A physics student wants to build a spring gun that when aimed horizontally will give a 35.0- g ball an initial acceleration of 'g' by releasing a spring that has been compressed 18.0 cm. What is the velocity of the ball as it emerges from the gun?
To find the velocity of the ball as it emerges from the gun, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring will be converted into the kinetic energy of the ball when it is released.
The potential energy stored in a spring is given by the equation:
PE = (1/2) k x^2
where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
In this case, the spring has been compressed by 18.0 cm, which is equal to 0.18 m. The ball has a mass of 35.0 g, which is equal to 0.035 kg. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Since we want the ball to have an initial acceleration of 'g' when released, we can equate the gravitational force acting on the ball to the force exerted by the compressed spring:
m * g = k * x
where m is the mass of the ball.
Solving for k, we get:
k = (m * g) / x
Substituting the known values, we have:
k = (0.035 kg * 9.8 m/s^2) / 0.18 m
Simplifying this expression, we find:
k = 1.91 N/m
Now that we know the spring constant, we can find the potential energy stored in the compressed spring:
PE = (1/2) * k * x^2
PE = (1/2) * 1.91 N/m * (0.18 m)^2
PE ≈ 0.0614 J
Next, we equate the potential energy to the kinetic energy of the ball at the instant of release:
PE = KE
0.0614 J = (1/2) * m * v^2
Solving for v (the velocity of the ball), we get:
v = sqrt((2 * PE) / m )
v = sqrt((2 * 0.0614 J) / 0.035 kg)
v ≈ 5.44 m/s
Therefore, the velocity of the ball as it emerges from the gun is approximately 5.44 m/s.