1000 x2 A small canon projects a 100-gram bullet via a chemical process that causes a gas to rapidly expand behind the bullet, propelling the bullet forward. Assuming that the force the gas exerts on the bullet can be described by the equation F = (with appropriate SI units): How much work will the gas do on the bullet in propelling it from/to the two points shown below? (15pts) b. Neglecting any sort of dissipative forces, what will be the speed of the bullet as it exits the canon? (10pts) a. X+, cm 0 2
The Correct Answer and Explanation is :
To determine the work done by the expanding gas on the bullet and its exit speed from the cannon, we need to analyze the problem using principles of physics.
a. Work Done by the Gas on the Bullet
The work WW done by a force FF over a displacement dd is given by:
W=F×dW = F \times d
In this scenario, the force exerted by the expanding gas on the bullet varies along the barrel’s length. To calculate the total work, we integrate the force over the displacement from the initial position x1x_1 to the final position x2x_2:
W=∫x1x2F(x) dxW = \int_{x_1}^{x_2} F(x) \, dx
The specific form of F(x)F(x) depends on the details provided in the problem, such as the pressure of the gas and the area of the barrel. Without the exact equation for F(x)F(x), we cannot compute the precise work done.
b. Speed of the Bullet as It Exits the Cannon
Neglecting dissipative forces like friction and air resistance, the work done by the gas translates into the kinetic energy of the bullet. The kinetic energy KEKE is related to the bullet’s mass mm and velocity vv by:
KE=12mv2KE = \frac{1}{2} m v^2
Equating the work done by the gas to the kinetic energy:
W=12mv2W = \frac{1}{2} m v^2
Solving for vv:
v=2Wmv = \sqrt{\frac{2W}{m}}
Given that the bullet’s mass mm is 100 grams (0.1 kg), we can calculate the exit speed once we know the work WW done by the gas.
Conclusion
To accurately determine the work done by the gas and the bullet’s exit speed, we need the specific equation for the force exerted by the gas on the bullet along the barrel’s length. This information is essential for performing the integration to find the total work and subsequently calculating the bullet’s velocity.