Warm Up: Which of the following has the greatest momentum? a) A 100 g bullet moving at 300 m/s b) A 5000 kg train car moving at 1.5 m/s c) A 75 kg motorcyclist moving at 35 m/s
Linear Momentum What is Linear Momentum? “Trains are big. They’re hard to stop.”
p ≡ mv TRAIN
p = Mv
More on Momentum • Units: kg·m/s • Momentum is a vector: px=mvx, py=mvy, etc. • Like KE affected by mass and velocity
BULLET…Also hard to stop
p = mv
Why Study Momentum? Another powerful way to more efficiently analyze systems of interacting objects:
• UNLIKE KE increases linearly with both mass
and velocity • UNLIKE KE, can be negative or positive!
Force and Momentum: Derivation Recall Newton’s Second Law:
If mass is constant, build it into the derivative.
So force and momentum are related… F=
What if ΣF=0?
dp =0 dt
p = constant
Conservation of Linear Momentum pi = pf
Systems with no net external forces have constant momentum. (Impulse approximation helps here.)
2-Particle System p1 m1
Conservation of Linear Momentum dp1 dt dp2 F12 = dt
If there is no net external force, then
Newton’s 3rd Law gives: F12 = −F21
ΣFinternal = 0
Fint = 0
dp1 dp2 + dt dt
d ( p1 + p2 ) dt
ptot = p1 + p2
Example two particle system:
ΣFinternal = 0 Ftot =
A bullet with mass 35 g moving at 320 m/s embeds itself in a 2.40 kg wooden block sitting on a frictionless surface. How fast do the bullet and block move together after the collision? demo
Define system: bullet and block, so
Σpinit = Σpfinal = constant
pi = pf mbulletvbullet,i= (mbullet+mblock)vf
vf = 4.6 m/s
Conservation of Linear Momentum
Note: this is 3 equations: px, py, pz
What if force 0?
dp = F dt
Consider a very large force acting over a very short time. ex: hitting a hockey puck with a stick.
F dt ≡ I
∆p = pf − pi = i
Ι = ∆p An Impulse is a change in momentum: - The effect of a force acting over a time - The agent of momentum transfer. - Handy for time-varying forces.
During the short time of the impulse from the large force, we can safely ignore other forces acting on the puck.
∆t Ι = F ∆t
Useful to approximate the impulse as due to a constant force F acting over the same time ∆t. Applications: collisions & explosions!
A 60.0 g tennis ball is initially horizontally east at 50.0 m/s. A player hits the ball with a racket. The ball is in contact with the racket for 35 ms. After the hit, the ball is moving west at 40.0 m/s. What average force did the racket exert on the ball?
Systems of particles interacting via impulsive forces (large, short time)
Impulse = p=pf- pi =.060 (-40) - .060 (50) =-5.4 kg m/s But, Impulse also =F t => F(0.035 s)=-5.4 kg m/s => F= -154 N Note: Compare with force/impulse to stop ball… Negative => Westward
1D collision: final m2 v1i
Only consider large contact forces; Instants just before & after collision
Apply linear momentum conservation.
Types of collisions:
Linear momentum conservation:
ΣPi = ΣPf m1v1i – m2v2i = m1v1f + m2v2f Given v1i and v2i we have one equation, but two unknowns.
Elastic linear momentum is conserved total kinetic energy is conserved Inelastic linear momentum is conserved total kinetic energy is NOT conserved Totally inelastic collisions: particles stick together. Where does the energy go? demo
1D Perfectly Inelastic Collisions initial m1
One more inelastic example: Explosion! 5.0 kg vi=0
v1f = v2f = vf
2.0 kg, v1f =3.6 m/s
m1v1i – m2v2i = m1v1f + m2v2f m1v1i – m2v2i = (m1+m2)vf
3.0 kg, v2f =?
pi=pf 0 = 2(-3.6) +3(v2f) => v2f = 2.4 m/s
vf = (m1v1i - m2v2i)/(m1+m2) Caution with signs!
1D Elastic Collisions initial m1
1D Elastic Collisions cont.
Some algebra shows: m2
v1i – v2i = − (v1f − v2f) v2f
m1v1i – m2v2i = m1v1f + m2v2f
Kinetic Energy is conserved:
½m1v21i + ½m2v22i = ½ m1v21f + ½ m2v22f Two equations
can solve for two unknowns.
v1 f =
m1 − m2 2m2 v1i + v 2i m1 + m2 m1 + m2
v2 f =
m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2