ANSWER TO CHAPTER 9 HOMEWORK

1.  The change in momentum is

                 Dp = p₂ -  p₁ = mvj - mvi

                       = (0.145kg)(30m/s)j - (1.45kg)(30m/s)i =  -(4.35kg.m/s)i + (4.35kg.m/s)j.

2.  During the throwing we use momentum conservation for the one-dimensional motion:

                  0 = (m_boat + m_child)v_boat + m_packagev_package

                  0 = (55.0kg +26.0kg)v_boat + (5.40kg)(10.0m/s), which gives

                  v_boat =  -0.667m/s (opposite to the direction of the package).

3.  In the reference frame of the capsule before the push, we take the positive direction in the direction the capsule will move.

         (a) Momentum conservation give us          

                    mv_astronaut + Mv_satellite = mv_astronaut' + Mv'_satellite

                         0 + 0 = (140kg)(-2.50m/s) + (1800kg)v'_satellite, which gives v'_satellite=  0.194m/s

         (b) We find the force on the satellite from

                    F_satellite = Dp /Dt = m_satelliteDv_satellite/Dt   

                                 = (1800kg)(0.194m/s - 0)/(0.500s) = 700N.

                There will be an equal but opposite force on the astronaut.

          (c)  The kinetic energies are:

                    K_astronaut = 1/2mv'_astronaut² = 1/2(140kg)(2.50m/s)² = 438J.

                   K_satellite = 1/2Mv'_satellite² = 1/2(1800kg)(0.194m/s)² = 33.9J.

4.  For the elastic collision of the two pucks, we use momentum conservation for this one-dimensional motion:

                   m₁v₁ + m₂v₂ =m₁v₁' + m₂v₂';

                  (0.450kg)(4.20m/s) + (0.900kg)(0) = (0.450kg)v₁' + (0.900kg)v₂'.

          Because the collision is elastic, the relative speed does not change:

                   v₁ - v₂ = -(v₁' - v₂'), or 4.200m/s - 0 = v₂' - v₁'.

          Combining these two equations, we get

                   v₁' = -1.40m/s (rebound),    and   v₂' = 2.80m/s.

5.  We let V be the speed of the block and bullet immediately after the collision and before the pendulum swings. For this perfectly inelastic collision, we use momentum conservation:                   mv + 0 = (M + m)V;                  (0.018kg)(180m/s) = (0.018kg + 3.6kg)V'            which gives V = 0.896m/s.            Because the tension does no work, we can use energy conservation for the swing:                   1/2(M + m)V2 = (M+ m)gh, or V2 = 2gh;                   (0.896m/s)2 = 2(9.80m/s2)h, which gives h = 0.0409m.             We find the horizontal displacement from the triangle:                   L2 = (L - h)2 + x2.                   (2.8m)2 = (2.8m - 0.0409m)2 + x2, which gives x =  0.48m.

6.  Using the coordinate system shown, for momentum conservation we have                   x:  0 + mv2 = mv1'cosa + 0;                        3.7m/s = v1'cosa;                   y:   mv1 + 0 = mv1'sina + mv2';                        2.0m/s = v1'sina + v2', or                        v1'sina = 2.0m/s - v2'           For the conservation of kinetic energy, we have                        1/2mv12 + 1/2mv22 = 1/2mv'12 + 1/2mv'22;                        (2.0m/s)2 + (3.7m/s)2 = v'12 + v'22.           We have three equations in three unknowns: a, v1', v2'.           We eliminate a by squaring  and adding the two momentum results, and then combine this with the energy equation,           with the results:                        a = 0°, v1' = 3.7m/s, v2' = 2.0m/s.            The two billiard balls exchange velocities.

7.  We know from the symmetry that the center of mass lies on the y-axis: xCM = 0           We treat the plate as an infinite number of semicircular wires. For the wire with radius r and thickness dr, we choose a differential length rdq, shown in the diagram, which has a mass dm = (2m/pR2)rqdr.           We integrate over the plate to find yCM:                        yCM = òòydm/òdm = [òòrsinq(2m/pR2)rdqdr]/m = (2/pR2)ò0Rr2drò0psinqdq                               = (2/pR2)(R3/3)(-cosq)|0p = 2R/3p[-(-1) + (1)] = 4R/3p.            Thus the center of mass is at  xCM = 0, yCM = 4R/3p

8.  We find the velocity of tahe center of mass from

                        v_CM = (m₁v₁ + m₂v₂)/(m₁ + m₂)

                                = {(35kg)[(12m/s)i - (16m/s)j] + (35kg)[(-20m/s)i + (14m/s)j]}/(35kg + 35kg)

                                =  (-4m/s)i + (-1m/s)j.

*9 (a) At the maximum compression of the spring, there will be no relative motion of the two blocks. Because there is no friction, we can use momentum conservation:

                        m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂';

                        m₁v₁ + 0 = (m₁ + m₂)V, or V = m₁v₁/(m₁ + m₂).

Energy is also conserved, so we we have

                       1/2m₁v₁² + 0 = 1/2(m₁ + m₂)V² + 1/2kx²;

                       1/2m₁v₁² = 1/2(m₁ + m₂)[m₁v₁/(m₁ + m₂)]² + 1/2kx², or

                        x² =  m₁m₂v₁²/k(m₁ + m₂) = (2.0kg)(4.5kg)(80m/s)²/(850N/m)(2.0kg + 4.5kg),

which gives x = 0.32m.

(b)  From the initial motion of the first block to the final separation, all horizontal forces are internal to the system of the two blocks and are conservative. For momentum conservation we have

                          m₁v₁ + m₂v₂ = m₁v₁'' + m₂v₂'';

                          m₁v₁ + 0  = m₁v₁'' + m₂v₂''.

Because the collision is elastic, the relative speed does not change:

                         v₁ - v₂ = - (v₁'' - v₂''), or v₂'' = v₁ - 0 + v₁'' .

When we combine the equations, we get

                         v₁'' = (m₁ - m₂)v₁/(m₁ + m₂)

                               = (2.0kg - 4.5kg)(8.0m/s)/(2.0kg + 4.5kg) = -3.1m/s (rebound).

For v₂'' we get

                         v₂'' =  v₁ + v₁'' = 8.0m/s + (-3.1m/s) = 4.9m/s.

(c) Yes, because the force by the spring is conservative, the collision is elastic.