a 0.22 calibre rifle shoots a bullet of mass 1.8 g with a muzzle velocity of 500 m/s. if the barrel is 25 cm long, what is the average force exerted on the bullet while it is in the barrel?
The Correct Answer and Explanation is :
To calculate the average force exerted on the bullet while it is in the barrel of the rifle, we can use the principles of physics, specifically Newton’s second law and the work-energy principle.
Given Data:
- Mass of the bullet (m) = 1.8 g = 0.0018 kg (since 1 g = 0.001 kg)
- Muzzle velocity (v) = 500 m/s
- Length of the barrel (d) = 25 cm = 0.25 m (since 1 cm = 0.01 m)
Steps to Calculate Average Force:
- Calculate the Acceleration (a):
Using the kinematic equation:
[
v^2 = u^2 + 2ad
]
where:
- (v) = final velocity (500 m/s)
- (u) = initial velocity (0 m/s, since the bullet starts from rest)
- (a) = acceleration
- (d) = distance (0.25 m) Plugging in the values:
[
(500)^2 = (0)^2 + 2a(0.25)
]
[
250000 = 0.5a \implies a = \frac{250000}{0.5} = 500000 \text{ m/s}^2
]
- Calculate the Average Force (F):
Using Newton’s second law:
[
F = ma
]
where:
- (F) = force
- (m) = mass of the bullet (0.0018 kg)
- (a) = acceleration (500000 m/s²) Plugging in the values:
[
F = 0.0018 \times 500000 = 900 \text{ N}
]
Conclusion:
The average force exerted on the bullet while it is in the barrel is 900 N.
Explanation:
The force exerted on the bullet is a result of the rapid acceleration it undergoes as it travels through the barrel. When the trigger is pulled, the gunpowder ignites, creating gas that pushes the bullet forward. The bullet starts from rest and reaches a high speed of 500 m/s in a short distance (25 cm), resulting in a very high acceleration. According to Newton’s second law, the force can be calculated by multiplying the bullet’s mass by its acceleration. The calculated force of 900 N illustrates the significant energy transfer involved in firing a bullet, which is essential for the bullet’s high velocity and effectiveness.