Question

A toy gun uses a spring with a force constant of 285 N/m to propel a 9.5-g steel ball. Assuming the spring is compressed 7.6 cm and friction is negligible, answer the following questions: (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?

Answer 1

(a) The force needed to compress the spring is 21.66 N. (b) The maximum height the ball can be shot is 8.16 m. (c) The child can aim at an angle of 17.8 degrees above the horizontal to hit the target. (d) The gun's maximum range on level ground is 3.62 m.

Step-by-step explanation:

(a) To calculate the force needed to compress the spring, we can use Hooke's Law, which states that force is equal to the product of the force constant and the displacement of the spring. F = kx, where F is the force, k is the force constant, and x is the displacement. Substituting the given values, we have F = (285 N/m)(0.076 m) = 21.66 N.

(b) The maximum height the ball can be shot can be calculated using the principle of conservation of mechanical energy. The potential energy at the maximum height is equal to the initial potential energy stored in the compressed spring. The potential energy can be calculated using the equation PE = 1/2kx^2, where PE is the potential energy, k is the force constant, and x is the displacement. Substituting the given values, we have PE = (1/2)(285 N/m)(0.076 m)^2 = 0.776 J. Since potential energy is equal to mgh, where m is the mass, g is the acceleration due to gravity, and h is the height, we can rearrange the equation to solve for h. h = PE/(mg) = (0.776 J)/(0.0095 kg)(9.8 m/s^2) = 8.16 m.

(c) To hit a target 3.00 m away at the same height as the gun, we can use projectile motion equations. The horizontal distance traveled by the projectile is given by x = v_initial*cos(theta)*t, where v_initial is the initial velocity, theta is the angle of projection, and t is the time of flight. The vertical distance traveled by the projectile is given by y = v_initial*sin(theta)*t - (1/2)g*t^2, where g is the acceleration due to gravity. Since the target is at the same height as the gun, the vertical distance traveled is zero, and we can solve for t. 0 = v_initial*sin(theta)*t - (1/2)g*t^2. Solving for t, we get t = 2*v_initial*sin(theta)/g. Substituting the values, we have 3 = v_initial*cos(theta)*[2*v_initial*sin(theta)/g]. Simplifying the equation, we get 3 = 2*v_initial^2*sin(theta)*cos(theta)/g. Rearranging the equation, we have v_initial^2*sin(2*theta) = 3g. Therefore, sin(2*theta) = 3g/v_initial^2. Given that v_initial = sqrt(2*g*h), we can substitute the value of h from part (b) to get v_initial = sqrt(2*9.8 m/s^2*8.16 m) = 9.50 m/s. Substituting the values, we have sin(2*theta) = 3(9.8 m/s^2)/(9.50 m/s)^2. Solving for theta, we get theta = (1/2)*sin^(-1)(3(9.8 m/s^2)/(9.50 m/s)^2) = 17.8 degrees.

(d) The gun's maximum range on level ground can be calculated using the equation R = (v_initial^2*sin(2*theta))/g, where R is the range. Substituting the values, we have R = (9.50 m/s)^2*sin(2*17.8 degrees)/(9.8 m/s^2) = 3.62 m.

Answer 2

a) F = 21.7 N , b) h = 8.84 m , c) θ = 58.5º d) R = 15.75 m

Step-by-step explanation:

a) Hook's law is

x = 7.6 cm (1m / 100cm) = 0.076 m

m = 9.5 g (1 kg / 1000g) = 9.5 10-3 kg

F = K x

F = 285 0.076

F = 21.7 N

b) We use energy conservation

Eo = Ke = ½ kx²

Ef = U = mg h

Eo = Ef

½ k x² = mg h

h = ½ k x² / mg

h = ½ 285 0.076² / 9.5 10⁻³ 9.8

h = 8.84 m

c) Let's calculate the speed with which it leaves the gun

Ke = K

½ k x² = ½ m v²

v = √ (k/m x²)

v = √( 285 / 9.5 10-3 0.076²)

v = 13.16 m / s

v₀ = v = 13.16 m/s

As we have the horizontal distance we can calculate the travel time

x = vox t

t = vox / x

t = vo cos θ / x

t = 13.16 cos θ / 3

t = 4.39 cos θ

y =
`v_(oy)`t - ½ g t²

0 = vo sin θ t-1/2 9.8 t²

0 = 13.16 sin θ t-4.9t²

0 = 13.16 sin θ (4.39 cos θ) - 4.9 (4.39 cos θ)²

0 = 57.78 sin θ cos θ - 94.43 cos² θ

0 = 57.78 sin θ - 94.43 cos θ

tan θ = 94.43 / 57.78

θ = 58.5º

d) The range maximum is

R = vo² sin 2θ / g

R = 13.16² sin 2 58.5 /9.8

R = 15.75 m