The spring in the muzzle of a child's spring gun has a spring constant of 780 N/m. To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 28° to the horizontal, a 59 g ball is shot to a maximum height of 1.74 m above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.
To solve this problem, we can use the principle of conservation of mechanical energy. The mechanical energy of the ball at the maximum height is equal to the mechanical energy of the spring when it is released.
Let's calculate the spring launch speed of the ball first.
(a) To find the speed at which the spring launches the ball, we need to calculate the potential energy and the kinetic energy of the ball at the maximum height.
The potential energy of the ball at the maximum height can be calculated using the equation:
PE = m * g * h
where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the maximum height.
PE = 0.059 kg * 9.8 m/s² * 1.74 m
PE = 0.098 J
The kinetic energy of the ball when it is launched can be calculated using the equation:
KE = 1/2 * m * v²
where KE is the kinetic energy, m is the mass of the ball, and v is the launch velocity.
We can equate the potential energy and the kinetic energy to find the launch velocity:
0.098 J = 1/2 * 0.059 kg * v²
Rearranging the equation, we get:
v² = (2 * 0.098 J) / 0.059 kg
v² = 3.322 m²/s²
Taking the square root of both sides, we find:
v = √(3.322 m²/s²)
v ≈ 1.82 m/s
Therefore, the spring launches the ball at a speed of approximately 1.82 m/s.
(b) To find the spring's initial compression distance, we can use the equation for the potential energy stored in the spring:
PE = 1/2 * k * x²
where PE is the potential energy, k is the spring constant, and x is the compression distance.
We can rearrange the equation to solve for the compression distance:
x = √(2 * PE / k)
Substituting the values into the equation, we get:
x = √(2 * 0.098 J / 780 N/m)
x ≈ 0.080 m
Therefore, the spring's initial compression distance is approximately 0.080 m.
To find the speed at which the spring launches the ball, we can use the principle of conservation of energy. When the ball is at its maximum height, its kinetic energy is zero, and its potential energy is at its maximum.
The potential energy at the maximum height is given by the formula:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the maximum height.
In this case, the maximum height is 1.74 m and the mass of the ball is 59 g, which can be converted to 0.059 kg.
PE = 0.059 kg * 9.8 m/s^2 * 1.74 m
PE = 0.100 J
Since the ball is not subject to air drag and no other external forces are acting on it, the potential energy at the maximum height is equal to the initial kinetic energy.
KE = 0.100 J
The kinetic energy can be expressed as:
KE = (1/2)mv^2
0.100 J = (1/2) * 0.059 kg * v^2
Rearranging the equation to solve for v:
v^2 = (2 * 0.100 J) / 0.059 kg
v^2 = 3.39 m^2/s^2
v ≈ 1.84 m/s
Therefore, the speed at which the spring launches the ball is approximately 1.84 m/s.
Now, let's determine the spring's initial compression distance.
To do this, we can use the potential energy stored in the spring when it is compressed.
The potential energy stored in a spring can be expressed as:
PE = (1/2)kx^2
where k is the spring constant and x is the displacement or compression distance of the spring.
We already know the spring constant, which is 780 N/m. We need to find the compression distance, x.
By equating the potential energy to the value we found earlier, 0.100 J, we can solve for x.
0.100 J = (1/2) * 780 N/m * x^2
Rearranging the equation to solve for x:
x^2 = (2 * 0.100 J) / (780 N/m)
x^2 = 0.0002564 m^2
x ≈ 0.016 m
Therefore, the spring's initial compression distance is approximately 0.016 m.
a. Y^2 = Yo^2 + 2g*h = 0.
Yo^2 - 19.6*1.74 = 0.
Yo^2 = 19.6*1.74 = 34.1.
Yo = 5.84 m/s = Ver. component of Vo.
Vo=Yo/sin28 = 5.84 / sin28=12.44 m/s.=
Initial velocity.
b. Wb = mg = 0.059kg * 9.8N/k=0.578 N =
Wt. of ball.
d = 0.578N / 780N/m = 7.4*10^-4 m.