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Example 9-6: Karate blow Estimate the impulse p & the average force F avg , 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.

Example 9-6: Karate blow

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Example 9-6: Karate blow. Estimate the impulse p & the average force F avg , 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. Example. Just before he hits the ground. - PowerPoint PPT Presentation

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Page 1: Example 9-6: Karate blow

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.

Page 2: Example 9-6: Karate blow

Example

Opposite the person’s momentum

Just before he hits the ground

Just after he hits the ground

The advantage of bending your knees when landing! a) m = 70 kg, h =3.0 m

Impulse: p = ? By definition, Ft = p = m(0-v)

First, we need to find v (just before landing). Use: K + U = 0

m(v2 -0) + mg(0 - h) = 0Gives v = 7.7 m/s

Impulse: p = -540 N s

v = 7.7 m/s

v = 0

Page 3: Example 9-6: Karate blow

• 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 )(7.7 + 0) = 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!!!

Page 4: Example 9-6: Karate blow

• 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 )(7.7 + 0) = 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!!!

Page 5: Example 9-6: Karate blow

Example: Crash TestA 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.

p = mv = -2.3 104 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

v = - 15 m/s

v = 2.6 m/s