Question

You have an internship working for the National Park Service. They start you off in the laboratory which tests possible new equipment. Your first job is to test a small cannon used to prevent avalanches in populated areas by shooting down heavy snow concentrations. You want to find out the velocity of the projectile as it leaves the cannon (muzzle speed) so that the range and trajectory of the cannon can be determined. The cannon you are testing weighs 260 kg and shoots a 15 kg projectile. During the lab tests the cannon is held horizontal in a rigid support so that it cannot move. Under those conditions, you use video analysis of the projectile to determine that the muzzle speed is 300 m/s. When fired in the field, the cannon is not rigidly attached to the ground, but it is free to move (recoil) when it is fired. There are ropes attached to anchors so that the cannon can only recoil 0.5 m. You must predict the projectile's muzzle speed under field conditions, when it recoils. The cannon is fired horizontally using cannon shells which are identical (same mass, same amount of gunpowder, etc.) to those in the laboratory. What is the projectile's speed in the field? Please explain how you arrived at the answer.

Answer 1

First we consider the condition when canon can not move.
In this case the energy of the explosion
= Kinetic energy of the canon+ kinetic energy of the projectile
since canon can not move therefore it's kinetic energy must be zero.
= 0 + (1/2)m_P*V_p² = (1/2)*15*300² = 675000 J
Where m_P is the mass of projectile = 15 kg
V_P is the velocity of projectile = 300 m/s
Now we will take the condition in field.
Kinetic energy after the explosion
=  Kinetic energy of the canon+ kinetic energy of the projectile
= (1/2)m_cV_c² + (1/2)m_PV_P²
Where m_c is mass of canon = 260 kg
V_C is velocity of canon
And this must be equal to the energy calculated initially
(1/2)m_cV_c² + (1/2)m_PV_P² = 675000 -------------(1)
Now applying conservation of momentum
m_cV_c + m_PV_P =0
V_C = (m_P/m_c)V_P = (15/260)V_P = 0.058V_P
Now putting the value in equation 1
(1/2)*260*(0.058V_P)² +(1/2)*15*V_P² = 675000
V_P = 291.618 m/s