You have been asked to design a \"ballistic spring system\" to measure the speed

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.

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

2. What was the speed of a 1.8 g bullet if the block's mass is 1.5 kg and if the spring, with k = 48 N/m, was compressed by 17 cm ?

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

Explanation / Answer

1) When bullet hits the bock collision is perfectly inelastic hence we use law of conservation of energy

Pi = Pf

mvBi + Mvbi = (m + M)vf

Since vbi = 0 m/s

mvBi = (m + M)vf

vBi = [(m + M)vf ]/m-----------(1)

When combined mass hits the spring

SPE = KE of combined mass

1/2kd^2 = 1/2(m+M)vi^2

kd^2 = (m+M)vi^2

where vi= vf

kd^2 = (m+M)vf^2

vf = dsqrt[k/(m+M)] ----------------(2)

Plug (2) in (1),

vBi = [(m + M)*dsqrt[k/(m+M)]]/m

vBi = [dsqrt[k(m+M)]]/m

2. vBi = [dsqrt[k(m+M)]]/m

vBi = [0.17sqrt[48(0.0018+1.5)]]/0.0018 = 802 m/s

3.

KEi = 1/2mvBi ^2 = 1/2*0.0018*802^2 = 579 J

KEi = SPEf = 1/2kd^2 = 1/2*48*0.17^2 = 0.7 J

Energy lost = KEi – KEf = 579 – 0.7 = 578.3 J

Percentage = (lost/KEi)*100 = (578.3/579)*100 = 99.9 %

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