Chapter 9

# Chapter 9 - 9.1. Model: Model the car and the baseball as...

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9.1. Model: Model the car and the baseball as particles. Solve: (a) The momentum p mv = () ( ) 1500 kg 10 m/s = 4 1.5 10 kg m/s. (b) The momentum p mv = ( ) 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, car bicycle pp = car car car bicycle bicycle bicycle car bicycle m mv m v v v m ⇒= = () 1500 kg 5.0 m/s 75 m/s 100 kg ⎛⎞ == ⎜⎟ ⎝⎠ Assess: This is a very high speed ( 168 mph). This problem shows the importance of mass in comparing two momenta.
9.3. Visualize: Please refer to Figure EX9.3. Solve: The impulse x J is defined in Equation 9.6 as () f i t xx t J Ftd t == area under the x Ft curve between i t and f t ( ) ( ) 1 4 ms 1000 N 6 4 ms 1000 N 4 Ns 2 x J =+ =

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9.4. Model: The particle is subjected to an impulsive force. Visualize: Please refer to Figure EX9.4. Solve: Using Equation 9.6, the impulse is the area under the curve. From 0 s to 2 ms the impulse is () ( ) 3 1 2 500 N 2 10 s 0.5 N s Fdt =− × = From 2 ms to 8 ms the impulse is ( ) 1 2 2000 N 8 ms 2 ms 6.0 N s Fdt =+ = + From 8 ms to 10 ms the impulse is ( ) 1 2 500 N 10 ms 8 ms 0.5 N s Fdt = Thus, from 0 s to 10 ms the impulse is 0.5 6.0 0.5 N s 5.0 N s. −+ − =
9.5. Visualize: Please refer to Figure EX9.5. Solve: The impulse is defined in Equation 9.6 as () f i t xx t J Ftd t == area under the x Ft curve between i t and f t ( )( ) 3 1 max max 2 6.0 N s = 8 ms 1.5 10 N FF ⇒⇒ = ×

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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 fi , x xx p pJ =+ 2 kg m/s 6 kg m/s x J −= ++ 8 kg m/s 8 N s x J ⇒= = Since avg , x J Ft we have avg 8 N s Δ=− 2 avg 8 N s 81 0 N 10 ms F = × The force is ( ) 2 0 N , l e f t. F G
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: Using the equations fi x xx p pJ =+ and f i () t t J Ftd t = = area under force curve ( ) f 2.0 kg 2.0 kg 1.0 m/s x v (area under the force curve) ( ) f 1 1.0 m/s 1.0 s 2.0 N 2.0 m/s 2.0 kg x v ⇒= + = Becaue f x v is positive, the object moves to the right at 2.0 m/s. Assess: For an object with positive velocity, a positive impulse increases the object’s speed. The opposite is true for an object with negative velocity.

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9.8. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.8. Solve: Using the equations fi x xx p pJ =+ and f i () t t J Ftd t = = area under force curve ( ) f 2.0 kg 2.0 kg 1.0 m/s x v (area under the force curve) ( ) f 1 1.0 m/s 2.0 N 0.50 s 0.50 m/s 2.0 kg x v ⎛⎞ ⇒= + = ⎜⎟ ⎝⎠ Assess: For an object with positive velocity, a negative impulse slows the object. The opposite is true for an object with negative velocity.
9.9.

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## This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 9 - 9.1. Model: Model the car and the baseball as...

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