Physics Solution Manual for 1100 and 2101

C using the same notation as in part b above we have

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Unformatted text preview: impulse-momentum theorem (Equation 7.4) to find the velocity of the first ball just before the collision. SOLUTION According to the impulse-momentum theorem, F ∆t = mvf − mv0 , and setting v0 = 0 m/s and solving for vf , we find that the velocity of the first ball after it is struck by the pool stick and just before it hits the second ball is vf = F ∆t +1.50 N ⋅ s = = 9 .09 m / s m 0.165 kg Substituting values into Equation 7.8b (with v01 = 9.09 m/s), we have 2m1 2m vf2 = v= v = v = +9.09 m/s m + m 01 m + m 01 01 1 2 38. REASONING a. Since air resistance is negligible as the ball swings downward, the work done by nonconservative forces is zero, Wnc = 0 J. The force due to the tension in the wire is perpendicular to the motion and, therefore, does no work. Thus, the total mechanical energy, which is the sum of the kinetic and potential energies, is conserved (see Section 6.5). The Chapter 7 Problems 369 conservation of mechanical energy will be used to find the speed of the ball just before it collides with the block. b. When the ball collides with the stationary block, the collision is elastic. This means that, during the collision, the total kinetic energy of the system is conserved. The horizontal or x-component of the total momentum is conserved, because the horizontal surface is frictionless, and so the net average external force acting on the ball-block system in the horizontal direction is zero. The total kinetic energy is conserved, since the collision is known to be elastic. SOLUTION a. As the ball falls, the total mechanical energy is conserved. Thus, the total mechanical energy at the top of the swing is equal to that at the bottom: m1 gL { Total mechanical energy at top of swing, all potential energy = 2 1 2 m1v 123 Total mechanical energy at bottom of swing, all kinetic energy In this expression L is the length of the wire, m1 is the mass of the ball, g is the magnitude of the acceleration due to gravity, and v is the speed of the ball just before the collision. We have chosen the zero-level for the potential energy to be at ground level, so the initial potential energy of the ball is m1gL. Solving for the speed v of the ball gives v = 2 gL = 2 ( 9.80 m/s 2 ) (1.20 m ) = 4.85 m/s At the bottom of the swing the ball is moving horizontally and to the right, which we take to be the +x direction (see the drawing in the text). Thus the velocity of the ball just before impact is vx = +4.85 m/s . b. The conservation of linear momentum and the conservation of the total kinetic energy can be used to describe the behavior of the system during the elastic collision. This situation is identical to that in Example 7 in Section 7.3, so Equation 7.8a applies: m − m2 vf = 1 v m1 + m2 x where vf is the final velocity of the ball after the collision, m1 and m2 are, respectively, the masses of the ball and block, and vx is the velocity of the ball just before the collision. Since vx = +4.85 m/s, we find that 370 IMPULSE AND MOMENTUM 1.60 kg − 2.40 kg vf = ( +4.85 m/s ) = −0.97 m/s 1.60 kg + 2.40 kg The minus sign indicates that the ball rebounds to the left after the collision. 39. REASONING AND SOLUTION The total linear momentum of the two-car system is conserved because no net external force acts on the system during the collision. We are ignoring friction during the collision, and the weights of the cars are balanced by the normal forces exerted by the ground. Momentum conservation gives (m1 + m2 )vf = m1v01 + m2v02 14 3 24 14244 4 3 Total momentum after collision Total momentum after collision where v02 = 0 m/s since the 1900-kg car is stationary before the collision. a. Solving for vf , we find that the velocity of the two cars just after the collision is vf = m1v01 + m2 v02 m1 + m2 = ( 2100 kg ) ( +17 m/s ) + (1900 kg ) ( 0 m/s ) = 2100 kg + 1900 kg +8.9 m/s The plus sign indicates that the velocity of the two cars just after the collision is in the same direction as the direction of the velocity of the 2100-kg car before the collision. b. According to the impulse-momentum theorem, Equation 7.4, we have F ∆t + + { = (m1 4m2 )vfinal − (m1 4m2 )vafter 14 244 3 14 244 3 Impulse due to friction Final momentum when cars come to a halt Total momentum just after collision where vfinal = 0 m/s since the cars come to a halt, and vafter = vf = +8.9 m/s . Therefore, we have F ∆t = ( 2100 kg + 1900 kg ) ( 0 m/s ) – ( 2100 kg + 1900 kg ) ( 8.9 m/s ) = – 3.6 × 104 N ⋅ s The minus sign indicates that the impulse due to friction acts opposite to the direction of motion of the locked, two-car system. This is reasonable since the velocity of the cars is decreasing in magnitude as the cars skid to a halt. c. Using the same notation as in part (b) above, we have from the equations of kinematics (Equation 2.9) that 2 2 vfinal = vafter + 2ax Chapter 7 Problems 371 where vfinal = 0 m/s and vafter = vf = +8.9 m/s . From Newton's second law we have that a = – f k / ( m1 + m2 ) , where f is the force of kinetic friction that acts on the cars as they skid k to a halt. Therefore, – fk 0 = v +2 x m +m 1 2 2 f or x= ( m1 + m2 ) vf2 2 fk According to Equation 4.8, f...
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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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