Chapter09 - Solve(a The momentum p = mv =(1500 kg(10 m/s = 1.5 10 4 kg m/s(b The momentum p = mv =(0.2 kg 40 m/s = 8.0 kg m/s 9.1 Model Model the

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9.1. Model: Model the car and the baseball as particles. Solve: (a) The momentum pm v = = () ( ) 1500 kg 10 m/s 15 10 4 . kg m/s. (b) The momentum v = = ( ) 0.2 kg 40 m/s = 8.0 kg m/s.
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9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle. Solve: From the definition of momentum, pp car bicycle = ⇒= = mv m v v m m v car car bicycle bicycle bicycle car bicycle car = () = 1500 kg 100 kg 5.0 m/s m/s 75 0 . Assess: This is a very high speed ( 168 mph). This problem shows the importance of mass in comparing two momenta.
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9.3. Visualize: Please refer to Figure Ex9.3. Solve: The impulse is defined in Equation 9.6 as JF t d t xx t = () = i f t area under the F x ( t ) curve between t i and t f () ⇒= 6 0 8 1500 1 2 . N s = ms N max max FF
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9.4. Model: Model the rubber ball and the steel ball as particles, and their interaction as a collision. Solve: When objects collide, Newton’s third law tells us that the forces between the objects are equal and opposite. The time of contact is the same for each object. Therefore, the impulse on each will be the same.
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9.5. Model: The particle is subjected to an impulsive force. Visualize: Please refer to Figure Ex9.5. Solve: Using Equation 9.5, the impulse is the area under the curve. From 0 s to 2 ms the impulse is Fdt =− () × 1 2 3 21 0 500 N s 0.5 N s From 2 ms to 8 ms the impulse is Fdt =+ 1 2 2000 N 8 ms 2 ms 6.0 N s From 8 ms to 10 ms the impulse is Fdt 1 2 500 N 10 ms 8 ms 0.5 N s Thus, from 0 s to 10 ms the impulse is −+ − 05 60 05 . . . N s = 5.0 N s.
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9.6. Model: Model the object as a particle and the interaction as a collision. Visualize: Please refer to Figure Ex9.6. Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative momentum). Using the impulse-momentum theorem ppJ xx x fi =+ , −= ++ 26 kg m/s J x ⇒= = J x 8 kg m/s 8 N s Since JFt x = avg , we have Ft avg 8 N s ∆=− =− F avg 8 N s 10 ms 800 N The force is r F = () 800 N, left .
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9.7. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure Ex9.7. Solve: (a) Using the equations p f x = p i x + J x JF t d t xx t t = () i f = area under force curve ⇒= + v x f 1.0 m/s 2.0 kg 1 (area under the force curve) = + = 2.0 kg 1.0 N s m/s 1 15 . (b) Likewise, v x f 2.0 kg = + 1 (area under the force curve) = + = 2.0 kg 1.0 N s 1 05 . Assess: For an object with positive velocity, a negative impulse slows an object, whereas a positive impulse increases an object’s speed. The opposite is true for an object with negative velocity.
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9.8. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum theorem. Visualize: Note that the force of kinetic friction f k imparts a negative impulse to the sled. Solve: Using pJ xx = , we have p p F t dt f dt f t x t t t t fi k k i f i f = =− () ∆ ⇒ mv mv n t mg t k k µµ ∆∆ We have used the model of kinetic friction f k = µ k n , where k is the coefficient of kinetic friction and n is the normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken
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This note was uploaded on 04/02/2008 for the course PHYS 131-133 taught by Professor All during the Spring '08 term at Cal Poly.

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Chapter09 - Solve(a The momentum p = mv =(1500 kg(10 m/s = 1.5 10 4 kg m/s(b The momentum p = mv =(0.2 kg 40 m/s = 8.0 kg m/s 9.1 Model Model the

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