Asked by JasonJones 3 years ago

**1.** A revolutionary war cannon, with a mass of 2000 kg, fires a 15 kg ball horizontally. The cannonball has a speed of 120 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? The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same?

Conservation of

JasonJones

Answered by lancer 3 years ago

The law of conservation of momentum says that the total momentum of a closed system remains constant.

the equation of momentum is

$ p = mv $

and the conservation of momentum says that

$ (p_1+p_2+...)_i = (p_1+p_2+...)_f $

Given,

$ m $ = 15 kg

$ M $ = 2000 kg

$ v $ = 120 m/s

using the law of conservation of momentum, where the initial momentum are equal to the final momentum of the closed system:

$ 0=MV+mv \,\,\to\,\, V=\frac{m}{M} v = \frac{15\,kg}{2000\,kg}(120\,m/s) = 0.9 \, m/s $

$ m $ = 15 kg

$ M $ = 2000 kg

$ v $ = 120 m/s

also, let $ v \prime $ be the velocity of the cannon ball when the ball is fixed

In this case we will use the conservation of energy to find the solution. since the total energy of the system from problem 1 must be the combined energy of the cannonball of the cannon,

$ \frac{1}{2} mv \prime^2 = \frac{1}{2}(mv^2+MV^2) \,\,\to\,\, v\prime = v\sqrt{1+\frac{M}{m}} = (120)\sqrt{1+\frac{15}{2000}} = 120.5 $

and the difference in velocity = 120.5 m/s - 120 m/s = 0.5 m/s

Surround your text in `*italics*`

or `**bold**`

, to write a math equation use, for example, `$x^2+2x+1=0$`

or `$$\beta^2-1=0$$`

Stats

Views: 52

Asked: 3 years ago