Two speeding lead bullets, one of mass 15.0 g moving to the right at 325 m/s and
ID: 2252867 • Letter: T
Question
Two speeding lead bullets, one of mass 15.0 g moving to the right at 325 m/s and one of mass 7.85 g moving to the left at 390 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0
Two speeding lead bullets, one of mass 15.0 g moving to the right at 325 m/s and one of mass 7.85 g moving to the left at 390 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0 degree C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3 degree C, and a latent heat of fusion of 2.45 times 104 J/kg.) What analysis model is appropriate for the system of two bullets for the time interval from before to after the collision? isolated system (momentum) particle in equilibrium particle under a net force nonisolated system (momentum) From the model, what is the speed of the combined bullets after the collision? m/s How much of the initial kinetic energy has transformed to internal energy in the system after the collision? Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. J (d) Does all the lead melt due to the collision? Yes No What is the temperature of the combined bullets after the collision? degree C (f) What is the phase of the combined bullets after the collision?Explanation / Answer
a) isolated system
b) 15*325-7.85*390=22.85*v
v=79.36m/s
c) change in kinetic energy= 0.5*0.015(325^2-79.36^2) + 0.5*0.00785*(390^2-79.36^2)=744.95+572.27=1317.2252J
d) energy required for raising the temperature to melting point
=0.02285*128*(327.3-30)=869.5430J
energy left for melting = 447.6822J
m*2.45*10^4=447.6822
m=0.01827kg
=18.27gm will be converted to liquid
lead in solid state = 22.85-18.27=4.58g
Hence, No
e) 327.3C
f) m(solid)=4.58g
m(liq.)=18.27g
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.