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 PresentationTRANSCRIPT
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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](https://reader035.vdocuments.us/reader035/viewer/2022062520/5681596e550346895dc6b001/html5/thumbnails/2.jpg)
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](https://reader035.vdocuments.us/reader035/viewer/2022062520/5681596e550346895dc6b001/html5/thumbnails/3.jpg)
• 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](https://reader035.vdocuments.us/reader035/viewer/2022062520/5681596e550346895dc6b001/html5/thumbnails/4.jpg)
• 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](https://reader035.vdocuments.us/reader035/viewer/2022062520/5681596e550346895dc6b001/html5/thumbnails/5.jpg)
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