A 2.43-g lead bullet with an initial speed of 351 m/s strikes a wooden target an

Question

A 2.43-g lead bullet with an initial speed of 351 m/s strikes a wooden target and burrows to a stop.
    The bullet, which had an initial temperature of 40 °C, absorbs 75% of the heat generated by the collision.
    Lead has a melting point of 327°C, a specific heat of 128 J/kg·°C, and its latent heat of fusion is Lf = 24.5 kJ/kg.

    (a) How much heat did the bullet absorb?

Qbullet =  J

    (b) How much heat is required to warm the bullet to its melting point?

Qwarm =  J

    (c) How much of the lead bullet melted?

m =  g

    (d) How much work was done on the bullet?

Wnet =  J

    (e) What was the magnitude of the net impulse delivered to the bullet by the target?

Jnet =  kg·m/s

Explanation / Answer

(a)
Initial Velocity = 351 m/s
Final Velocity = 0

Heat Energy = Change in Kinetic Energy
Heat Energy = 1/2 *m (v1^2 - v2^2)
Heat Energy = 1/2 * 2.43/1000 * 351^2
Heat Energy = 149.7 J

Heat Absorbed by Bullet = 75 % of 149.7
Heat Absorbed by Bullet = 0.75 * 149.7 J
Heat Absorbed by Bullet = 112.3 J


b) How much heat is required to warm the bullet to its melting point?
q = m*c *T
q = 2.43/1000 * 128 * (327- 40)
q = 89.3 J
heat required to warm the bullet to its melting point, q = 89.3 J

(c)
How much of the lead bullet melted?
112.3 - 89.3 = m*Lf
m = 23/(24.5*10^3)
m = 0.000938 Kg
m = 0.94 g

Mass of Lead bullet melted, m = 0.94 g

(d)
Work done = Change in Kinetic Energy
Work done = 112.3 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.