Question

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired? Answer in units of m/s.

Answer 1

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s