Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 1.60 m from the edge of the table, see the figure. Bobby compresses the spring 1.10 cm, but the center of the marble falls 25.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?
To solve this problem, we can use the principles of projectile motion. The motion of the marble fired from the spring-loaded gun can be considered as a projectile.
First, let's define some variables:
- Distance from the edge of the table to the target box: D = 1.60 m
- Initial compression of the spring by Bobby: h1 = 1.10 cm = 0.011 m
- Horizontal distance between the center of the marble and the center of the target box: x = 0.25 m
We are trying to find the compression of the spring needed for Rhoda to score a direct hit, which we'll call h2.
Now, let's analyze the situation. When the spring is compressed by h1, the marble is launched and follows a parabolic trajectory. It has a horizontal component of motion and a vertical component of motion.
The horizontal component of motion is constant and can be described by the equation:
x = (v0 * t)
where v0 is the initial horizontal velocity of the marble and t is the time of flight.
The vertical component of motion is affected by gravity and can be described by the equation:
y = (v0 * sin(θ) * t) - (0.5 * g * t^2)
where y is the vertical displacement, v0 is the initial vertical velocity of the marble, θ is the launch angle from the horizontal, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At the maximum height of the projectile, the vertical velocity becomes zero. We can use this information to find the time of flight (t) and the launch angle (θ).
To find the time of flight (t):
0 = (v0 * sin(θ) * t) - (0.5 * g * t^2)
Simplifying the equation:
t * (v0 * sin(θ) - (0.5 * g * t)) = 0
Since we are looking for a non-zero time of flight, we can set the expression in parentheses equal to zero:
v0 * sin(θ) - (0.5 * g * t) = 0
Rearranging the equation to solve for t:
t = (2 * v0 * sin(θ)) / g
Now, let's find the launch angle (θ):
To do this, we can use the fact that the horizontal distance between the center of the marble and the center of the target box (x) is equal to the horizontal component of motion (v0 * t).
x = (v0 * t)
Substituting the expression for t we derived earlier:
x = (v0 * ((2 * v0 * sin(θ)) / g))
Rearranging the equation to solve for v0:
v0 = sqrt((x * g) / (2 * sin(θ)))
Now, we have an expression for v0 in terms of x and θ. We can use this to find the v0 needed for Rhoda to score a direct hit, and then use that to find the compression of the spring (h2).
Since the horizontal component of motion is constant, the compression of the spring is directly proportional to the initial horizontal velocity.
Therefore, the compression of the spring needed for Rhoda to score a direct hit (h2) can be found using the following equation:
h2 = (v0_Rhoda / v0_Bobby) * h1
where v0_Rhoda is the initial horizontal velocity needed for Rhoda to score a direct hit and v0_Bobby is the initial horizontal velocity achieved by Bobby with the given spring compression h1.
To summarize, we need to calculate the initial horizontal velocity for Rhoda (v0_Rhoda) using the expression for v0 derived earlier, and then substitute into the equation for h2 to find the compression of the spring needed for Rhoda to score a direct hit.