions page 23/24 Qu 1 to 6
Purple book Ex 2.9
Momentum = mass x velocity
momentum = mv
Small mass, fast
Or
Large mass, slow
Which would do the most damage?
Momentum is a vector quantity measured in kgm/s or kgms-1
Tutorial questions page 24 Ou1
m1
kgu1
ms-1
Momentum beforem1u1 kgm/s
m2
kgv
m/s(m1 + m2) v
Total momentum after kgm/s
0.5 0.5 x = 0.5 (0.5 + 0.5) x =
0.5 0.5 x = 1 (0.5 + 1) x =
0.5 0.5 x = 1.5 (0.5 + 1.5) x =
1 1 x = 0.5 (1 + 0.5) x =
1 1 x = 1 (1 + 1) x =
Collisions – trolleys stick together afterwards
At start trolley 2 is stationary.
before after
The Principle of Conservation of momentum
Total momentum remains the same provided there are no outside forces.
Total momentum before = total momentum after
Elastic collisions – where the total kinetic energy is conserved.
Inelastic collisions – where the kinetic energy is not conserved eg. Some of the kinetic energy is changed into heat and sound energy
1. Toy car A with a mass of 2kg and a velocity of 1m/s, collides with stationary car B, mass 1kg. Velcro causes them to stick together.
(a) What is their speed after the collision?(b) Is it an elastic or inelastic collision?
AB
Tutorial Questions Page 24/25 Qu 2 to 4
2. Two rubber balls collide head on as shown.The red ball rebounds at 1m/s.
(a)What is the velocity of the blue ball?
(b) Is it an elastic or inelastic collision?
mass 4kg mass 2kg
velocity 2m/s velocity -3m/s
Tutorial Questions Page 25/26 Qu 5 to 9
A B
Momentum before
m1
kg
v1
m/s
m1v1
kgm/s
m2
kg
v2
m/s
m2v2
kgm/s
Total momentum
after0 0.5 0.5
0 0.5 1
0 0.5 1.5
0 1 0.5
0 1 1
Explosion – velocity before explosion is zero.
3. A firework explodes and breaks into two pieces, 500g and 250g. (a)If the 500g part travels at 15 ms-1, what will the velocity of the other part be immediately after the explosions?
(b)Why does it not stay at this velocity?
(c)Can kinetic energy ever be conserved in explosions?
Tutorial Questions Page 27 Qu 10 to 13
4. A field gun of mass 1000kg fires a shell of mass 5kg with a velocity of 100m/s.
Calculate the recoil velocity of the gun.
When you jump, what’s the least painful way to land?
If you fall off something, what sort of surface would you prefer
to land on? Explain!
So the relationship F × t is an important one.
Impulse = Ft
In each case you have the same momentum to loose to come to a stop.
Increasing the time it takes you to stop decreases the force.
Impulse – on one object
1. Impulse = average force x time
= Ft
Impulse is a vector quantity measured in newton seconds (Ns)
2. Impulse = change in momentum of the one object
= mv - mu
Impulse is also measured in kgm/s or kgms-1
3. Impulse = area under the force time graph
t (s)0
F (N)
This means that
Impluse = Ft = mv – mu = area under force time graph
Things to beware of
F is average force, not the maximum.
Direction and sign of velocities ie rebound
Impact time is often very short and can be given in milliseconds (ms)
Mass is often given in grams
1. A car with mass 600 kg and velocity of 40 m/s skids and crashes into a wall.
The car comes to rest 50 ms after hitting the wall. Calculate the average force on the car during the collision.
2. During a game of hockey a stationary ball of mass 150 g is struck by a player. The graph shows how the force on the ball varies with time.
t/ms
F/N
1200
3 6
(a) Calculate the impulse on the ball.
(b) Calculate the speed which it leaves the stick.
(c) A softer ball is hit and leaves the stick with the same velocity. Sketch its force time graph.
Tutorial Questions Page 28/29 Qu 14 to 20
Aim: To find the average force exerted by a cue on a snooker ball.
Measurementsd, diameter of ball = mt1, contact time = st2, time to go through light gate = sm, mass of ball = kg
CalculationsVelocity of ball after collision, v = d/t2 =Average force, F = (mv – mu)/t1 =
Tutorial Questions Page 28/29 Qu 14 to 20
Which ball would you prefer to be dropped on your foot and why?
Use the words force, time, change in momentum and impulse in your
explanation!
Open Ended Question
Wearing no seatbelt
The person continues to move forward at a constant speed. Newton’s first law
Until they collide with the dashboard etc, stopping them suddenly.
F = (mv – mu)/t so short time means large average force
Wearing a seat belt
The person is brought to a stop at the same time as the car is stopping. The stopping time is increased as the car crumples and the seatbelt has some give. The force is also on the parts of the body where it will do the least harm
F = (mv-mu)/t the same change in momentum but a longer time means a smaller average force.
1. The rate the fuel is ejected at is 30 kg/s.
This means in a time of 1 second there is a mass of 30 kg
2. change in momentum of the rocket =
- change in momentum of fuel
Newton’s Third Law
Tutorial Questions Page 30/31 Qu 21 to 26
Points to note
A careless school pupil drops a 1p coin from his pocket at the top of the Eiffel Tower.Find the average force exerted by the coin on the ground. You will need to estimate some values.
324 m
Calculate the velocity of the coin on impact on the ground. (Estimate the mass of the coin and use equations of motion)
Open Ended Question
324 m
Calculate the velocity of the coin on impact on the ground. (Estimate the mass of the coin and use equations of motion)
Open Ended Question
Calculate change in momentum. What value would you give to the rebound velocity?
Calculate average force. What value would you give to the contact time?
