246_PartUniversity Physics Solution

246_PartUniversity Physics Solution - Momentum, Impulse,...

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Momentum, Impulse, and Collisions 8-5 (b) 22 2 2 111 1 222 2 (1750 kg)(1.50 m/s) (1450 kg)(1.10 m/s) 2846 J AA BB Km v m v =+= + = . 2 2 1 2 (1750 kg)(0.250 m/s) (1450 kg)(0.409 m/s) 176 J v m v + = . The change in kinetic energy is 21 176 J 2846 J 2670 J KK K Δ= − = = . EVALUATE: The total momentum of the system is constant because there is no net external force during the collision. The kinetic energy of the system decreases because of negative work done by the forces the cars exert on each other during the collision. 8.19. IDENTIFY: Since the rifle is loosely held there is no net external force on the system consisting of the rifle, bullet and propellant gases and the momentum of this system is conserved. Before the rifle is fired everything in the system is at rest and the initial momentum of the system is zero. SET UP: Let + x be in the direction of the bullet&s motion. The bullet has speed 601 m/s 1.85 m/s 599 m/s −= relative to the earth. 2rbg x xxx Pppp =++ , the momenta of the rifle, bullet and gases. r 1.85 m/s x v =− and b 599 m/s x v =+ . EXECUTE: 0 xx PP == . rbg 0 ppp ++= . gr b (2.80 kg)( 1.85 m/s) (0.00720 kg)(599 m/s) x pp p and g 5.18 kg m/s 4.31 kg m/s 0.87 kg m/s x p = . The propellant gases have momentum 0.87 kg m/s , in the same direction as the bullet is traveling. EVALUATE: The magnitude of the momentum of the recoiling rifle equals the magnitude of the momentum of the bullet plus that of the gases as both exit the muzzle. 8.20. IDENTIFY: In part (a) no horizontal force implies x P is constant. In part (b) use the energy expression, Eq. 7.14, to find the potential energy intially in the spring. SET UP: Initially both blocks are at rest. Figure 8.20 EXECUTE: (a) 11 2 2 AAx BBx mv +=+ 0 3.00 kg ( 1.20 m/s) 3.60 m/s 1.00 kg B Ax Bx A m vv m ⎛⎞ + ⎜⎟ ⎝⎠ Block A has a final speed of 3.60 m/s, and moves off in the opposite direction to B. (b) Use energy conservation: 1 1 other 2 2 K UW KU ++ = + . Only the spring force does work so other el 0 and . WU U 1 0 K = (the blocks initially are at rest) 2 0 U = (no potential energy is left in the spring) 2 2 11 1 1 22 2 2 (1.00 kg)(3.60 m/s) (3.00 kg)(1.20 m/s) 8.64 J v m v + = 11 , e l UU = the potential energy stored in the compressed spring. Thus 1,el 2 8.64 J UK EVALUATE: The blocks have equal and opposite momenta as they move apart, since the total momentum is zero. The kinetic energy of each block is positive and doesn&t depend on the direction of the block&s velocity, just on its magnitude. 8.21. IDENTIFY: Since friction at the pond surface is neglected, there is no net external horizontal force and the horizontal component of the momentum of the system of hunter plus bullet is conserved. Both objects are initially at rest, so the initial momentum of the system is zero. Gravity and the normal force exerted by the ice together produce a net vertical force while the rifle is firing, so the vertical component of momentum is not conserved. SET UP: Let object A be the hunter and object B be the bullet. Let + x be the direction of the horizontal component of velocity of the bullet. Solve for 2 v .
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8-6 Chapter 8 EXECUTE: (a) 2 965 m/s Bx v =+ .
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246_PartUniversity Physics Solution - Momentum, Impulse,...

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