An 11g bullet is fired with a muzzle velocity of 679m/s. What is the kinetic energy of the bullet? If the bulle hits a target that is attached to a spring with a spring constant of 1500N/m how far will the spring compress when struck by the bullet?

Question

anonymous · Answer

m=11g = 0.011kg, v = 679 m/s,
KE =  (1/2)mv^2=(1/2)*0.011*(679)^2=2535.7255 Joules
PE of spring  = (1/2)kx^2, here k = spring constant, x = compression distance
Assuming conservation of energy and all KE of the bullet is converted to the potential energy of the spring when it hits the spring.
so KE of bullet=PE of spring
(1/2)mv^2=(1/2)kx^2
=>(1/2)kx^2=2535.7255
=>kx^2=5071.451
=>x^2=5071.451/1500=3.38096
=> x = square root of 3.38096=1.8387 meters = 183.87 cm