Unformatted text preview: st to 25 m/s. He makes contact with the ball for 0.02 seconds. What is the average force on the ball ? t Δp = I = ∫ F (t ) dt I = Favg Δt
f ti Δp = p f − pi = (60 × 10 −3 kg )(25 m/s) − 0 = 1.5 kg m/s 1.5 kg m/s Favg = = 75 N 0.02s Example A parachu;st falls 370 m from an airplane. His parachute does not open. Miraculously he survives the impact when he hits a snow bank. Assume that his speed at impact is 56 m/s (terminal speed) and that his mass is 85 kg. If the maximum survivable force is 1.2 x 105 N, what is the minimum snow depth that would stop him safely? Magnitude of impulse: I = F Δt = Δp.......(1) Change in momentum: Δp = p f − pi = 0 − (−85 kg ⋅ 56 m/s) = 4760 Ns v +v 2d 2d Assume constant accelera6on: d = vavg Δt = i f Δt ⇒ Δt = = ............( 2)
2 vi + v f vi (1) & (2) Δt = I v ( 4760 Ns)(56m/s) 2d I = ⇒d = i = = 1.1m vi F 2F 2(1.2 × 105 N ) Example A racecar hits a wall as shown. If vi = 70 m...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.
 Spring '08
 BLANK
 Momentum

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