11- Question 7 by Kambiz Ehsani - Machine Design - 4/26/2018 - Solved

Design the K factor of the old revolver so that the initial speed of the bullet before it exit the gun become Vb2.
The known values are as following:

1. The length of the spring at the normal situation is L1
2. The length of the spring at the normal situation is L2
3. The Weight of the hammer piece is Mh
4.  Bullet initial speed is zero or Vb1 = 0
5. Vb2 is the velocity of the bullet after the impact is known
6. At the time of impact the velocity of the hammer will be Vh1
7. Bullet mass is know and that is Mb

Solution: (Parametric)

Start from the conservation of energy for the spring and hammer system:

Potential Energy absorbed in the Spring before pulling the trigger = 0.5 * K* L1^2

Potential Energy absorbed in the Spring after the impact = 0.5 * K * L2^2

Kinetic Energy at the time of the impact = 0.5 * Mh * Vh1^2 

Energy Conservation => Total energy before pulling the trigger  = Total energy at the time of impact

0.5 * K* L1^2 = 0.5 * K * L2^2 + 0.5 * Mh * Vh1^2         Eq (1)

K = (Mh * Vh1 ^2) / (L2 ^2 - L1^2)                                    Eq (2)

Conservation of the momentum => 

1. Vb1 is the velocity of the bullet before the impact was zero 
2. Vb2 is the velocity of the bullet after the impact is known
3. Vh1 is unknown
4. Vh2 is assumed to be zero after the impact

Mh * Vh1 = Mb2 * Vb2                                                    Eq (3)

Rearranging :

Vh1 = Mb2 * Vb2 / Mh1                                                     Eq (4)

Substituting Vh1 from Eq (4) in Eq (2)

K = Mh * (Mb2 * Vb2 / Mh) ^ 2 / (L2^2 - L1^2)