A toy dart gun has a spring with a spring constant of 315 N/m. An 12 gram dart is loaded in the gun and the spring is compressed a distance of 7.8 cm and then shot into the air. When the dart is 5.8 meters high in the air, what is its speed?



If the maximum height the dart can reach with the spring fully compressed is given by h, what fraction of this height will the dart reach when the spring is compressed by 1/2 of its maximum compression?

1. 6.78 m/s at 5.8 meters high

2. 1/4 Height

To find the speed of the dart when it is 5.8 meters high, we can use the principle of conservation of mechanical energy. The potential energy stored in the compressed spring is converted to kinetic energy when the dart is released and launched into the air. At the maximum height, all of this potential energy is converted back into potential energy, and at this point, the kinetic energy is zero.

Let's break down the problem step by step:

Step 1: Calculate the potential energy stored in the compressed spring.
To do this, we can use Hooke's Law, which states that F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the compression distance.

In this case, we need to convert the compression distance from centimeters to meters:
x = 7.8 cm = 0.078 m

The force applied by the spring is given by:
F = -kx = -315 N/m * 0.078 m = -24.57 N

The potential energy stored in the compressed spring is given by:
PE = 1/2 * k * x^2 = 1/2 * 315 N/m * (0.078 m)^2 = 1.1126 J

Step 2: Calculate the velocity of the dart at the maximum height.
At the maximum height, all of the potential energy stored in the spring is converted into gravitational potential energy.
The gravitational potential energy at height h is given by:
PE_grav = m * g * h

where m is the mass of the dart and g is the acceleration due to gravity. Since the kinetic energy is zero at the maximum height, the total mechanical energy (potential energy + kinetic energy) is also zero.

Using the equation PE + KE = 0, we can write:
PE_grav = -PE = m * g * h
m * g * h = 1.1126 J

We need to find the velocity, so we rearrange the equation to solve for it:
v = sqrt(2 * PE_grav / m)

v = sqrt(2 * 1.1126 J / 0.012 kg) = 25.09 m/s

So, the speed of the dart when it is 5.8 meters high in the air is approximately 25.09 m/s.

Now, let's move on to the second question:

The maximum height the dart can reach with the spring fully compressed is given by h. We need to determine what fraction of this height the dart reaches when the spring is compressed by 1/2 of its maximum compression.

Let's assume that the maximum compression of the spring is given by x_max. The fraction of the maximum height reached when the spring is compressed by 1/2 of its maximum compression will be equal to the fraction of the potential energy stored in the spring at this compression distance compared to the potential energy at maximum compression.

Using the formula for potential energy stored in the spring (PE = 1/2 * k * x^2), we can write:

PE_1/2 = 1/2 * k * (x_max/2)^2
PE_max = 1/2 * k * (x_max)^2

The fraction of the maximum height reached is then:
fraction = PE_1/2 / PE_max = (1/2 * k * (x_max/2)^2) / (1/2 * k * (x_max)^2)
fraction = (1/2 * (x_max/2)^2) / (1/2 * (x_max)^2) = (1/2 * 1/4) / (1/2) = 1/4

So, the fraction of the maximum height the dart reaches when the spring is compressed by 1/2 of its maximum compression is 1/4.