Physics Solution Manual for 1100 and 2101

8a as m m2 vf 1 1 v m m 01 1 2 where m1 and m2 are

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Unformatted text preview: site directions and are both slowing down. Finally, we are given that skater 1 glides twice as far as skater 2. Thus, the displacement of skater 1 is related to that of skater 2 by x1 = −2x2 , where, the minus sign denotes that the skaters move in opposite directions. Substituting these values into Equation (2) yields m1 = m2 ( 0 m/s ) 2 ( 0 m/s ) 2 − 2 ( −a1 )( x2 ) − 2a1 ( −2 x2 ) = 1 = 0.707 2 26. REASONING AND SOLUTION Since no net external force acts in the horizontal direction, the total horizontal momentum of the system is conserved regardless of which direction the mass is thrown. The momentum of the system before the mass is thrown off the wagon is mWvA, where mW and vA are the mass and velocity of the wagon, respectively. When ten percent of the wagon’s mass is thrown forward, the wagon is brought to a halt so its final momentum is zero. The momentum of the mass thrown forward is 0.1mW(vA + vM), where vM is the velocity of the mass relative to the wagon and (vA + vM) is the velocity of the mass relative to the ground. The conservation of momentum gives 360 IMPULSE AND MOMENTUM 0.1mW ( vA + vM ) = 144 2444 4 3 Momentum after mass is thrown forward mv 1W 3 4A 24 Momentum before mass is thrown forward Solving for vM yields vM = 9vA (1) When the direction in which the mass is thrown is reversed, the velocity of the mass relative to the ground is now (vA − vM). The momentum of the mass and wagon is, therefore, 0.1mW ( vA − vM ) + 144 2444 4 3 Momentum of the mass 0.9m vB 14 W 3 24 Momentum of the wagon where vB is the velocity of the wagon. The conservation of momentum gives 0.1mW ( vA − v ) + 0.9mW vB = 144444 M 244444 3 Momentum after mass is thrown off mv 1W 3 4A 24 (2) Momentum before mass is thrown off Substituting Equation (1) into Equation (2) and solving for vB/vA gives vB/vA = 2 . 27. SSM WWW REASONING The cannon and the shell constitute the system. Since no external force hinders the motion of the system after the cannon is unbolted, conservation of linear momentum applies in that case. If we assume that the burning gun powder imparts the same kinetic energy to the system in each case, we have sufficient information to develop a mathematical description of this situation, and solve it for the velocity of the shell fired by the loose cannon. SOLUTION For the case where the cannon is unbolted, momentum conservation gives m1vf1 + m2vf2 = 14 244 40 3 14 244 Initial momentum 4 3 Total momentum after shell is fired (1) of system where the subscripts "1" and "2" refer to the cannon and shell, respectively. In both cases, the burning gun power imparts the same kinetic energy to the system. When the cannon is bolted to the ground, only the shell moves and the kinetic energy imparted to the system is 2 KE = 1 mshellvshell = 1 (85.0 kg)(551 m/s) 2 = 1.29 ×107 J 2 2 The kinetic energy imparted to the system when the cannon is unbolted has the same value and can be written using the same notation as in equation (1): Chapter 7 Problems 2 2 KE = 1 m1vf1 + 1 m2vf2 2 2 361 (2) Solving equation (1) for vf1, the velocity of the cannon after the shell is fired, and substituting the resulting expression into Equation (2) gives KE = 22 m2 vf2 2m1 2 + 1 m2vf2 2 (3) Solving equation (3) for vf2 gives vf2 = 2KE 2(1.29 × 107 J) = = +547 m/s m2 85.0 kg (85.0 kg) + 1 m2 + 1 m1 5.80 × 103 kg 28. REASONING Together, Ashley and Miranda constitute an isolated system, since their combined weight is balanced by an upward normal force, and friction is negligible. The total momentum of the system is, therefore, conserved when Miranda hops onto the tube. We will use the momentum conservation principle m1vf1 + m2 vf2 = m1v01 + m2 v02 (Equation 7.7b) to analyze this one-dimensional collision. We are ignoring the mass and momentum of the inner tube. SOLUTION After Miranda (m2 = 58 kg) jumps onto the inner tube, she and Ashley (m1 = 71 kg) both have the same final velocity: vf = vf1 = vf2. Making this substitution in Equation 7.7b, and solving for their common final velocity, we obtain m1vf + m2vf 14243 Total momentum after Miranda hops on = m1v01 + m2v02 14 244 4 3 or vf = Total momentum before Miranda hops on m1v01 + m2v02 m1 + m2 Their common velocity after Miranda hops on is, therefore, vf = m1v01 + m2 v02 ( 71 kg ) ( +2.7 m/s ) + ( 58 kg ) ( +4.5 m/s ) = = +3.5 m/s m1 + m2 71 kg + 58 kg The common speed is the magnitude of this value or 3.5 m/s . 29. REASONING Since the collision is an elastic collision, both the linear momentum and kinetic energy of the two-vehicle system are conserved. The final velocities of the car and van are given in terms of the initial velocity of the car by Equations 7.8a and 7.8b. 362 IMPULSE AND MOMENTUM SOLUTION a. The final velocity vf1 of the car is given by Equation 7.8a as m − m2 vf 1 = 1 v m + m 01 1 2 where m1 and m2 are, respectively, the masses of the car and van, and v01 is the initial velocity of the car. Thus, 715 kg − 1055 kg vf 1 = ( +2.25 m/s ) = −0.432 m/s 715 kg + 1055 kg b. The final velocity of the van is given by Equa...
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