Looking for Best Answer -- Physics Problem, please explain.?

Question

Your BB gun can shoot a BB 40 m horizontally when fired frm a height of 2 m. What is the spring constant of the srping in the BB gun? Use energy conservation methods.

Answer 1

Simple to solve. E initial = E final
E initial = KE spring + KE + PE
E initial = .5kx^2 + .5 mv^2 + mgh
Since you start at a hight of h of 2 meters, you got PE. Since you start at rest, v= 0 so KE = 0. And assuming the spring is compressed a distance x, then
E initial = .5 kx^2 + mgh

E final is at the ground level. At ground level h=0, you got a certain value v, x = is back to its resting position = no KE of spring.
E final = .5 mv^2

Answer 2

well, actually it depends on some more factores that you haven't menthioned them, like the distance that the Gun spring is pressed, and the mass of a bullet.
Suppose the dx is the distance that the spring is pressed from equilibrium state and m is bullet's mass.
According to conservation of mechanical energy:
0.5*m*v^2=k*dx (k: spring constant)
=> k=0.5*m*v^2/dx
It is important to know that the height which the gun is placed doesn't effect the energy of spring, so it doesn't effect the velocity of bullet.

Answer 3

Common sense would say it travels 40 m (horizontally) before the vertical pull of gravity pulls it to the ground.

Gravity pulls at approx 10m/s so it travelled for 1/5th of a second

One cannot tell you the force that was used to expel the bullet till we know the weight of the bullet though.

Answer 4

the optimal displacement takes position at the same time as it strikes to x=A. The initial position is x=0. Then it has to strikes the area a at the same time as the area it strikes in a era is 4A. So we've the Ta is: Ta= T/4 =.25*T=0.5 s