A spring-loaded toy gun is used to shoot a ball of mass m=1.5 kg straight up in the air. The spring has spring constant k=667 N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position and then released, the ball reaches a maximum height (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.
Find the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y=0).
Ui=mgh+.5kh^2
ui=(1.5)(9.81)(-.25)+(.5(667)(-.25)^2)
ki=0 Ui=17.165 Kf=.5mv^2
ki+Ui=.5mv^2 solve for v
v=4.78 m/s
Why did the ball go to therapy?
Because it was under a lot of spring tension!
To find the muzzle velocity of the ball, we can use conservation of energy. The total mechanical energy of the ball-spring system is conserved, assuming no energy loss due to air resistance or other dissipative forces. At the equilibrium position, all the potential energy stored in the compressed spring is converted into kinetic energy of the ball. Therefore, the initial potential energy of the spring is equal to the final kinetic energy of the ball at the spring's equilibrium position.
The potential energy stored in a spring can be calculated using the formula:
Potential energy = (1/2) * k * x^2,
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
In this case, the spring is compressed a distance of 25.0 centimeters, which is equivalent to 0.25 meters.
Potential energy = (1/2) * 667 N/m * (0.25 m)^2
= 20.875 J
At the spring's equilibrium position, this potential energy is completely converted into kinetic energy:
Kinetic energy = (1/2) * m * v^2,
where m is the mass of the ball and v is its velocity.
By equating the calculated potential energy to the kinetic energy, we can solve for the muzzle velocity:
20.875 J = (1/2) * 1.5 kg * v^2
Simplifying the equation:
v^2 = (2 * 20.875 J) / 1.5 kg
v^2 = 27.8333 m^2/s^2
Taking the square root of both sides, we find:
v = √(27.8333 m^2/s^2)
v ≈ 5.279 m/s
Therefore, the muzzle velocity of the ball is approximately 5.279 m/s.
To find the muzzle velocity of the ball, we need to apply the principle of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant if there are no external forces acting on it.
In this case, the mechanical energy is given by the sum of the potential energy and kinetic energy of the ball:
E = PE + KE
At the maximum height, the ball is momentarily at rest, so its kinetic energy is zero. Therefore, the total mechanical energy is equal to the potential energy.
At the equilibrium position of the spring (y = 0), the potential energy is given by the formula:
PE = (1/2)kx^2
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
In this case, the displacement (x) is given as 25.0 centimeters, which is equivalent to 0.25 meters.
So, the potential energy at the equilibrium position is:
PE = (1/2)(667 N/m)(0.25 m)^2 = 20.875 J
Since the mechanical energy is conserved, the potential energy at the equilibrium position is equal to the initial kinetic energy of the ball when it is shot out of the gun.
At the equilibrium position, the potential energy is entirely converted into kinetic energy, so:
KE = PE = 20.875 J
The kinetic energy is given by the formula:
KE = (1/2)mv^2
where m is the mass of the ball and v is its velocity.
Rearranging the equation, we can solve for the velocity:
v^2 = 2KE / m
v^2 = (2 * 20.875 J) / 1.5 kg
v^2 = 27.8333333 m^2/s^2
Taking the square root of both sides, we find:
v = √(27.8333333 m^2/s^2) ≈ 5.28 m/s
Therefore, the muzzle velocity of the ball is approximately 5.28 m/s.
Energy in spring= 1/2 k .25^2
gravitational energy gained by ball going to equilibrium position 1.5*g*.25
KE at equilibrium= 1/2k .25^2-1.5*g*.25
and of course, KE is 1/2 1.5*v^2
solve for v.