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LESSON 36 – MOMENTUM & IMPULSE By: Fernando Morales September 18, 2013 LEARNING GOALS Familiarize with Cornell Style of Note Taking Define Momentum and Impulse Derive Impulse from Newton’s Third Law Solve Problems related to momentum, impulse, and area under the curve of a Force vs. time graph THE CORNELL NOTE-TAKING SYSTEM MOMENTUM Linear momentum is the product of an object’s mass and velocity p = mv Linear momentum is a vector direction is that of the velocity & magnitude is mv The value of the momentum depends on reference frame (since velocity depends on the reference frame) The units are kgm/s MOMENTUM & NEWTON’S 2ND LAW: NET FORCE Mathematical derivation: F = Fnet = ma 2nd Law v =m t vf - vi =m t pf - pi = t p = t conclusion: Def’n of acceleration pf = mvf; pi = mvi Force is needed to change momentum The rate of change of momentum equals the net force applied p F = t EXAMPLE: A RAIN STORM Assuming the rain comes to rest upon striking the car (vf=0m/s), find the average force exerted by the raindrop. 6 Rain comes straight down with velocity of v0=15m/s and hits the roof of a car perpendicularly. Mass of rain per second that strikes the car roof is 0.06kg/s. [A] F m vf m v0 t m ( ) v 0 t F = -(0.06kg/s)(15m/s)=0.9 N According to action-reaction law, the force exerted on the roof also has a magnitude of 0.9 N points downward: -0.9N 7 IMPULSE AND 3RD LAW FORCES Forces of Interaction (collisions) cause deformations of objects – the forces vary over time Forces of Interaction rise from 0 to a maximum and then fall to 0 again IMPULSE & MOMENTUM Fnet p = t Fnet t = p The Impulse of a force is Ft and equals the change in momentum p The area under a Force-time curve is the Impulse X2 -3 EXAMPLE What is the value of airbags in a car? Explain in terms of impulse and momentum. [T, C] By slowing the passengers down gradually as they move into the windscreen or dashboard, airbags increase the time of contact. The momentum to be dissipated is fixed, whether the airbags are there or not. Momentum change = impulse = force x time, so if the time of impact increases then the average force will decrease. A smaller average force means less injury. X2 -3 EXAMPLE In the early days cars were made very rigid, so that damage to the vehicle would be minimized. Now cars are built with “crumple zones”: the drive train folds up, and engine compartment bends like an accordion. What is the advantage? [T, C] The folding of the crumple zone increases the amount of time of the impact, which decrease the average force. A lower average force means that passengers are more likely to survive, although the car will be more severely damage. A good trade-off. EXAMPLE X7 -5 Impulse and Momentum Find the impulse, velocity change and average acceleration when a force as graphed is applied to a 33 kg object. [K] 9N Solution (a) Impulse = Area under the curve = (9 N x 3 s) + (9 N x 8 s) + (9 N x 3 s) 3s = 99 Ns, the impulse, momentum change is 99 Ns (b) Velocity change = total momentum change divided by mass = 99 Ns / 33 kg = 3 m/s, in the direction of the force. (c) Acceleration = velocity change divided by time = 3 m/s / 14 s = 0.21 m/s² , in the direction of the force. 11 s 14 s RUNNING SHOE ANALYSIS Which causes a greater impulse, heel running or forefoot running? Is it safer for you to run with shoes that cause high impulse or low impulse? [T] Impulse can be calculated from Force vs. Time graphs Impulse is the area under the F-T graph EXAMPLE Bend your knees when landing! [A] (a) find the impulse felt by a 70 kg man landing on hard ground after jumping from 3.0 m; estimate the average force exerted on him if (b) he lands stiff-legged and (c) he lands with bent legs. In (b) assume the body moves 1.0 cm on landing, and in (c) about 50 cm Solution (a) Impulse = Change in Momentum; Ft = mv We need the speed change from Conservation of energy: KE = PE v = (2gh) = (2 · 9.8 ms² · 3.0 m) = 7.67 ms Ft = mv = 70 kg (0 - 7.67 ms ) = -539 Ns Is the Ns a unit of momentum? kgm 1 Ns = 1 kgm · 1s = 1 s² s EXAMPLE X7 -5 Bend your knees when landing! (cont) Solution (b) stiff-legged – comes to rest in 1 cm v vavg = = 3.84 ms 2 d 10-2 m The collision lasts t = = = 2.60 (10-3) s vavg 3.84 ms F= p t = 539 Ns 2.60 (10-3) s 5 = 2.07 (10 ) N F is the net force on the man; it is the sum of the normal force, Fgrd, the ground exerts upward and the weight of the man, 690 N downward: F = Fgrd - mg so Fgrd = F + mg = 2.1(105) N + 0.0069 (105) N 2.1 (105) N EXAMPLE X7 - 5 (cont) Bend your knees when landing! Solution (c) springy – comes to rest in 50 cm v vavg = = 3.8 ms 2 d 0.5 m The collision lasts t = = vavg 3.8 ms p 540 Ns F= = = 4.2 (103) N t 0.13 s = 130 (10-3) s F is the net force on the man; it is the sum of the normal force, Fgrd, the ground exerts upward and the weight of the man, 690 N downward: F = Fgrd - mg so Fgrd = F + mg = 4.2 (103) N + 0.69 (103) N 5 (103) N So bending the legs reduces the force considerably from about 210 (103) N to 5 (103) N – a factor of 40, thus saving broken legs and hips – the bones could not stand the force in (b) EXAMPLE Washing a car – momentum and force [A] X7 - 1 Water leaves a hose at 1.5 kg/s with a speed of 20 m/s, and hits the side of a car which stops it (we ignore splashing). Find the force exerted by the car on the water Solution Take x +ve right. Each second 1.5 kg hits the car with a speed of 20 m/s; the momentum change is -30 kgm/s. The force required is p t This is -30 N, since t = 1 s Washing a car – momentum and force EXAMPLE X7 - 1 water splashes back? Is the force greater? What if the Solution The change in momentum is now increased The force required is p but p is larger so the force must be larger t REQUIRED BEFORE NEXT CLASS Read Section 5.2 from Nelson Textbook Pg. 227 # 1, 6, 7, 9, 12