In a coilgun system, a magnetic field accelerates a 1-gram projectile over a 2-meter distance. Calculate the speed of the projectile at the end of the coilgun.
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To calculate the speed of the projectile at the end of the coilgun, we need to know the acceleration of the projectile and the time it takes to travel the 2-meter distance.
The first step is to calculate the acceleration. The formula for acceleration in a coilgun system is:
a = (v_f - v_i) / t
where:
- a is the acceleration
- v_f is the final velocity
- v_i is the initial velocity (which in this case is 0 as the projectile starts from rest)
- t is the time taken to travel the distance (which we need to calculate)
Since the projectile starts from rest, v_i = 0. So the formula can be simplified to:
a = v_f / t
The next step is to calculate the time it takes to travel the distance. We can use the formula:
d = v_i * t + (1/2) * a * t^2
where:
- d is the distance (2 meters in this case)
- v_i is the initial velocity (0 in this case)
- t is the time taken to travel the distance (which we need to calculate)
- a is the acceleration (which we also need to calculate)
Plugging in the values, we get:
2 = 0 * t + (1/2) * a * t^2
2 = (1/2) * a * t^2
4 = a * t^2
Now we have two equations with two unknowns:
a = v_f / t
4 = a * t^2
Rearranging the first equation, we get:
t = v_f / a
Substituting this into the second equation, we get:
4 = a * (v_f / a)^2
4 = v_f^2 / a
Rearranging, we get:
v_f^2 = 4 * a
Finally, substituting back into the first equation, we get:
t = v_f / a = v_f / (1/4 * v_f^2)
Simplifying, we get:
t = 4 / v_f
Now we have an equation for t that only depends on v_f. Plugging this into the first equation:
a = v_f / (4 / v_f)
a = v_f^2 / 4
Now we can solve for v_f. Rearranging the equation:
v_f^2 = 4 * a
v_f = sqrt(4 * a)
Plugging in the given mass of 1 gram, the acceleration due to gravity (9.8 m/s^2), and converting the mass to kilograms:
a = (0.001 kg) * (9.8 m/s^2)
a = 0.0098 m/s^2
Finally, substituting this into the equation:
v_f = sqrt(4 * 0.0098)
v_f = sqrt(0.0392)
v_f ≈ 0.198 m/s
Therefore, the speed of the projectile at the end of the coilgun is approximately 0.198 meters per second.
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