Don't stand directly behind a cannon

The canon is at a height of 4 metres and canonball projected goes for a maximum range of 500 metres therefore,

1.vertical velocity of canon ball =0 using,

s=ut+1/2gt^2 4=1/2 9.8 t^2 t=sqrt{8 /9.8}

after getting time ,we can easily get the horizontal velocity of canonball ie v*sqrt{8 /9.8}=500 v=500/sqrt{8 /9.8}..........(i)

2.now the second concept of conservation of momentum in horizontal direction as centre of mass was earlier at rest,v(cm)=0 then after projection, velocity of canon=u Given, Mass of canon =3000 kg Mass of canon ball=10 kg applying conservation of momentum,

3000 u=10 v u=v/300...............(ii)

3.Applying v^2=u^2+2as,for canon, final velocity of canon =0 initial velocity of cannon=v/300(v from (i)) s=1m

Hence we can easily find a ,a represents retardation given by the ropes on the canon, a=-1.7013 meter per second square

hence force= -ma

resistive force=3000*1.7013=5104.1667N

Let $v$ be the velocity of cannon and $v_b$ be the velocity of cannonball after the fire and $F$ be the force.

Using the principle of conservation of momentum: $0=3000v+10v_b$ .

$v=\frac{-10v_b}{3000}$ ....(1) (The minus sign indicates the opposite direction of velocities).

For the cannonball:

$x=500=v_bt$ and $y=4=0.5gt^2$ .

Solving, $v_b=553.398$ m/s.

Putting in $(1)$ :

$v=1.84466$ m/s.

Now the work done by the force to stop the cannon is equal to the change in kinetic energy (Neglecting friction)

$0.5mv^2=F\times 1$

Plugging in values $m=3000$ kg $v=1.84466$ m/s we get $\boxed{F= 5104.16}$ N.

$a(t) = [0, g]$

$v(t) = \int{a(t) dt} = [v0_x, v0_y+g*t]$

$p(t) = \int{v(t) dt} = [p0_x + t*v0_x, p0_y + t*v0_y - \frac{g}{2}*t^2]$

From the problem we know that $p0_x = 0$ (setting the ship as the origin) and $p0_y = 4$ , $v0_y = 0$ (velocity has only a horizontal component).

$p(t) = [t*v0_x, 4 - \frac{g}{2}*t^2]$

We also know that when $p_y=0$ , $p_x=500$ . Solving for v0_x:

$t0 = 2*sqrt(2*g)$

$t0*v0_x = 2*sqrt(2*g)*v0_x = 500$

$v0_x = \frac{250}{sqrt(2*g)}$

Now we know the kinetic energy of the ball (in J) as it is fired:

$E_b = \frac{1}{2}m_b*v0_x^2 = 15625*g*m_b$

Applying Newton's third law:

$m_c * v_c = -m_b * v_b$

$v_c = \frac{-m*b * v_b}{m_c}$

$E_c = \frac{1}{2}*m_c*v_c^2 = \frac{1}{2}*m_c* \frac{m_b^2 * v_b^2}{m_c^2}$

$E_c = \frac{E_b * m_b}{m_c} = \frac{15625*g*m_b^2}{m_c}$

From the work energy principle:

$W = F * d = \Delta E$

Since the cannon is coming to a complete stop, $\Delta E = E_c$ .

$F = \frac{W}{d} = \frac{E_c}{d} = \frac{15625*g*m_b^2}{m_c} \approx 5104 N$

Let, $M=3000 kg$ , $m=10 kg$ , $v_i$ =initial velocity of the cannon, $u_i$ =initial velocity of cannonball, Now, as the net force in the horizontal direction is zero so we can apply "Conservation of Linear Momemtum"

$M v_i=m u_i$

Now using the concept of projectile motion we can obtain the initial velocity of the cannonball,

$4=\frac{1}{2} *g*t^2$ and $500=v_i *t$

we get, $v_i=\frac{35}{3*\sqrt{40}}$

Now as the question uses the assumption that force becomes constant so we can apply equation of motion,

