Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 rn/s an

Question

Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 rn/s and one of mass 7.85 g moving to the left at 375 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0 degreeC. 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 X 10^4 J/kg.) (a) What analysis model is appropriate for the system of two bullets for the time interval from before to after the collision? isolated system (momentum) nonisolated system (momentum) particle in equilibrium particle under a net force (b) From the model, what is the speed of the combined bullets after the collision? m/s (c) How much of the initial kinetic energy has transformed to internal energy in the system after the collision? J (d) Does all the lead melt due to the collision? Yes No (e) What is the temperature of the combined bullets after the collision? OC (f) What is the phase of the combined bullets after the collision? mbullet ,solid = 9 mbullet,Iiquid = 9

Explanation / Answer

isolated system

Mass m1 = 15 g
Velocity v1 = 295 m/s

Mass m2 = 7.85 g
Velocity v2 = 375 m/s

Kinetic Energy of Bullet = 0.5 * m1 * v1^2 + 0.5 * m2 *v2^2
Kinetic Energy of Bullet = (0.5 * 15 * 295^2 + 0.5 * 7.85 *375^2)/1000 J
Kinetic Energy of Bullet = 1204.65 J

Initial Temperatue = 30 c
Final Temperature = 30 + 138.5 = 168.5 c

(a) Isolated System.
(b)
Using Momentum Conservation , m1 * v1 - m2 * v2 = (m1 + m2) *v
15 * 295 - 7.85 * 375 = (15 + 7.85) * v
v = 64.82 m/s
Speed of combined bullet after collision v = 64.82 m/s
(c)
Initial Kinetic Energy of Bullet = 0.5 * m1 * v1^2 + 0.5 * m2 *v2^2
Initial Kinetic Energy of Bullet = (0.5 * 15 * 295^2 + 0.5 * 7.85 *375^2)/1000 J
Initial Kinetic Energy of Bullet = 1204.65 J

Final Kinetic Energy = 0.5 * (m1+m2)*v^2
Final Kinetic Energy = 0.5 * 22.85 * 64.82^2 /1000
Final Kinetic Energy = 48 J

Kinetic Energy Transformed = 1204 - 48 = 1156 J

(d) No.

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