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.

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

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 )(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!!!

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!!!

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 = -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