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07-203_hmw-Ch9

# 07-203_hmw-Ch9 - Chapter 9 Linear Momentum and...

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Chapter 9: Linear Momentum and CollisionsSolutions to Problems 4. Picture the Problem : The two carts approach each other on a frictionless track at different speeds. Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting expression for 2 v . Then use equation 7-6 to find the total kinetic energy of the two-cart system. Let cart 1 travel in the positive direction. Solution: 1. (a) Set 0 = p arrowrightnosp and solve for 2 : v ( )( ) 1 1 2 2 1 1 2 2 0 0.35 kg 1.2 m/s 0.69 m/s 0.61 kg m m mv v m = + = = - = = p v v arrowrightnosp arrowrightnosp arrowrightnosp 2. (b) No, kinetic energy is always greater than or equal to zero. 3. (c) Use equation 7-6 to sum the kinetic energies of the two carts: ( )( ) ( )( ) 2 2 1 1 1 1 2 2 2 2 2 2 1 1 2 2 0.35 kg 1.2 m/s 0.61 kg 0.69 m/s 0.40 J K mv mv = + = + = Insight: If cart 1 is traveling in the positive ˆ x direction, then its momentum is ( ) ˆ 0.42 kg m/s x and the momentum of cart 2 is ( ) ˆ 0.42 kg m/s - x . 8. Picture the Problem : The force of the kick acting over a period of time imparts an impulse to the soccer ball. Strategy: Multiply the force by the time of impact to find the impulse according to equation 9-5. Solution: Apply equation 9-5 directly: ( ) ( ) 3 av 1450 N 5.80 10 s 8.41 kg m/s I F t - = Δ = × = Insight: The 1450-N force is equivalent to 326 lb! The same impulse could be delivered to the ball with 8.41 N (1.89 lb) of force acting over a period of one second. 17. Picture the Problem : The two skaters push apart and move in opposite directions without friction. Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for 2 m . Let the velocity 1 v arrowrightnosp point in the negative direction, 2 v arrowrightnosp in the positive direction.

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07-203_hmw-Ch9 - Chapter 9 Linear Momentum and...

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