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Published byChelsea Bunts Modified about 1 year ago

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Aim: How can we explain momentum and impulse? Do Now: Which is easier to do: Stop a skateboard traveling at 5 m/s or stop a car traveling at 5 m/s? Why?

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Which is easier to do: Stop a bullet fired from a gun or stop a bullet that is thrown at you? Why?

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Momentum (p) “Mass in motion” Momentum = mass * velocity Or p = mv Vector quantity What about the units? Units are kg·m/s

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What is the momentum of a 60 kg halfback moving 9 m/s, eastward? p = mv p = (60 kg)(9 m/s) p = 540 kg·m/s eastward

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If a 1 kg ball bounces off of a wall with the same velocity as shown below, is there a change in momentum? Yes because there has been a change in direction Momentum is a vector quantity v = 10 m/s

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So how do we define a change in momentum? Δp = p f – p i Δp = mv f – mv i Δp = m(v f – v i ) Δp = mΔv

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Calculate the change in momentum of the bouncing ball Δp = mΔv Δp = (1 kg)(-10 m/s – 10 m/s) Δp = (1 kg)(-20 m/s) Δp = -20 kg·m/s The negative sign indicates a change in direction

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Why do you follow through when: Swinging a baseball bat Swinging a tennis racket Swinging a golf club Kicking a football

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Impulse (J) A force must act on an object for a time in order to change its velocity Impulse (J) = Force * time Or J = Ft Vector quantity What about the units? The units are N·s

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Calculate the impulse on a baseball being hit by a baseball bat with a force of 1200 N over 0.02 s J = Ft J = (1200 N)(0.02 s) J = 24 N·s

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Egg Demo Why doesn’t the egg break when it hits the bed sheet?

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We know: Doesn’t J = Ft? Doesn’t Δp = mΔv? So J = Δp Impulse is a change in momentum!!

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mΔv is a constant mass has not changed initial and final velocities have not changed Time to slow down increased Therefore force has to decrease Hence, the egg does not break!

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Real-World Applications Baseball and tennis player’s ‘following through’ Boxer’s ‘riding the punch’ Airbags Padded dashboards

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What if force and time are constant but the mass changes? Astroblaster Demo If mass decreases, velocity must increase!

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A car with m=725 kg is moving at 32 m/s to the east. The driver applies the brakes for 2 s. An average force of 5.0 x 10 3 N is exerted on the car. What is the change in momentum? Δp = Ft Δp = (-5 x 10 3 N)(2 s) Δp = -1 x 10 4 N What is the impulse on the car? J = Δp J = -1 x 10 4 N What is the car’s final velocity? Δp = mΔv -1 x 10 4 N = (725 kg)(v f – 32 m/s) -1 x 10 4 = 725v f – 23,200 v f = 18.2 m/s

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An impulse of 30.0 N·s is applied to a 5.00 kg mass. If the mass had a speed of 100 m/s before the impulse, what is its speed after the impulse? J = mΔv 30 N·s = (5 kg)(v f – 100 m/s) 30 = 5v f – 500 v f = 106 m/s

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A car with a mass of 1.0 x 10 3 kg is moving with a speed of 1.4 x 10 2 m/s. What is the impulse required to bring the car to rest? J = mΔv J = (1 x 10 3 kg)(0 m/s – 1.4 x 10 2 m/s) J = -1.4 x 10 5 N·s

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