A bullet of a mass of $0.02 k g$ is fired from a rifle. The bullet takes 0.02 s to move through the barrel and leaves with a velocity of $200 m / s$ . What is the acceleration acting on the bullet?

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Solution

Step 1: Given data
The mass of the bullet is $m = 0.02 k g .$
The initial velocity of the bullet is zero.
The final velocity of the bullet is $200 m / s$ .
Time taken by the bullet is $0.02 s e c .$
Step 2: Formulae and concept
Acceleration is the time rate of change in velocity, i.e, $a = \frac{d v}{d t}$ , where, dv is the change in momentum and a is the acceleration.
From the formulae of kinematics, we know, $v = u + a t$ , where v and u are the initial and final velocity and a is the acceleration.
Step 3: Finding the acceleration
Let a be the acceleration of the bullet in the barrel.
Now, the acceleration of the bullet is,
$v = u + a t o r a = \frac{v - u}{t} = \frac{200 - 0}{0.02} = 10000 o r a = 10000 m / s^{2}$
The acceleration of the bullet is $10000 m / s^{2}$ .

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A bullet of a mass of 0.02kg is fired from a rifle. The bullet takes 0.02 s to move through the barrel and leaves with a velocity of 200m/s. What is the acceleration acting on the bullet?

A bullet of a mass of $0.02 k g$ is fired from a rifle. The bullet takes 0.02 s to move through the barrel and leaves with a velocity of $200 m / s$ . What is the acceleration acting on the bullet?