A cannon on a railroad car is facing in a direction parallel to the tracks (the figure below ). It fires a 99-kg shell at a speed of 105 m/s (relative to the ground) at an angle of 60 above the horizontal. If the cannon plus car have a mass of 4.6 104 kg, what is the recoil speed of the car if it was at rest before the cannon was fired?
change athe speed of the shell to horizontal speed, then use the conservation of momentum principle.
To find the recoil speed of the car, we need to use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. According to the conservation of momentum, the total momentum of a system before an event is equal to the total momentum after the event, provided there are no external forces.
In this case, the system consists of the cannon, the shell, and the railroad car. Initially, the car is at rest, so its momentum is zero. The final momentum of the system will be the sum of the momentum of the car and the momentum of the shell.
Let's break down the problem into components. The horizontal component of the velocity of the shell doesn't contribute to the horizontal momentum of the system because it is parallel to the tracks. However, the horizontal momentum of the cannon plus car must be equal in magnitude and opposite in direction to the horizontal momentum of the shell in order to satisfy conservation of momentum.
The horizontal momentum of the shell can be calculated using trigonometry. The horizontal component of the shell's velocity is given by Vx = V * cos(θ), where V is the shell's velocity and θ is the angle above the horizontal.
Vx = 105 m/s * cos(60°) = 52.5 m/s
Now, we can calculate the momentum of the shell in the horizontal direction. Momentum is given by P = m * v, where m is the mass and v is the velocity. The mass of the shell is 99 kg.
P(shell) = m * Vx = 99 kg * 52.5 m/s = 5197.5 kg*m/s
According to the conservation of momentum, the initial momentum of the cannon plus car must be equal and opposite to the momentum of the shell in the horizontal direction. Since the car is initially at rest, the initial momentum is zero. Therefore,
P(cannon+car) = -P(shell)
Let's denote the recoil velocity of the car as V(car). The initial momentum of the cannon plus car is given by the product of the mass of the cannon plus car and the recoil velocity.
P(cannon+car) = (mass of cannon+car) * V(car)
Equating the two momenta and rearranging the equation, we can solve for V(car):
(mass of cannon+car) * V(car) = -P(shell)
V(car) = -P(shell) / (mass of cannon+car)
Substituting the known values:
V(car) = -5197.5 kg*m/s / 4.6 * 10^4 kg = -0.113 m/s
The negative sign indicates that the car recoils in the opposite direction to the motion of the shell. Therefore, the recoil speed of the car is approximately 0.113 m/s.