![Momentum (Symbol = p) [Unit = kg m/s]
ďš Inertia in motion ď
Momentum = Mass ⢠Velocity
p = M ⢠V
Newtonâs 2nd Law
F = ma a= âv = v f â vi
t t
v f â vi
F = m( ) = mv f â mvi
t
Ft = Impulse
Impulse (Symbol = I)[Unit = Nsec]
ďš A force in a certain time
Concept: In order to change momentum,
we need a force in a certain amount of time!](https://image.slidesharecdn.com/ch6-momentumnotes-121109132437-phpapp02/85/Ch6-momentum-notes-2-320.jpg)
Ch6 - momentum notes
- 2. Momentum (Symbol = p) [Unit = kg m/s] ďš Inertia in motion ď Momentum = Mass ⢠Velocity p = M ⢠V Newtonâs 2nd Law F = ma a= âv = v f â vi t t v f â vi F = m( ) = mv f â mvi t Ft = Impulse Impulse (Symbol = I)[Unit = Nsec] ďš A force in a certain time Concept: In order to change momentum, we need a force in a certain amount of time!
- 3. I = Ft = âp = mv â mv f i Changing Momentum mV - 0 = T F mV - 0 = Ft Momentum is a vector, so direction is important! How is impulse affected when an object bounces? Impulse is greater! Why? Takes more to âcatchâ & âthrowâ than to just âcatchâ
- 4. A 4.0 x 104 N rocket is at rest when it fires its thrusters. If the thrusters put out an average force of 5.0 x 10 3 N over 20 sec, then (a) What is the impulse the rocket experiences? (b) What is the rocketâs final velocity? (a) Impulse I=Ft I = 5000 (20) I = 100,000 Nsec (b) v I = âp = mvf â mvi 100,000 = 4080 V â 4080(0) v = 24.5 m/s
- 5. collisions
- 6. Conservation of Momentum Law of Conservation of Momentum ďš Momentum cannot be created or destroyed Momentum In = Momentum out pin = pout This is for 2 Types of Collisions Collisions! Elastic ď Energy is conserved Inelastic ď Energy is lost to heat, sound, etc. Since we work in a happy, ideal world, we will deal with all elastic collisions.
- 7. 4 Types of Problems 1. They hit and all the momentum is transferred. Before After A B A B MA = 4.0 kg MB = 2.0 kg VA = 0 VB = ? VA = 3.0 m/s VB = 0 m/s pin = pout MA VA + MB VB = MA VA + MB VB (4)(3) + 2(0) = 4(0) + 2 VB VB = 6.00 m/s
- 8. 2. They hit and stick together. Before After 4 2 4 2 4.0 m/s -2.0 m/s VAB = ? p in = p out mv + mv = mv + mv = (m + m)v (4)(4) + 2(-2) = (4 + 2) VAB VAB = 2.00 m/s
- 9. 3. They hit and both bounce away with a velocity Before After ? ? 4 2 4 2 2.0 m/s -3.0 m/s -1.0 m/s VB = ? p in = p out mv + mv = mv + mv (4)(2) + 2(-3) = 4(-1) + 2 VB VB = 3.0 m/s
- 10. 4. They both start with zero velocity Before After 4 2 4 2 v=0 ? m/s VB = 20 p in = p out mv + mv = mv + mv 0 = 4(v) + 2 (20) v = -10.0 m/s