Physics Solution Manual for 1100 and 2101

# Since the y component of the final momentum of object

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the same for both cars. Momentum, 2 mv, is a vector quantity that has a magnitude and a direction. The two cars have different directions, so they have different momenta. 2. (d) Momentum is a vector quantity that has a magnitude and a direction. The magnitudes (m0v0) and directions (due north) are the same for both runners. 3. The magnitude of the total momentum is 47 700 kg ⋅ m/s . 4. (d) According to the impulse-momentum theorem, the impulse is equal to the final momentum minus the initial momentum. Therefore, the impulses are the same in both cases, since the final momenta are the same and the initial momenta are the same. The impulse is also the product of the net average force and the time of impact. Since the impulses are equal and the time of impact is larger for the air mattress than for the ground, the air mattress exerts a smaller net average force on the high jumper. 5. The impulse is +324 kg ⋅ m/s . 6. (c) According to the impulse-momentum theorem, the impulse is equal to the change in the particle’s momentum. The magnitude of the momentum change is the same in regions A and C; therefore, in these regions the particle experiences impulses of the same magnitude. The momentum in region B is constant. Therefore, the change in momentum is zero, and so is the impulse. 7. (b) According to the impulse-momentum theorem, the net average force is equal to the change in the particle’s momentum divided by the time interval. This ratio is greatest in region C. The ratio is equal to zero in region B, since the change in the particle’s momentum is zero there. 8. Magnitude of net average force in Region A = 1.0 N, Magnitude of net average force in Region C = 4.0 N 9. (d) Since there are no external forces acting on the rocket, momentum is conserved before, during, and after the separation. This means that the momentum of the rocket before the separation (+150 000 kg⋅m/s) is equal to the sum of the momenta of the two stages after the separation [+250 000 kg⋅m/s + (− 100 000 kg⋅m/s) = +150 000 kg⋅m/s]. 342 IMPULSE AND MOMENTUM 10. (a) The net external force acting on the ball/earth system is zero. The gravitational forces that the ball and earth exert on each other are internal forces, or forces that the objects within the system exert on each other. The space probe is also an isolated system, since there are no external forces acting on it. 11. (c) The net external force that acts on the objects is zero, so the total linear momentum is conserved. Therefore, the total momentum before the collision (+12 kg⋅m/s − 2 kg⋅m/s = +10 kg⋅m/s) equals the total momentum after the collision (−4 kg⋅m/s +14 kg⋅m/s = +10 kg⋅m/s). 12. (b) The collision between the objects is an elastic collision, so the total kinetic energy is conserved. Therefore, the total kinetic energy before the collision (8 J + 6 J = 14 J) is equal to the total kinetic energy after the collision (10 J + 4 J = 14 J). 13. (c) Since the net external force acting on the two objects during the collision is zero, the total linear momentum of the system is conserved. In other words, the total linear momentum before the collision (−3 kg⋅m/s + 4 kg⋅m/s = +1 kg⋅m/s) equals the total linear momentum after the collision (+1 kg⋅m/s). Furthermore, some kinetic energy is lost during a completely inelastic collision, which is the case here. The final kinetic energy (4 J) is less that the total initial kinetic energy (1 J + 6 J = 7 J). 14. (d) The net external force acting on the two objects during the collision is zero, so the total momentum of the system is conserved. In two dimensions this means that the x-component of the initial total momentum (+16 kg⋅m/s) is equal to the x-component of the final total momentum. Since the x-component of the final momentum of object 1 is +6 kg⋅m/s, the x-component of the final momentum of object 2 must be +10 kg⋅m/s (+16 kg⋅m/s − 6 kg⋅m/s = +10 kg⋅m/s). 15. (c) The net external force acting on the two objects during the collision is zero, so the total momentum of the system is conserved. In two dimensions this means that the y-component of the initial total momentum (0 kg⋅m/s) is equal to the y-component of the final total momentum. Since the y-component of the final momentum of object 1 is −5 kg⋅m/s, then the y-component of the final momentum of object 2 must be +5 kg⋅m/s [0 kg⋅m/s − (−5 kg⋅m/s) = +5 kg⋅m/s] . 16. x coordinate of the center of mass = −1.5 m, y coordinate of the center of mass = 0 m 17. Velocity of center of mass = 2.0 m/s, along the +x axis Chapter 7 Problems CHAPTER 7 343 IMPULSE AND MOMENTUM PROBLEMS 1. SSM REASONING According to Equation 7.1, the impulse J produced by an average force F is J = F∆t , where ∆t is the time interval during which the force acts. We will apply this definition for each of the forces and then set the two impulses equal to one another. The fact that one average force has a magnitude that is three times as large as that of the other average force will then be used to obtain the desired time interval. SOLUT...
View Full Document

## 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.

Ask a homework question - tutors are online