The cannon (Side view in Figure 2) has a mass of 2200 kg It uses 15 kg shells with an initial speed of 442 m/s at an angle of 35o relative to the horizontal. Two rigid beams (Trails), at the rear of the cannon prevent it to move backwards from its shooting position. The wheels are free to rotate. The end of each beam B is embedded in the ground making the recoil impossible. The dynamic friction coefficient between the cannon and the ground is 0.6. The height of the tube front end is at 1600 mm above ground. The wheel radius is 252 mm. Le recoil of the tube itself last 0.05 s.
1. The coordinates of the impulse reaction at each wheel A and at each beam end B.
2. The speed that the cannon would have if the trails were not embedded in the ground.
3. The distance that the cannon would move if the trails were not embedded in the ground.
4. The force pushing the shell at its start.
5. The maximum altitude that the shell will obtain by using the Energy Principle
6. The cannon range
the speed of the shell can be decomposed on horizontal and vertical axes ( x and y)
$V x =V*cos(alpha) =442*cos(35) =362.06 m/s$
$V y =V*sin(alpha) =442*sin(35) =253.52 m/s$
The momentum of the shell is ($m=15 kg$)
$P = m*v =m*V x*i + m*V y*j = 5430.9*i + 3802.8*j (kg*m/s)$
The reaction impulse is just this impulse (or momentum) taken with changed sign.
By taking the (0,0) of the axes at point B the coordinates of the point B are (0,0) and the coordinates of point A are (-2280,+252) mm. The reaction impulse is the same at both points A and B
2. On the vertical axis there is the reaction of the ground. The y momentum does not conserve.
On the horizontal axis the momentum stays the same (initial and final).
$P x i =P x f$
$5430.9 =M*V x f$
$V x f = 5430.9/2200 =2.469 m/s$
3. Friction force is
$Ff = mu*M*g = 0.6*2200*9.81 =12949.2 N$
The acceleration due to friction is
$a f = -Ff/M =-mu*g =-0.6*9.81 =-5.886 m/s^2$
The distance of recoil s comes from equation of speed
$0 = V x^2 + 2*a*s$
$s = – V x^2/(2*a) =2.469^2/2/5.886 =0.5177 m =0.518 m$
4. $F =P/time =m*v/t = 15*442/0.05 =132600 N$
P is the initial momentum of shell, t is time of shell recoil
5. $m V y^2/2 = m*g*h$
$h = Vy^2/2g =253.52^2/2/9.81 = 3275.9 m$
Or from the equation of motion
On the y axis the motion is uniform decelerated.
$0= V y^2 – 2*g*h$
$h = V y^2/2/g = 3275.9 m$
6. Range is
$D x = (V^2/g)*sin(2*35) =(442^2/9.81)*sin(70) =18713.8 m$