A toy gun uses a spring with a force constant of 300. N/m to propel a 10.0-g steel ball. If the spring

is compressed 7.00 cm and friction is negligible: (a) How much force is needed to compress the
spring? (b) To what maximum height can the ball be shot? (c) At what angles above the horizontal
may a child aim to hit a target 3.00 m away at the same height as the gun? (d) What is the gun’s
maximum range on level ground?

(a) To find the force needed to compress the spring, we can use Hooke's Law, which states that the force required to compress or extend a spring is proportional to the displacement:

Force = -k * x

where k is the force constant of the spring and x is the displacement.

Given:
k = 300 N/m
x = 7.00 cm = 0.07 m (convert from cm to m)

Substituting these values into the formula, we get:

Force = -300 N/m * 0.07 m
= -21 N (negative sign indicates that the force is being applied in the opposite direction of compression)

Therefore, the force needed to compress the spring is 21 N.

(b) To find the maximum height the ball can be shot, we can use the concept of mechanical energy. When the spring is compressed, it stores potential energy, which will be completely converted into kinetic energy when the ball is shot.

Potential energy stored in the spring = (1/2) * k * x^2

Given:
k = 300 N/m
x = 0.07 m

Potential energy stored in the spring = (1/2) * 300 N/m * (0.07 m)^2
= 0.735 J (joules)

The kinetic energy of the ball will be equal to the potential energy stored in the spring, assuming negligible friction. The formula for kinetic energy is:

Kinetic energy = (1/2) * m * v^2

where m is the mass of the ball and v is its velocity.

Given:
m = 10.0 g = 0.01 kg (convert from grams to kilograms)

Using the equation for kinetic energy, we can solve for v:

0.735 J = (1/2) * 0.01 kg * v^2

Rearranging the equation, we get:

v^2 = (2 * 0.735 J) / 0.01 kg
v^2 = 147 J/kg
v = √(147 J/kg)
v ≈ 12.12 m/s

The maximum height the ball can reach can be found using the formula:

Maximum height = (v^2) / (2 * g)

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Maximum height = (12.12 m/s)^2 / (2 * 9.8 m/s^2)
= 7.366 m

Therefore, the ball can be shot to a maximum height of 7.366 m.

(c) To find the angles above the horizontal at which a child can aim to hit a target 3.00 m away at the same height as the gun, we can use projectile motion equations.

Let's assume the launch angle is θ and the initial velocity is v.

Since the range is given, we can use the formula for horizontal displacement:

Range = (v^2 * sin(2*θ)) / g

Given:
Range = 3.00 m
g = 9.8 m/s^2

Rearranging the equation, we get:

3.00 m = (12.12 m/s)^2 * sin(2*θ) / 9.8 m/s^2

Simplifying, we have:

sin(2*θ) = (3.00 m * 9.8 m/s^2) / (12.12 m/s)^2
sin(2*θ) ≈ 0.6316

Now we can solve for θ:

2*θ = arcsin(0.6316)
θ = (1/2) * arcsin(0.6316)

Using a calculator, we find:

θ ≈ 38.5°

Therefore, a child can aim at an angle of approximately 38.5° above the horizontal to hit a target at the same height 3.00 m away.

(d) The maximum range on level ground can be calculated using the formula:

Range_max = (v^2 * sin(2θ_max)) / g

where θ_max is the angle that gives the maximum range.

Given that the maximum range occurs when θ_max is 45°, we can substitute the values into the formula:

Range_max = (12.12 m/s)^2 * sin(2 * 45°) / 9.8 m/s^2
≈ 14.86 m

Therefore, the gun's maximum range on level ground is approximately 14.86 m.

To solve this problem, we can use the principles of energy conservation and projectile motion. Let's go step by step to find the answers to each of the parts:

(a) How much force is needed to compress the spring?
To find the force required to compress the spring, we can use Hooke's Law. Hooke's Law states that the force required to compress or extend a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be represented as F = kx, where F is the force, k is the force constant, and x is the displacement.
Given that the force constant of the spring (k) is 300 N/m and the spring is compressed by 7.00 cm (convert to meters: 7.00 cm = 0.07 m), we can substitute these values into the equation to find the force required:
F = kx
F = (300 N/m)(0.07 m)
F = 21 N
Therefore, the force required to compress the spring is 21 N.

(b) To what maximum height can the ball be shot?
To determine the maximum height the ball can reach, we need to consider the potential energy of the ball when it is shot. At the maximum height, all of the potential energy will be converted from the initial kinetic energy when the ball was shot.
The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the steel ball (10.0 g = 0.01 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height.
Since the potential energy is equal to the initial kinetic energy, we can equate these two:
PE = KE
mgh = (1/2)mv^2
where v is the velocity of the ball when it is shot.
We know from Hooke's Law that the force required to compress a spring (21 N) is equal to the work done on the spring. Therefore, the work done on the spring is equal to the potential energy stored in it before being released. The work done is given by W = (1/2)kx^2.
We can equate the potential energy to the work done on the spring since it is converted to kinetic energy as the spring is released:
PE = W
mgh = (1/2)kx^2
Substituting the given values, we have:
(0.01 kg)(9.8 m/s^2)h = (1/2)(300 N/m)(0.07 m)^2
h = (1/2)(300 N/m)(0.07 m)^2 / (0.01 kg)(9.8 m/s^2)
h ≈ 0.0893 m
Therefore, the maximum height the ball can reach is approximately 0.0893 meters.

(c) At what angles above the horizontal may a child aim to hit a target 3.00 m away at the same height as the gun?
To find the angles above the horizontal at which the child can aim to hit the target, we need to consider the projectile motion of the ball. The horizontal range (R) is the total horizontal distance the ball travels before hitting the target.
The horizontal range (R) is given by the equation:
R = (v^2 * sin(2θ)) / g
where θ is the launch angle and v is the velocity of the ball at launch.
To calculate the launch angle, we can use the conservation of energy again. The initial kinetic energy is equal to the potential energy when the ball is at its maximum height, so we can equate these two:
(1/2)mv^2 = mgh
v^2 = 2gh
v = sqrt(2gh)
Substituting this value of velocity into the horizontal range equation, we can solve for the launch angle:
R = (v^2 * sin(2θ)) / g
3.00 m = [sqrt(2gh)^2 * sin(2θ)] / g
3.00 m = 2h * sin(2θ)
sin(2θ) = (3.00 m) / (2h)
Now, we substitute the values of g (9.8 m/s^2) and h (0.0893 m) we calculated previously to solve for the launch angle:
sin(2θ) = (3.00 m) / (2 * 0.0893 m)
sin(2θ) ≈ 16.83
This value exceeds the maximum possible value of sin, which is 1. Therefore, there is no launch angle that will result in a horizontal range of 3.00 m.

(d) What is the gun's maximum range on level ground?
The maximum range on level ground occurs when the projectile is launched at an angle of 45 degrees. At this angle, the projectile achieves the maximum horizontal distance.
To find the maximum range (R), we use the equation:
R = (v^2 * sin(2θ)) / g
Substituting the value of velocity we calculated earlier (v = sqrt(2gh)) and the launch angle (θ = 45 degrees), we have:
R = (sqrt(2gh)^2 * sin(2 * 45 degrees)) / g
R = (2gh * sin(90 degrees)) / g
Since sin(90 degrees) = 1, we get:
R = 2h
R = 2 * 0.0893 m
R ≈ 0.1786 m
Therefore, the gun's maximum range on level ground is approximately 0.1786 meters.

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