Solutions to Week 7 Assignment

1. Use the impulse-moment theorem.
               Dp = F_avDt
          p_f - p_i = mv_f - m(0)
                v_f = F_avDt/m
                    = (24N)(0.028s)/0.16kg = 4.2m/s.
2. (a) Use the impulse-momentum theorem.
          Dp_north = 0
            Dp_east = F_avDt = mDv_east = mv_east
         Find the magnitude and direction of v_f
         v_f = Öv_north² + v_east² = Öv_north² + (F_avDt/m)² = Ö(15m/s)² + [(15N)(4.0s)/(3.0kg)]² = 25m/s.
         q = tan^-1(v_north/v_east) = tan^-1(15m/s)/[(15N)(4.0s)/3.0kg] = 37º(north of east
        So, v_f = 25m/s at 37º north of east.
   (b) Let +y be north and +x be east.
           Dp = m(v_f - v_i)
          |Dp| = mÖ(v_fx - v_ix)² + (v_fy - v_iy)²
                 = mÖ(F_avDt/m - 0)² + (v_fy - v_iy)²
                 = F_avDt = (15N)(4.0s) = 60kg·m/s.
3. Use conservation of momentum.
   p_sf + p_tf = p_si + p_ti
            p_sf = 0 + 0 - p_tf
           m_sv_sf = -  m_tv_tf
          |v_sf| = m_t/m_s|-v_tf|
                 = 250kg/2.5 x 10⁶kg(100.0m/s) = 0.010m/s.
4. Find x_CM
         x_CM = (m_Ax_A+ m_Bx_B)/(m_A + m_B)
                 = (0 + m_Bx_B)/(m_A + m_B)
       m_Bx_B = (m_A + m_B)x_CM
         x_CM = (m_A + m_B)/m_B
                 = [(30.0g + 10.0g)/(10.0g)](2.0cm)
                 = 8.0cm
   Similarly
        y_CM = [(30.0g + 10.0g)/(10.0g)](5.0cm) = 20cm
   So, the coordinates of particle B are (x_B, y_B) = (8.0cm, 20cm)
5. p = Mv_CM = m_Av_A + m_Bv_B =  since p = p_A + p_B. Thus v_CM = (m_Av_A + m_Bv_B )/(m_A + m_B).
   Let east be in the positive direction.
   v_CM = [(5.0kg)(10m/s) + (15kg)(-10m/s)]/(5.0kg + 15kg) = -5m/s
   So,  v_CM = 5m/s west.
6. Use momentum conservation. The collision is perfectly inelastic, so v_1f = v_2f = v_f. Also, the block is initially at rest, so v_2i = 0.
   m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i
    (m₁ + m₂)v_f = m₁v_1i + m₂(0)
            v_f = [ m₁/(m₁ + m₂)]v_1i
                = [(0.050kg)/(0.050kg + 0.95kg)\(100.0m/s) = 5.0m/s.
7. Let right be +x and +y be in the initial direction of the puck. Use conservation of momentum.
   Find v_2fx
   m₁v_1fx + m₂v_2fx = m₁v_1ix + m₂v_2ix
     v_1fx + v_2fx = 0 + 0
                v_2fx = -v_1fx
   Find v_2fx
   m₁v_1fy + m₂v_2fy = m₁v_1iy + m₂v_2iy
   v_1fy + v_2fy = v_1iy + 0
              v_2fy = v_1iy - v_1fy
   Calculate v_2f.
    v_2f  = Öv_2fx² + v_2fy² = Ö(-v_1fx)² + (v_1iy - v_1fy)² = Ö(v_1fsinq₁)² + (v_1i - v_1fcosq₁)²
          = Ö[(0.36m/s)sin37°]² + [0.45m/s - (0.36m/s)cos37°]² = 0.27m/s
   Calculate the direction of the second puck.
   q  = tan^-1(v_2fx/v_2fy) = tan^-1[(-v_1fx)/(v_1iy - v_1fy)] = tan^-1[-(0.36m/s)sin37º(0.45m/s - 0.36m/scos37º] = 53ºto the left
   So, v_2f = 0.27m/s at 53ºto the left.