A metallic rod of length 2 m is moving longitudinally on a smooth horizontal surface with a velocity of 200 m/s. It undergoes a perfectly inelastic collision with a rigid wall and comes to rest. If 40 % of its initial kinetic energy is converted into heat and absorbed entirely by the rod, what will be the increase in the length of the rod? [Given: Specific heat capacity of the rod s = 500 J/(kg ^ C) and coefficient o

Step-by-step solution

Initial kinetic energy of the rod is given by: KE = 1 2 mv^2 = 1 2 m(200)^2 = 20000m J Heat absorbed by the rod is 40 % of the initial kinetic energy: Q = 0.40 20000m = 8000m J Using the calorimetry principle Q = ms T : 8000m = m 500 T T = 8000 500 = 16^ C The increase in the length of the rod is: L = L T L = 2 1.5 10⁻⁵ 16 = 48 10⁻⁵ m Converting to millimeters: L = 48 10⁻⁵ 10^3 mm = 0.48 mm Answer: 0.48 mm