A rifle shoots a 4.1g bullet out of it's barrel. The bullet has a muzzle velocity of 975 m/s just as it leaves the barrel.
Assuming a constant horizontal acceleration over a distance of 43.0 cm starting from rest, with no friction between the bullet and the barrel, what force does the rifle exert on the bullet while it is in the barrel?
How many g's of acceleration does the rifle give this bullet?
For how long is the bullet in the barrel?
dontknoq
To determine the force exerted by the rifle on the bullet, we can use Newton's second law of motion:
Force = Mass × Acceleration
The mass of the bullet is given as 4.1 grams, which equals 0.0041 kilograms (since 1 gram = 0.001 kilograms). The acceleration can be calculated using the following equation:
Acceleration = (Final Velocity - Initial Velocity) / Time
Given that the muzzle velocity is 975 m/s, the initial velocity is 0 m/s (since the bullet starts from rest), and the distance traveled is 43.0 cm (or 0.43 meters), we can rearrange the equation as follows:
Acceleration = (975 m/s - 0 m/s) / Time
To find the time, we can use the equation of motion:
Distance = (Initial Velocity × Time) + (0.5 × Acceleration × Time^2)
Plugging in the values:
0.43 m = 0 + (0.5 × Acceleration × Time^2)
Simplifying the equation:
0.43 m = 0.5 × Acceleration × Time^2
Solving for Time:
Time = sqrt((2 × Distance) / Acceleration)
Plugging in the given distance and the calculated acceleration:
Time = sqrt((2 × 0.43 m) / Acceleration)
Now we can substitute this time value back into the acceleration equation to find the force:
Force = Mass × Acceleration
Let's calculate the acceleration first:
Acceleration = (975 m/s - 0 m/s) / Time
Now we can substitute the acceleration back into the force equation:
Force = Mass × Acceleration
Lastly, to calculate the acceleration in terms of g's, we can use the relationship:
1 g = 9.8 m/s^2
To get acceleration in g's, divide the acceleration by 9.8.
Now, let's solve for each variable step by step.
To find the force exerted by the rifle on the bullet, we can use Newton's second law of motion, which states that the force applied to an object is equal to the object's mass multiplied by its acceleration.
1. First, let's calculate the acceleration of the bullet. We can use the formula:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Rearranging the formula, we have:
a = (v^2 - u^2) / (2s)
Plugging in the given values:
a = (975^2 - 0^2) / (2 * 0.43)
a ≈ 1057627 m/s^2
2. Now that we have the acceleration, we can find the force exerted by the rifle:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
Plugging in the given values:
F = 0.0041 kg * 1057627 m/s^2
F ≈ 4342.1 N
Therefore, the rifle exerts a force of approximately 4342.1 Newtons on the bullet while it is in the barrel.
To calculate the acceleration in g's, we need to convert the acceleration from meters per second squared to g's. One g is equal to the acceleration due to gravity, approximately 9.8 m/s^2.
3. Converting acceleration to g's can be done by dividing the acceleration by the acceleration due to gravity:
g = a / 9.8
Plugging in the calculated acceleration:
g ≈ 1057627 m/s^2 / 9.8 m/s^2
g ≈ 107936.8 g's
Therefore, the rifle imparts an acceleration of approximately 107936.8 g's to the bullet.
To find how long the bullet is in the barrel, we'll use one of the kinematic equations.
4. We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Rearranging the formula, we have:
t = (v - u) / a
Plugging in the given values:
t = (975 m/s - 0 m/s) / 1057627 m/s^2
t ≈ 0.00092 seconds
Therefore, the bullet is in the barrel for approximately 0.00092 seconds.