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1. some naval destroyer-class boats weight as much as 1500 tons(6.6 x 10^6 kg).

ID: 1285707 • Letter: 1

Question

1. some naval destroyer-class boats weight as much as 1500 tons(6.6 x 10^6 kg). suppose a 19 kg cannonball is launched at the ship, striking it directly as 250m/s.

a. If the cannonball embeds itself in the hull of the ship, what would the recoil velocity of the ship be? Assume that the boat is at rest for this one-dimensional collision,

b.What fraction of the system's kenetic energy is remaining after the collision?

c.Where did the lost kinetic energy go?

d. If the cannonball is stopped in the hull of the ship over a distance of 0.8m, what was the average force applied to the hull?

Explanation / Answer

a)

By conservationof momentum

m1 * v1 = m1 * v2

19 *250 = 6.6 * 10^ 6 * v2

V2 = 0.72 * 10^(-3) m/s =0.72 mm/s

b) Initial kinetic energy = 0.5 * 19 * 250*250 = 593750 J

Final kinetic energy = 0.5 * (0.72 * 10^(-3))^2 * (6.6*10^6 + 19)

= 1.71 J

c) The energy is lost in friction and sound.

d)

u = 250 m/s

v =0

s= 0.8m

we know

v^2 = u^2+ 2*a*s

substituting and solving we get

a= -39062.5 m/s^2 on the cannon

The force on the cannon = m*a = 19* 39062.5 = 742187.5 N =724.19kN

By newton's third law this is the force applied by cannon on the ship