1. Newton’s 2nd Law (again?!)
Impulse
Newton’s 2nd Law (again?!)

2. Impulse
A baseball being hit by a bat outlines impulse quite well: the ball has an initial velocity vi before contact and a final velocity vf after contact. The force exerted by the bat changes from zero (just before contact) to some maximum value and back to zero in a time Δt.
The size of the force and the time of impact are both factors for a well hit ball.
Impulse is the product of Force and time of contact.
J = FΔt
Impulse is a vector quantity and has the same direction as the average force.

3. Force versus time graphs
Analysis of Force versus time graphs is utilized by companies: tennis racket design, baseball bat design, etc. “Response”
Integration would give us the area under the curve = Impulse!
We can approximate by using the average force (treats the curve like a triangle).

4. Force versus time graphs
When a tennis ball is hit, large forces are exerted over milliseconds.
The tension of the strings can affect this, so beginners will use a looser net to provide larger forces by increasing the impact time.
Professionals want a short impact time as they can provide larger forces.
Pros will have alternate racquets to change strategies and will adjust their own net tensions to reflect their abilities.
Sometimes table-tennis paddles will have varied coated pads, and the type of material can vary on each side of the racquet in case they want to change strategies while playing!

5. Tennis ball impact on racquet

6. Newton’s 2nd Law revisited
Recall Newton’s 2nd Law:
F = ma
𝐹=𝑚 ∆𝑣 ∆𝑡
𝐹∆𝑡=𝑚∆𝑣
J = ∆p
Impulse = Change in Momentum
(Impulse-momentum theorem)
Note: As a net force is applied, the momentum will change in these questions.

7. Graph Examples F Constant Force Δt
1) Determine the impulse outlined by the provided Force versus time graph: F Δt
Constant Force
8.00 N
5.00 h

8. Graph Examples
1) Determine the impulse outlined by the provided Force versus time graph:
8.00 N F Δt
2.00 min

9. Numerical Examples
A baseball (m = 0.14 kg) has an initial velocity of – 38 m/s as it approaches the bat. (Direction of approach chosen as negative). The bat applies a force greater than the weight of the ball and the ball leaves at +58 m/s. If the contact time is 4.0 ms, what is the average force exerted on the ball?
J = ∆p
F∆t = m∆v

10. Numerical Examples
During a storm, rain falls at -15 m/s and hits the roof of a car at right angles. If the rain falls at an average rate of kg/s, and comes to rest upon contact with the roof (no bouncing), what is the average force exerted by the rain on the car roof?
J = ∆p
F∆t = m∆v
F1,2 =
The force of the roof on the rain is equal and opposite in direction to the force of the rain on the roof. (Newton’s _____ Law)
F2,1 =

11. Challenge Question
If hail (same weight as the rain) fell at the same rate and velocity but bounced, is the force exerted on the roof the same, greater or less than the rain?

12. Text Questions
4U: Irwin Text Page
4UB: Giancoli Text Page

13. Video links https://www.youtube.com/watch?v=QFlEIybC7rU
Fun video: