Final answer:
Using the conservation of energy and the formula for potential energy stored in the spring, the speed of the ball at release from the projectile launcher is calculated to be 2.74 meters per second.
Step-by-step explanation:
The subject of this question is specifically Physics within the category of mechanics. To find the speed at which the ball leaves the projectile launcher, we need to use the conservation of energy principle because the stored potential energy in the spring is converted into kinetic energy of the ball and launcher holder.
Firstly, the potential energy stored in the spring when compressed (Us) is given by the formula:
Us = (1/2)kx²
Where k is the spring constant and x is the compression distance. Here, k = 500 N/m and x = 0.0375 m (converting cm to m by multiplying by 0.01).
Substituting these values into the formula, we find:
Us = (1/2)(500 N/m)(0.0375 m)²
Us = 0.35156 Joules
The total mass of the ball and holder is the sum of their individual masses, which is 10 g + 75 g = 85 g = 0.085 kg (conversion from grams to kilograms by multiplying by 0.001).
Using the conservation of energy, the potential energy stored in the spring will be equal to the kinetic energy (K) of the ball and holder at the moment the spring is uncompressed, which can be written as:
K = (1/2)mv²
Setting the kinetic energy equal to the stored potential energy and solving for v, we have:
0.35156 Joules = (1/2)(0.085 kg)v²
After rearranging and solving for v, the result is:
v = 2.74 m/s
The ball would be moving at 2.74 meters per second at the instant it leaves the projectile launcher.