1. Momentum and Collisions

2. Goal of the class To understand the conservation of linear momentum.
To understand impulse and collisions.
Question of the day: How we do find the final velocity of two pool balls colliding?

3. Linear Momentum Reminder
Momentum is a property of all moving objects
It can be found by the formula:
Momentum (p) = mass (m) x velocity(v)
The most common unit for momentum is kg·m/s or N·s
P and v are both vectors

4. Linear Momentum P=mv Example:
Let’s say we have a wall and a ball of mass 0.1kg that hits the wall with velocity 10m/s. It bounces and moves back with a velocity of -10m/s.
P1 = (0.1)(10) = 1
P2 = (0.1)(-10) =-1
The change in the momentum is delta p= p2-p1 = 2kgm/s

5. Newton’s 2nd Law of Motion
F=ma but a = lim delta v / delta t = lim v(t+del t) – v(t) / delta t
F = mv ….. = p …. / delta t
F delta t = delta p
We call this the Impluse J = F delta t

6. Impulse A force applied for a short time Δt J = F Δt
Ball has mass 0.15kg and v1 (u) =-30m/s
After he hits the ball it travels at v2=50m/s
The force he hits the ball with it isn’t constant. In fact if we draw it like so:
When he first hits the ball, at that instant the force is 0. The same when the ball leaves the ball.
The time he hits it over t2-t1 = delta t which will be a few milliseconds only.
J = F_av delta t
Example baseball
Δt = 2x10-3 seconds
What is the average force on the ball
F = Δp/Δt or Δp = FΔt
Find Δp = 12kgm/s
Find F_av = 6x10-3N

7. Conservation of Linear Momentum
Momentum is a property that is always conserved
The amount of momentum is always the same unless acted on by outside forces
The amount of momentum before a collision is the same as the momentum after

8. Transfer of momentum
Moving objects can transfer momentum to other objects
When an object collides with another object the momentum is transferred
The final momentum must equal the initial momentum
momentum

9. Collisions 2 types Elastic collisions – mechanical energy is conserved
Inelastic collisions – heat is produced
Ex of elastic = pool balls draw on balls going towards, then contact at on point, then move away from each other.
Contact surfaces has area of 0. So there is no heat produced. All energy remains as KE.
Inelastic collision, there is some deformation and some heat produced eg rubber balls.
If 2 objects stick together then this is the extreme case of inelastic (completely inelastic)

10. Completely Inelastic Collision
M1 = 1kg m2=0.5kg
V1 = 5m/s v2=-2m/s
V’=?
Mom before = mom after
M1v1+m2v2 = (m1+m2)v’
V’ = 2.67m/s
KE in = 1/2m1v1^2 +1/2m2v2^2 =13.5J
KE out = 8/1.5J
Heat = about 8J

11. Ballistic Pendulum Draw on height h
Cons of mom to find v’ in terms of v
V’= mv/(m+M)
KE -> PE to find v’^2
V = (m+M)/m (2gh)^0.5

12. Problem
A 2 kg ball strikes a wall with a speed of 10m/s at an angle of 53 degrees with the surface. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.1 s. What is the average force exerted by the ball on the wall?
Pi = mvi vix = vi sin 53 = 0.8vi = 8m/s
viy = vi cos 53 = 0.6 vi = 6m/s
Pi = m(vix, viy) = m(8,6) = (16i + 12j) kgm/s
Pf = (-16i+12j)
Δp = -32i
F_av Δt = Δp = -32i/0.1 = -320N by wall
By Newton’s 3rd 320N by ball