Physics Solution Manual for 1100 and 2101

# The procedure is discussed in example 7 in the text

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Unformatted text preview: tion 7.8b: 2m1 2 ( 715 kg ) vf 2 = v01 = ( +2.25 m/s ) = +1.82 m/s m +m 715 kg + 1055 kg 2 1 30. REASONING Since momentum is conserved, the total momentum of the two-object system after the collision must be the same as it was before the collision. Momentum is mass times velocity. Since one of the objects is at rest initially, the total initial momentum comes only from the moving object. Let m1 and v01 be, respectively, the mass and initial velocity of the moving object before the collision. In addition, m and vf are the total mass and final velocity of the two objects (which stick together) after the collision. The conservation of linear momentum can be written as = mvf { Total momentum after the collision m1v01 123 Total momentum before the collision Solving this equation for vf, the final velocity of the two-object system gives vf = m1v01 m SOLUTION Large-mass object (8.0 kg) moving initially: Assume that, before the collision, the object is moving in the + direction so that v01 = +25 m/s. Then, vf = m1v01 (8.0 kg ) ( +25 m/s ) = = +18 m/s m 3.0 kg + 8.0 kg Chapter 7 Problems 363 The final speed is 18 m/s . Small-mass object (3.0 kg) moving initially: vf = m1v01 ( 3.0 kg ) ( +25 m/s ) = = +6.8 m/s m 3.0 kg + 8.0 kg The final speed is 6.8 m/s . 31. SSM REASONING We obtain the desired percentage in the usual way, as the kinetic energy of the target (with the projectile in it) divided by the projectile’s incident kinetic energy, multiplied by a factor of 100. Each kinetic energy is given by Equation 6.2 as 1 mv 2 , 2 where m and v are mass and speed, respectively. Data for the masses are given, but the speeds are not provided. However, information about the speeds can be obtained by using the principle of conservation of linear momentum. SOLUTION We define the following quantities: KETP = kinetic energy of the target with the projectile in it KE0P = kinetic energy of the incident projectile mP = mass of incident projectile = 0.20 kg mT = mass of target = 2.50 kg vf = speed at which the target with the projectile in it flies off after being struck v0P = speed of incident projectile The desired percentage is KE TP Percentage = ×100 = KE 0P 1 2 ( mT + mP ) vf2 1 m v2 2 P 0P ×100 % (1) According to the momentum-conservation principle, we have mT vf (14+ mP )3 4 244 Total momentum of target and projectile after target is struck = 0 + mP v0P 14 244 4 3 Total momentum of target and projectile before target is struck Note that the target is stationary before being struck and, hence, has zero initial momentum. Solving for the ratio vf /v0P, we find that 364 IMPULSE AND MOMENTUM vf mP = v0P mT + mP Substituting this result into Equation (1) gives Percentage = = 1 2 ( mT + mP ) vf2 1 m v2 2 P 0P ×100 % = 1 2 ( mT + mP ) 1 m 2P 2 × 100 % m +m P T mP mP 0.20 kg × 100 % = ×100 % = 7.4 % mT + mP 2.50 kg + 0.20 kg 32. REASONING The weight of each vehicle is balanced by the normal force exerted by the road. Assuming that friction and other resistive forces can be ignored, we will treat the twovehicle system as an isolated system and apply the principle of conservation of linear momentum. SOLUTION Using v0, car and v0, SUV to denote the velocities of the vehicles before the collision and applying the principle of conservation of linear momentum, we have 0 1 24 43 = mcar v0, car + mSUV v0, SUV 1444 24444 4 3 Total momentum before collision Total momentum after collision Note that the total momentum of both vehicles after the collision is zero, because the collision brings each vehicle to a halt. Solving this result for v0, SUV and taking the direction in which the car moves as the positive direction gives v0, SUV = − mcar v0, car mSUV = − (1100 kg ) ( 32 m/s ) 2500 kg = −14 m/s This result is negative, since the velocity of the sport utility vehicle is opposite to that of the car, which has been chosen to be positive. The speed of the sport utility vehicle is the magnitude of v0, SUV or 14 m/s . 33. SSM WWW REASONING The system consists of the two balls. The total linear momentum of the two-ball system is conserved because the net external force acting on it is zero. The principle of conservation of linear momentum applies whether or not the collision is elastic. Chapter 7 Problems 365 m v f1 + m2 v f2 = m1 v 01 + 0 11 4 44 42 3 14 4 23 Total momentum after collision Total momentum before collision When the collision is elastic, the kinetic energy is also conserved during the collision 2 2 1 1 m v 2 + 1 m v f2 = 2 m1 v 01 + 0 2 1 f1 144 22 2 3 14243 4 444 Total kinetic energy after collision Total kinetic energy before collision SOLUTION a. The final velocities for an elastic collision are determined by simultaneously solving the above equations for the final velocities. The procedure is discussed in Example 7 in the text, and leads to Equations 7.8a and 7.8b. According to Equation 7.8: v f1 = F − m Iv m G +m J m HK 1 2 01 1 and v f2 = 2 F 2 m Iv G +m J m HK 1 01 1 2 Let the initial direction of motion of the 5.0...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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