A 20.0 gram bullet exits the rifle at a speed of 985 m/s. The gun barrel is 0.80 m long. What was the force(N) during the rocket’s launch? (Hint: note the units of the bullet!)
To find the force during the rocket's launch, we can use Newton's second law of motion, which states that force is equal to the rate of change of momentum. The formula is:
Force = (mass) x (acceleration)
To calculate the acceleration, we need to find the change in velocity. Given that the bullet exits the rifle at a speed of 985 m/s, we can assume that it starts from rest. Therefore, the change in velocity is equal to the final velocity.
Acceleration = (final velocity - initial velocity) / time
Since the bullet starts from rest, the initial velocity is zero. So, the acceleration simplifies to:
Acceleration = final velocity / time
To find the time, we can use the formula:
Distance = velocity x time
Since the barrel length is given as 0.80 m and the bullet travels that distance, we can rearrange the formula to solve for time:
Time = Distance / velocity
Now, we can plug in the values and calculate the time:
Time = 0.80 m / 985 m/s
Simplifying this calculation, we get:
Time = 0.000812 s
Now, we have all the required information to calculate the force:
Mass of the bullet = 20.0 g = 0.020 kg
Acceleration = (final velocity) / (time) = 985 m/s / 0.000812 s
Simplifying this calculation, we get:
Acceleration = 1,211,330.04 m/s^2
Finally, we can calculate the force:
Force = (mass) x (acceleration) = 0.020 kg x 1,211,330.04 m/s^2
Simplifying this calculation, we get:
Force = 24,226.60 N
Therefore, the force during the rocket's launch is approximately 24,226.60 N.
a = v^2/(2x)
F = ma