In Fig. 8.22b, the kinetic energy of the Ping-Pong ball is larger after its interaction with the bowling ball than before. From where does the extra energy come? Describe the event in terms of conservation of energy.
Solution 13DQ Step 1 of 3: Before collision the the bowling ball of mass M was moving with speed v and the ping b1 pong ball is at rest with velocity zero v = 0 as shown in the figure below, p1 The energy of the system before collision, E 1 Kinetic energy of bowling ball + potential energy of ping pong ball E1= mv2 2 + PE ………….1 b1 Step 2 of 3: After collision with the ping pong ball with mass m, the bowling ball speed reduces to a value v b2d the fraction of initial kinetic energy of bowling ball and initial potential energy of the ping pong ball will add up to give some speed (v ) orp2inetic energy for the ping pong ball(which was at rest before) as shown in the figure below, The total energy of the system after collision, E = Kinetic energy of bowling ball + Kinetic energy of ping pong ball 2 1 2 1 2 E 2 mv2 b2+ 2mv …p2…..2