a cannon with mass of 2000kg fires a 10kg shell at an angle of 60 degrees above the horizontal. if the recoil velocity of the cannon across the level ground is 0.5 m/s what is the velocity of the cannon ball.
To find the velocity of the cannonball, we can use the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event.
Given:
Mass of the cannon (m1) = 2000 kg
Mass of the shell (m2) = 10 kg
Recoil velocity of the cannon (V1) = 0.5 m/s
Launch angle of the shell (θ) = 60 degrees
Let's assume the initial velocity of the cannonball is V2. To apply the conservation of momentum, we need to consider both vertical and horizontal components of momentum separately.
Vertical Component:
Since there is no vertical force acting on the cannonball after it is fired, the vertical component of momentum remains constant.
m2 * V2 * sin(θ) = 0
Horizontal Component:
Considering the horizontal component of momentum, we have:
(m1 + m2) * V1 = m2 * V2 * cos(θ)
Now, let's plug in the given values and solve for V2:
m2 * V2 * sin(60°) = 0 (since sin(60°) = √3/2)
10 kg * V2 * (√3/2) = 0
V2 = 0 m/s (since sin(60°) ≠ 0)
(m1 + m2) * V1 = m2 * V2 * cos(60°) (since cos(60°) = 1/2)
(2000 kg + 10 kg) * 0.5 m/s = 10 kg * V2 * (1/2)
2010 kg * 0.5 m/s = 5 kg * V2
V2 = (2010 kg * 0.5 m/s) / 5 kg
V2 = 201 m/s
Therefore, the velocity of the cannonball is approximately 201 m/s.
0=mvcosα -Mu
v=Mu/ mcosα=
=2000•0.5/10•cos60º= 200 m/s