$(v_i)^2=(0^2)+2*a*(1))$

where, $a$ =acceleration of cannon

So, $a=\frac{25*1225}{6}$ then $F_{req.}=M*a \approx 5104$

The cannon ball will be following a ballistic trajectory, so the distance travelled by the missile, $d$ can be given by:

$d = \frac{v \cos \theta}{g}(v \sin \theta + \sqrt{(v \times \sin \theta)^2+2gy_0}$

We know that the angle of launch is parallel to the ground, so $\theta$ is therefore equal to 0:

$d = \frac{v}{g}\sqrt{2gy_0}$

which implies:

$v = \frac{dg}{\sqrt{2gy_0}}$

Plugging in the values we know: Distance Travelled: $d = 500m$

Acceleration due to gravity: $g = 9.8 ms^-1$

Initial Height: $y_0 = 4m$

As a result, we end up with:

$v = \frac{500\times9.8}{\sqrt{2\times9.8\times4}}$

$v = 553.399 ms^-1$

From this we can find the momentum of the cannonball, just as it leaves the cannon:

$p = 553.399\times10 = 5533.99 kg m s^-1$

As momentum is maintained before and after the launch, the cannon must now have:

$p = -5533.99 kg m s^-1$

From this, we can find the new velocity of the cannon:

$u = \frac{p}{m} = \frac{-5533.99}{3000}$

We also know that the cannon travels 1m backwards, before coming to a stop, so we know that over this metre, the cannon will be decelerating and, as $F = ma$ the rate of deceleration will determine the force that we want to find.

$v^2 = u^2 +2as$

$v = 0$ , $u = \frac{-5533.99}{3000}$ , $s = 1$

Therefore,

$a = 0.5 \times \frac{5533.99}{3000}^2 = 1.70 m s^-1$

$F = ma = 3000 \times 1.70 = 5104 N$ (0 decimal places)

When the cannon fires the cannonball, it gives the cannonball an initial velocity . Hence, by the law of conservation of momentum (1), the cannon will also receive an initial velocity in the opposite direction , making it recoil. When the cannon recoil, the braking force of the rope decelerates the cannon(2) to rest after travelling 1 m(3). Meanwhile, the cannonball travels 500 m horizontally and 4 m vertically downwards.

Workings:

M and m are the masses of the cannon and the cannonball, respectively. (known)

V and v are the velocities of the cannon and the cannonball, respectively.

S and s are the horizontal displacements of the cannon and the cannonball, respectively. (known)

F is the braking force the rope exerts on the cannon.

a is the deceleration rate of the cannon.

t is the time it takes from firing for the cannonball to hit the water.

h is the vertical displacement of the cannonball.

Equations for the events described above:

(1) MV = mv

(2) F = Ma

(3) V_{rest}^{2} - V^{2} = 2aS

V_{rest} = 0 , therefore:

- V^{2} = 2aS

(4) Vertically: h = \frac{1}{2} gt^{2} (a)

Horizontally: s = v \times t (b)

We want to find F in equation (2), for which we need the value a , which we can calculate from equation (3) if provided with V , which we can find out from equation (1) if v is known. To find v , we have to use the pair of equations (4a) and (4b). So let's start with (4a):

h = \frac{1}{2} gt^{2}

t = \sqrt{\frac{2h}{g}}

Substitute t into (4b):

s = v \times t

v = \frac{s}{t}

v = \frac{s}{\sqrt{\frac{2h}{g}}}

Substitute v into (1):

MV = mv

V = \frac{mv}{M}

V = \frac{m \times \frac{s}{\sqrt{\frac{2h}{g}}}}{M}

Substitute V into (3):

- V^{2} = 2aS

a = \frac{-V^{2}}{2S}

a = \frac{-(\frac{m \times \frac{s}{\sqrt{\frac{2h}{g}}}}{M})^{2}}{2S}

Then finally, substitute a into (2):

F = Ma

F = M \times \frac{-(\frac{m \times \frac{s}{\sqrt{\frac{2h}{g}}}}{M})^{2}}{2S}

`F = 3000 \times \frac{-(\frac{10 \times \frac{500}{\sqrt{\frac{2 \times -4}{-9.8}}}}{3000})^{2}}{2 \times 1}

F = -5104.17 N

Because we only want the magnitude of the force, we will take the value 5104 as the answer.