Question #152388

A cannon from the Napoleonic wars was able to launch cannon balls at a velocity of 540 m s–1. The cannon is fired at the top of a cliff with the barrel in the horizontal position.
(a) Neglecting air resistance, what is the range of the cannon ball, to the point where it hits the water 100 m below?
(4 marks)
(b) Calculate the initial horizontal component of the momentum of the cannon ball, if it has a mass of 5.44 kg. (2 marks)
(c) If the cannon has a mass of 5000 kg and is on a trolley on the horizontal surface, what is its initial recoil velocity?
(2 marks)

Expert's answer

(a)Time of falling tt can be found from the kinematic equation (if there is no vertical initial velocity component):



h=gt22h = \dfrac{gt^2}{2}

where h=100mh = 100m is the height, and g=9.8m/s2g = 9.8m/s^2 is the gravitational acceleration. Thus, obtain:


t=2hg=21009.84.5st = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2\cdot 100}{9.8}} \approx 4.5s

In this time the stone covers the following horizontal distance:


d=v0t=540m/s4.5s2439md = v_0t = 540m/s\cdot 4.5s\approx 2439m

Here v0=540m/sv_0 = 540m/s is the initial horizontal velocity of the ball.


(b) The initial horizontal component of the momentum of the cannon ball of mass m=5.44kgm = 5.44kg is given by the expression:


pball=mv0pball=5.44540=2937.6 kgm/sp_{ball} = mv_0\\ p_{ball} = 5.44\cdot 540 = 2937.6\space kg\cdot m/s



(c) The initial recoil momentum is equal to the initial horizontal component of the momentum of the cannon ball (according to the momentum conservation):


precoil=pballmrecoilvrecoil=pballp_{recoil} = p_{ball}\\ m_{recoil}v_{recoil} = p_{ball}\\

where mrecoil=5000kgm_{recoil} = 5000kg is the mass of the cannon. Then the velocity of the recoil is:


vrecoil=pballmrecoilvrecoil=2937.650000.59m/sv_{recoil} = \dfrac{p_{ball}}{m_{recoil}}\\ v_{recoil} = \dfrac{2937.6}{5000} \approx 0.59m/s

Answer. (a) 2439 m, (b) 2937.6 kg*m/s, (c) 0.59 m/s.


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