Answer to chapter 7

1. 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 =  -0.720m/a (opposite to the direction of the package).

2. (a) We find the impulse on the ball from

                Impulse = Dp =mDv = (0.045kg)(45m/s - 0) =  2.0N.s

        (b) the average force is

                F =Impulse/Dt = (2.0N.s)/(5.0 x 10^-3s) =  4.0 x 10²N.

3. For the elastic collision of the two billiard balls, we use momentum conservation:

                m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂';

               m(2.00m/s) + m(-3.00m/s) = mv₁' + mv₂'.

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

                v₁ - v₂ = -(v₁' - v₂'), or 2.00m//s - (-3.00m/s) = v₂' - v₁'.

         Combining these two equations, we get

                v₁' = - 3.00m/s (rebound),  and  v₂' = 2.00m/s.

          Note that the two billiard balls exchange velocities.

4. Momentum conservation gives

                 0 = m₁v₁ + m₂v₂';

                 0 = mv₁' + 1.5mv₂', or v₁' = -1.5v₂'.

         The kinetic energy of each piece is

                 KE₂ = 1/2m₂v₂'²;

                 KE₁ = 1/2m₁v₁'² = 1/2(m₂/1.5)(-1.5v₂') = (1.5)1/2m₂v₂'² = 1.5KE₂.

         The energy supplied by the explosion produces the kinetic energy:

                 E = KE₁ + KE₂ = 2.5KE₂;

                 7500J = 2.5KE₂, which give KE₂ = 3000J.

         For the other piece we have

                 KE₁ = E - KE₂ = 7500J - 3000J = 4500J.

         Thus

           KE (heavier) = 3000J;  KE (lighter) = 4500J.

5. Because the cubes are made of the same material, their masses will be proportional to the volumes:                 m1, m2 = 23m1 = 8m1, m3 = 33m1 = 27m1.      From symmetry we see that yCM = 0.      We choose the x-origin at the outside edge of the small cube:                 xCM = (m1x1 + m2x2 +m3x3)/(m1 + m2 + m3)                         = {m1(1/2l0) + 8m1[l0 + 1/2(2l0)] +                               27m[l0 + 2l0 + 1/2(3l0)]}/(m1 + 8m1 + 27m1)                         = 138l0/36                         = 3.83l0 from the outer edge of the small cube.