The conservation of momentum is one of the basic principles we have learned in class so far. Here we shall see how it applies in a gun shot. Even though it is not a common everyday event, a gun shot can be used to analyse the application of conservation of momentum in our surrounding. An unfortunate manifestation of this concept is shown in the video below.

Here we see that the person held the gun very close to his face which resulted in him being hit in the face due to the recoil of the gun. This can be explained by the use of conservation of momentum as follows:

- Before the gun was shot, both the gun and the bullet inside are at rest so their momentum is 0.
- When the gun is shot the bullet that has a mass of 5 grams will attain a speed of 965 m/s. This will result on the movement of the rifle in the opposite direction as shown below:M1V1 = M2V2(0.005 Kg)*(965 m/s) = (3.5 Kg)*V2V2 = [(0.05 Kg)*(965 m/s)]
**/**(3.5 Kg)V2 = 13.78 m/s, towars the man's face - Assuming the guy held the gun 30 cm away from his face, we can get the acceleration of the gun as it reaches that speed while traveling the 30 cm distance to hit the guy in the face.(Vf)^2 = (Vi)^2 + 2*a*0.3m(13.78 m/s)^2 = (0)^2 + 2*a*0.3ma = [(13.78 m/s)^2]
**/**(0.6 m) = 316.74 m/s^2 - Therefore, the amount of force exerted on the mans face by the rifle of mass of 3.5 Kg can be calculated as follows:F = m*aF = (3.5 Kg)*(316.74 m/s^2)F = 1108.60 N

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