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 = 2.15 m from the edge of the table, see the figure. Bobby compresses the spring 1.10 cm, but the center of the marble falls 23.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?

kx²/2=mv²/2

v=x•sqrt(k/m)
t=sqrt(2h/g)
s=v•t= x•sqrt(k/m) • sqrt(2h/g) =>
s1= x1•sqrt(k/m) • sqrt(2h/g)
s2= x2•sqrt(k/m) • sqrt(2h/g)
s1/s2=x1/x2
x2=x1•s2/s1 =1.1 (2.15/(2.15-0.23) =1.23 cm

Well, it sounds like Bobby's aim could use a little work! But don't worry, Rhoda is here to save the day.

Let's break down the problem. Bobby compressed the spring by 1.10 cm and missed the box by 23.0 cm. So, we can say that there is a proportional relationship between the compression of the spring and the distance the marble travels.

If we let x represent the compression of the spring that Rhoda needs to achieve a direct hit, we can set up the following proportion:

1.10 cm / 23.0 cm = x / D

Now, all we need to do is substitute the values into the equation and solve for x:

1.10 cm / 23.0 cm = x / 2.15 m

To make our lives easier, let's convert everything to the same unit of measurement. That way, we don't mix apples and oranges (or in this case, centimeters and meters).

1.10 cm = 0.0110 m
23.0 cm = 0.23 m

0.0110 m / 0.23 m = x / 2.15 m

Doing the math, we find that:

x = (0.0110 m)(2.15 m) / 0.23 m

x ≈ 0.1026 m

So, Rhoda should compress the spring by approximately 0.1026 meters to score a direct hit. And let's hope her aim is a little better than Bobby's!

To solve this problem, we need to use the principle of conservation of energy. The potential energy stored in the compressed spring is converted into the kinetic energy of the marble as it leaves the gun.

Let's start by calculating the potential energy stored in the spring when Bobby compresses it by 1.10 cm. The formula for the potential energy of a spring is given by:

Potential energy (U) = (1/2) * k * x^2

Where k is the spring constant and x is the displacement from the equilibrium position.

Given that Bobby compresses the spring by 1.10 cm (or 0.011 m), we can substitute these values into the formula:

Potential energy (U) = (1/2) * k * (0.011^2)

Now, let's find the speed of the marble when it leaves the gun. Using the principle of conservation of energy, we can equate the potential energy stored in the spring to the kinetic energy of the marble:

Potential energy (U) = Kinetic energy (K)

(1/2) * k * (0.011^2) = (1/2) * m * v^2

Where m is the mass of the marble and v is its final velocity.

Next, let's consider the horizontal motion of the marble. The horizontal distance traveled by the marble can be calculated using the formula:

Distance (D) = velocity (v) * time (t)

Since the vertical distance is given as 23.0 cm (or 0.23 m), we can derive the time taken by the marble to fall using the formula:

Vertical distance (h) = (1/2) * g * t^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2). By substituting the given values, we can solve for t.

Once we have the time, we can substitute it back into the equation for the horizontal distance to solve for the final velocity v.

Now, to find the distance Rhoda should compress the spring to score a direct hit, we will use the concept of projectile motion. The horizontal distance traveled by the marble is given by:

D = (v initial * t) + (1/2) * a * t^2

Where v initial is the initial horizontal velocity of the marble (which is zero), a is the horizontal acceleration (which is also 0), and t is the time taken by the marble to reach the target.

Since we know the initial horizontal velocity is zero and the acceleration is zero, the equation simplifies to:

D = 0 + 0 = 0

Therefore, to score a direct hit, Rhoda should compress the spring to a distance of zero.

I hope this helps! Let me know if you have any further questions.

To solve this problem, we need to use the principles of projectile motion. The horizontal distance D is given as 2.15 m, and the vertical displacement (height difference) is given as 23.0 cm.

Let's start by finding the initial velocity of the marble. We can use the vertical displacement and the acceleration due to gravity (9.8 m/s^2) to solve for the time it takes for the marble to reach the box.

The vertical displacement (height difference) is 23.0 cm, which is equivalent to 0.23 m. The equation to calculate the time of flight (T) is:

T = sqrt((2*h) / g)

Where h is the vertical displacement and g is the acceleration due to gravity. Plugging in the values, we get:

T = sqrt((2*0.23) / 9.8) = 0.482 seconds

Now that we have the time of flight, we can use it to find the initial horizontal velocity (Vx) of the marble. The equation to calculate the initial horizontal velocity is:

Vx = D / T

Plugging in the values, we get:

Vx = 2.15 / 0.482 = 4.459 m/s

Now, we need to find the compression (C) of the spring that Rhoda should use to score a direct hit. We can use the formula for the potential energy stored in the spring:

PE = (1/2) * k * C^2

Where PE is the potential energy, k is the spring constant, and C is the compression of the spring. Since the spring constant is not given, we can cancel it out by taking the ratio of the potential energies of both shots:

(1/2 * k * Cb^2) / (1/2 * k * Cr^2) = Mb / Mr

Where Cb is the compression used by Bobby, Cr is the compression used by Rhoda, Mb is the mass of the marble, and Mr is the mass of Rhoda's marble (assuming both marbles have the same mass).

Since the masses are the same, we can simplify the equation to:

Cb^2 / Cr^2 = 4.459^2 / Vrx^2

Where Vrx is the initial horizontal velocity of the marble shot by Rhoda.

We know that Cb is 1.10 cm since that is the given compression used by Bobby. Rearranging the equation, we can solve for Cr:

Cr = Cb * Vrx / 4.459

Plugging in the values, we get:

Cr = 0.011 m * 4.459 / 4.459 = 0.011 m

Therefore, Rhoda should compress the spring by 0.011 m to score a direct hit on the target box.