1. Example 9-6: Karate blow
Estimate the impulse p & the average force Favg, delivered by a karate blow that breaks a board a few cm thick. Assume the hand moves at roughly 10 m/s when it hits the board.
Figure 9-11.
Solution: Take the mass of the hand plus a reasonable portion of the arm to be 1 kg; if the speed goes from 10 m/s to zero in 1 cm the time is 2 ms. This gives a force of 5000 N.

2. Opposite the person’s momentum
Example
Just before he
hits the ground
The advantage of bending your knees when landing! a) m = 70 kg, h =3.0 m
Impulse: p = ?
By definition, Ft = p = m(0-v)
First, we need to find v
(just before landing).
Use: K + U = 0
m(v2 -0) + mg(0 - h) = 0
Gives  v = 7.7 m/s
Impulse: p = -540 N s
v = 7.7 m/s
v = 0 
Opposite the person’s momentum
Just after he
hits the ground

3. Enough to fracture leg bone!!!
Advantage of bending knees when landing!
Impulse: p = -540 N s, m = 70 kg
h = 3.0 m, F = ?
b) Stiff legged:
v = 7.7 m/s to v = 0 in d = 1 cm (0.01m)!
v = (½ )( ) = 3.8 m/s
Time: t = d/v = 2.6  10-3 s
F = |p/t| = 2.1  105 N
(Net force upward on the person)
From the free body diagram,
F = Fgrd - mg  2.1  105 N
Enough to fracture leg bone!!!

4. Advantage of bending knees when landing!
Impulse: p = -540 N s, m = 70 kg
h = 3.0 m, F = ?
c) Knees bent:
v = 7.7 m/s to v = 0 in d = 50 cm (0.5m)!
v = (½ )( ) = 3.8 m/s
Time: t = d/v = 0.13s
F = |p/t| = 4.2  103 N
(Net force upward on the person)
From the free body diagram,
F = Fgrd - mg  4.9  103 N
Leg bone does not break!!!

5. Impulse J = Δp = p – p = 2.69  104 kg m/s (N s)
Example: Crash Test
A Crash Test
A car, m = 1500 kg, hits a wall. +x is to the
right. Before the crash, v = -15 m/s. After
the crash, v = 2.6 m/s. The collision lasts
Δt = 0.15 s.
Find: The impulse the car receives & the
average force on the car.
Assume: The force exerted by the wall is large
compared to other forces on the car. Gravity &
normal forces are perpendicular & don’t effect
the horizontal momentum.
v = - 15 m/s
v = 2.6 m/s
p = mv =  104 kg m/s, p = mv = 3.9  103 kg m/s
Impulse J = Δp = p – p = 2.69  104 kg m/s (N s)
Favg = (Δp/Δt) = 1.76  105 N