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251_PartUniversity Physics Solution

# 251_PartUniversity Physics Solution - 8-10 8.35 8.36...

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8-10 Chapter 8 8.35. I DENTIFY : Neglect external forces during the collision. Then the momentum of the system of the two cars is conserved. S ET U P : S 1200 kg m = , L 3000 kg m = . The small car has velocity S v and the large car has velocity L v . E XECUTE : (a) The total momentum of the system is conserved, so the momentum lost by one car equals the momentum gained by the other car. They have the same magnitude of change in momentum. Since m = ! ! p v and Δ ! p is the same, the car with the smaller mass has a greater change in velocity. S S L L m v m v Δ = Δ and L S L S 3000 kg 2.50 1200 kg m v v v v m Δ = Δ = Δ = Δ . (b) The acceleration of the small car is greater, since it has a greater change in velocity during the collision. The large acceleration means a large force on the occupants of the small car and they would sustain greater injuries. E VALUATE : Each car exerts the same magnitude of force on the other car but the force on the compact has a greater effect on its velocity since its mass is less. 8.36. I DENTIFY : The collision forces are large so gravity can be neglected during the collision. Therefore, the horizontal and vertical components of the momentum of the system of the two birds are conserved. S ET U P : The system before and after the collision is sketched in Figure 8.36. Use the coordinates shown. Figure 8.36 E XECUTE : There is no external force on the system so 1 2 x x P P = and 1 2 y y P P = . 1 2 x x P P = gives raven-2 (1.5 kg)(9.0 m/s) (1.5 kg) cos v φ = and raven-2 cos 9.0 m/s v φ = . 1 2 y y P P = gives raven-2 (0.600 kg)(20.0 m/s) (0.600 kg)( 5.0 m/s) (1.5 kg) sin v φ = + and raven-2 sin 10.0 m/s v φ = . Combining these two equations gives 10.0 m/s tan 9.0 m/s φ = and 48 φ = ° . E VALUATE : Due to its large initial speed the lighter falcon was able to produce a large change in the raven°s direction of motion. 8.37. I DENTIFY : Since friction forces from the road are ignored, the x and y components of momentum are conserved. S ET U P : Let object A be the subcompact and object B be the truck. After the collision the two objects move together with velocity 2 ! v . Use the x and y coordinates given in the problem. 1 1 0 A y B y v v = = . 2 (16.0 m/s)sin24.0 6.5 m/s x v = = ° ; 2 (16.0 m/s)cos24.0 14.6 m/s y v = = ° . E XECUTE : 1 2 x x P P = gives 1 2 ( ) A A x A B x m v m m v = + . 1 2 950 kg 1900 kg (6.5 m/s) 19.5 m/s 950 kg A B A x x A m m v v m + + = = = . 1 2 y y P P = gives 1 2 ( ) A B y A B y m v m m v = + . 1 2 950 kg 1900 kg (14.6 m/s) 21.9 m/s 1900 kg A B B y y A m m v v m + + = = = . Before the collision the subcompact car has speed 19.5 m/s and the truck has speed 21.9 m/s. E VALUATE : Each component of momentum is independently conserved. 8.38. I DENTIFY : Apply conservation of momentum to the collision. Apply conservation of energy to the motion of the block after the collision.

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Momentum, Impulse, and Collisions 8-11 S ET U P : Conservation of momentum applied to the collision between the bullet and the block: Let object A be the bullet and object B be the block. Let A v be the speed of the bullet before the collision and let V be the speed of the block with the bullet inside just after the collision.
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