2014_SP08_Sol06

2014_SP08_Sol06 - Impulse and Momentum 9-1 9.28. Model:...

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Impulse and Momentum 9-1 9.28. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations. Visualize: Solve: To find the ball’s velocity just before and after it hits the floor: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 1 0 1 0 1 2 2 2 2 2 2 3 2 3 2 2 2 2 0 m /s 2 9.8 m/s 0 2.0 m 6.261 m/s 2 0 m /s 2 9.8 m/s 1.5 m 0 m 5.422 m/s y y y y y y y y y v v a y y v v v a y y v v = + - = + - - = - = + - = + - - = The force exerted by the floor on the ball can be found from the impulse-momentum theorem: 2 1 1 area under the force curve y y y mv mv Fdt mv = + = + ( 29 ( 29 ( 29 ( 29 ( 29 3 1 max 2 0.2 kg 5.422 m/s 0.2 kg 6.261 m/s 5 10 s F - = - + × max 935 N F = Assess: A force of 935 N exerted by the floor is reasonable. 9.31. Model: Use the particle model for the ball of clay (C) and the 1.0 kg block (B). The two objects are a system and it is a case of a perfectly inelastic collision. Since no significant external forces act on the system in the x -direction during the collision, momentum is conserved along the x -direction.
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Impulse and Momentum 9-2 Visualize: Solve: (a) The conservation of momentum equation
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2014_SP08_Sol06 - Impulse and Momentum 9-1 9.28. Model:...

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