a cannon with mass of 2,000 kg fires a 10 kg shell with a velocity of 200 m/s at an angle of 60 degrees above the horizontal. What is the recoil velocity of the cannon across the level ground?
2000 u = 10 (200 cos 60 )
u = cos 60
To find the recoil velocity of the cannon, we can use the concept of conservation of momentum.
1. Firstly, we need to calculate the momentum of the shell. The momentum of an object is given by the product of its mass and velocity.
Momentum of the shell = mass of the shell × velocity of the shell
= 10 kg × 200 m/s
= 2000 kg·m/s (in the horizontal direction)
2. Since the cannon and the shell are initially at rest, the total momentum before the firing is zero. However, after the firing, the momentum of the shell is transferred to the cannon.
Momentum of the cannon + Momentum of the shell = 0
Momentum of the cannon = - Momentum of the shell
Therefore, the momentum of the cannon is equal in magnitude, but opposite in direction to the momentum of the shell.
3. We can calculate the recoil velocity of the cannon by using the formula for momentum:
Momentum = mass × velocity
Rearranging the formula to solve for velocity:
Velocity = Momentum / mass
Recoil velocity of the cannon = (momentum of the shell) / (mass of the cannon)
Recoil velocity of the cannon = (2000 kg·m/s) / (2000 kg)
= 1 m/s
Therefore, the recoil velocity of the cannon across the level ground is 1 m/s, in the opposite direction to the fired shell.
To find the recoil velocity of the cannon, we can use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. In a closed system where no external forces act, the total momentum before and after an event remains constant.
In this case, let's consider the cannon and the shell as a closed system.
The momentum of the cannon before firing can be found using the formula:
Momentum (cannon) = mass (cannon) x velocity (cannon)
Momentum (cannon) = 2000 kg x 0 m/s
= 0 kg⋅m/s
The momentum of the shell before firing can be determined by breaking it down into horizontal and vertical components:
Momentum (shell) = horizontal momentum (shell) + vertical momentum (shell)
The horizontal momentum of the shell before firing is given by:
Horizontal momentum (shell) = mass (shell) x velocity (shell) x cos(angle)
Horizontal momentum (shell) = 10 kg x 200 m/s x cos(60 degrees)
= 10 kg x 200 m/s x (1/2)
= 1000 kg⋅m/s
The vertical momentum of the shell before firing is:
Vertical momentum (shell) = mass (shell) x velocity (shell) x sin(angle)
Vertical momentum (shell) = 10 kg x 200 m/s x sin(60 degrees)
= 10 kg x 200 m/s x (√3/2)
= 1000 kg⋅m/s x (√3/2)
= 1000 kg⋅m/s⋅√3 / 2
≈ 866 kg⋅m/s
After firing the shell, the cannon and the shell move in opposite directions. Therefore, the momentum of the cannon after firing can be expressed as:
Momentum (cannon) = - mass (cannon) x velocity (recoil)
Since momentum is conserved, we have:
Momentum (cannon) + Momentum (shell) = 0
- mass (cannon) x velocity (recoil) + (Horizontal momentum (shell) + Vertical momentum (shell)) = 0
- 2000 kg x velocity (recoil) + 1000 kg⋅m/s + (866 kg⋅m/s) = 0
Simplifying the equation:
- 2000 kg x velocity (recoil) = - 1866 kg⋅m/s
Dividing both sides of the equation by -2000 kg:
velocity (recoil) = (-)1866 kg⋅m/s / (-)2000 kg
≈ 0.933 m/s
Therefore, the recoil velocity of the cannon across the level ground is approximately 0.933 m/s in the opposite direction of the shell's motion.