Question

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A bullet of mass m is fired into a block of mass M. The block, with the embedded bullet, then slides across a frictionless table and collides with a horizontal spring whose spring constant is k. The opposite end of the spring is anchored to a wall. The spring's maximum compression d is measured.

Find an expression for the bullet's initial speed vB in terms of m, M, k, and d.

What was the speed of a 2.0 g bullet if the block's mass is 1.2 kg and if the spring, with k = 33 N/m , was compressed by 12 cm ?

What percentage of the bullet's energy is "lost"?

Answer 1

here,

mass of bullet is m

mass of block is M

the spring constant is K

spring compression is d

let the velocity if block when bullet embedded into it be V

using conservation of momentum

m * vB = (m + M) * V

V = m * vB / (m + M) ......1

using conservation of energy

change in kinetic energy = change in spring energy

0.5 * m * v^2 = 0.5 * k * x^2

(m + M) * V^2 = k * d^2

putting value of V from equation 1

(m + M) * m * (vB/ (m + M))^2 = k * d^2

vB = sqrt( k * d^2 * (m + M) / m) ..........2

here,

m = 2.0 g

m = 0.002 kg

M = 1.2 kg

k = 33 N/m

d = 12 cm

d = 0.12 m

putting the values

vB = sqrt(33 * 0.12^2 * 1.202 / 0.002)

vB = 16.89 m/s

initial kinetic energy , KEi = 0.5 * m * vB^2

KEi = 0.5 * 0.002 * 16.89^2

KEi = 0.285 J

final kinetic energy , KEf = 0.5 * k * d^2

KEf = 0.5 * 33 * 0.12^2

KEf = 0.237 J

loss in kinetic energy = (KEi - KEf) * 100 / KEi

loss in kinetic energy = (0.285 - 0.237) * 100 / 0.285

loss in kinetic energy is 16.84 %