Question
Solve:
During the American Civil War, cannons and mortars were often mounted on railroad flatcars (see a photo here). Usually, the wheels of the flatcar were chocked before the mortars were fired (i.e. the wheel brakes were effectively locked), but we will investigate firing a mortar from a flatcar that is free to roll on level tracks. Assume that the car's horizontal motion on the tracks will effectively be entirely free of friction. Assume a mortar of mass 127 kg is mounted on a short railroad flatcar of mass 545 kg (our mortar is about one tenth the mass of the mortar in the photograph). The shell to be fired by the mortar has a mass of 9.3 kg. The railroad tracks are level and run North and South. When the mortar is fired, the flatcar has been disconnected from the rest of the train and is free to roll on the tracks with negligible friction. The mortar is elevated to an angle of 40 degrees, so as to lob shells at an enemy to the South of the train's position. The shell speed, immediately upon exiting the barrel of the mortar, is 111 m/s. What is the horizontal component of the momentum of the flatcar-mortar-shell system AFTER the mortar is fired? What is the size of the horizontal component of the velocity of the center-of-mass of the flatcar-mortar-shell system AFTER the mortar is fired? What is the horizontal component of the momentum of the shell immediately after being fired? What is the horizontal component of the momentum of the mortar and flatcar immediately after the shell has been fired? What is the velocity of the mortar and flatcar immediately after the shell has been fired?
Explanation / Answer
Mass of mortar = 127 kg
Mass of railroad car = 545 kg
Total mass = 545 + 127 = 672 kg
Mass of shell = 9.3 kg
(e)Horizontal momentum of flatcar-mortar-shell system = 0
(f) Velocity of center of mass of flatcarmortarshell system after firing = 0
(g) Horizontal component of momentum of shell = 9.3*111*cos(40) = 791kgm/s south
(h) Horizontal component of momentum of flatcar-mortar = Horizontal component of momentum of shell = 9.3*111*cos(40) = 791kgm/s north
(i)Horizontal component of velocity = horizontal momentum(from part (h))/total mass = 791/672 = 1.18 m/s north
For part of (j) you need to know the time of explosion.
