QUIZ 7

1. Two balls of mass m₁ = 2kg and m₂ = 5kg undergo a perfectly elastic head-on collision. If the speed of m₁ was initially 4.0m/s, m₂ at rest, what will be their speeds after the collision?

Solution. From the conservation of momentum we have

                   m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.

               Because two balls undergo a perfectly elastic collision we have

                   v₁ -  v₂ = v₂'  -  v₁' .

               Then we have

                  (2kg)(4.0m/s) + (5kg)(0) = (2kg)v₁' + (5kg)v₂';

                 ( 4.0m/s) - 0 = v₂' - v₁'.

              Solving the two equations we have

                  v₁' = -1.7m/s, and  v₂'= 2.3m/s.

2. A 15-g bullet is fired horizontally into a 5-kg wood block originally at rest with speed of 150m/s and imbedded in the block. The coefficient of kinetic friction between the block and the ground is 0.25. How far will the block slide on the ground after the collision?

Solution. We find the speed of the block immediately after the perfect inelastic collision from

                m₁v₁ + m₂v₂= (m₁ + m₂)V;

               (15 x 10^-3kg)(150m/s) + 0 = (15 x 10^-3kg + 5kg)V, which gives V = 0.449m/s.

               We find the friction force on the block  F_fr = m_k(m₁ + m₂)g.

               From conservation of energy we have

                  DE = W_NC;

                  1/2(m₁ + m₂)V_f² - 1/2(m₁ + m₂)V_f² = m_k(m₁ + m₂)gDx;

                  0 - 1/2(15 x 10^-3kg + 5kg)(0.449m/s)² = -(0.25)(15 x 10^-3kg + 5kg)(9.80m/s²)Dx, which gives Dx = 0.041m = 4.1cm.