Answer to Question #152388 in Physics for Amelia

Question

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)

Answer 1

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

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

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

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

In this time the stone covers the following horizontal distance:

$d = v_0t = 540m/s\cdot 4.5s\approx 2439m$

Here $v_0 = 540m/s$ is the initial horizontal velocity of the ball.

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

$p_{ball} = mv_0\\ p_{ball} = 5.44\cdot 540 = 2937.6\space kg\cdot m/s$

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

$p_{recoil} = p_{ball}\\ m_{recoil}v_{recoil} = p_{ball}\\$

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

$v_{recoil} = \dfrac{p_{ball}}{m_{recoil}}\\ v_{recoil} = \dfrac{2937.6}{5000} \approx 0.59m/s$

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