Question

An interesting exhibit at Gettysburg is two lead balls that are stuck/melted together....

2. An interesting exhibit at Gettysburg is two lead balls that are stuck/melted together. During the battle of Gettysburg, the two balls (bullets had a head-on collision. The balls stopped and dropped to the ground melted together For this problem assume the balls are identical and traveling at same speed in opposite directions. During the collision all of their kinetic energy is converted to heat. Also assume of the mass of each ball is melted. The initial temperature is 20.00 oct. Find the speed of the balls just before their collision to four significant figures. You can use the following constants for lead 2.50x104 328 C c 128 67

Answer 1

Calculate the total kinetic energy of the system before collision as follows: K -my? +-mv2 2 = mv2 As the total energy is converted into heat, therefore, the equation becomes, 40 my? = (2m) SAT + (2m); 100 v2 = 2SAT +=(2), (2)L, Here S is the specific heat, Lc is the latent heat of fusion and AT is the change in temperature. Calculate the change in temperature as follows, AT = 328°C -20°C = 308°C Substitute 128 J/kg K for S. 308°C for AT and 2.5x10* J/kg for LE 40 mv? = (2m)SAT+ 6}(2m)L, 100 = 2SAT +(L) 5 v2 = 2(128)(308) +$(2.5*10) = 98848 V=V98848 = 314.4 m/s Therefore, velocity before collision is 314.4 m/s.