Need Help! Assignment.. Dont Know Where To Start

Question

A canon shoots a 2 kg cannonball from rest. In 0.7 s, the canon ball is shot out of the canon. If the canon’s output power is 571 W and its barrel has a length of 2.5 m, find the ball’s launch velocity and the launch force.

Answer 1

Hi Kaylyn A.,

That question is unfortunately phrased awkwardly. When the explosive ignites, it generates a volume of hot gas of great pressure. As the cannonball is accelerated down the barrel, that gas expands and cools adiabatically, hence provides less and less propulsive force as the ball changes position. So it is false to state a power of the cannon, as power is an instantaneous measure and this changes during the acceleration phase.

However, if you consider "average" power as work/time, then 399.7 kg m^2 s^-2 / 0.7 s is equivalent to 571 W power. That work of 399.7 kg m^2 s^-2 occurred over 2.5 m, so the "average" force was 399.7/2.5 = 159.88 kg m s^-2 . Since the mass was 2 kg and F=ma , then "average acceleration" a = F/m = 79.94 m s^-2 . You should be able calculate v = at ?

-- Cheers, -- Mr. d.