Question

The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 31° to the horizontal, a 53 g ball is shot to a maximum height of 1.90 m above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.

Answer 1

The speed at which the spring launches the ball is approximately 6.12 m/s. The initial compression distance of the spring is approximately 0.024 m or 2.4 cm, calculated using the conservation of energy principle and given values for mass, height, gravity, and spring constant.

Step-by-step explanation:

Calculating the Launch Speed and Spring Compression

To find the speed at which the spring launches the ball, we apply the conservation of energy principle. The potential energy stored in the spring is converted into the gravitational potential energy at the ball's highest point. Given the maximum height h of 1.90 m, the gravitational potential energy at the highest point is mgh, where m is the mass of the ball and g is the acceleration due to gravity (9.81 m/s²). This energy is equal to the kinetic energy at launch, which is (1/2)mv².

Now, to find the initial spring compression, we again use energy conservation. The potential energy stored in the spring when compressed is (1/2)kx², where k is the spring constant and x is the compression distance. This is equal to the gravitational potential energy at the highest point, so we can solve for x.

Let's calculate the speed at which the spring launches the ball (v):

• 
  
• m = 53 g = 0.053 kg (mass of the ball)
• 
  
• g = 9.81 m/s² (acceleration due to gravity)
• 
  
• h = 1.90 m (maximum height)
• 
  
• mgh = (1/2)mv²
• 
  
• v = √(2gh)
• 
  
• v = √(2*9.81*1.90)
• 
  
• v ≈ 6.12 m/s

Now, let's calculate the initial compression distance of the spring (x):

• 
  
• k = 730 N/m (spring constant)
• 
  
• (1/2)kx² = mgh
• 
  
• x² = (2mgh)/k
• 
  
• x² = (2*0.053*9.81*1.90)/730
• 
  
• x = √[(2*0.053*9.81*1.90)/730]
• 
  
• x ≈ 0.024 m or 2.4 cm

Answer 2

a. V=11.84 m/s

b.x=0.052m

Step-by-step explanation:

a).

Given

`K=730 N/m`,
`m=0.053kg`,
`h=1.90m`.

`v_f^2=v_i^2+2*g*h`

`v_i^2=2*g*h=2*9.8m/s^2*1.9m`

`v_i=√(2*9.8m/s^2*1.9m)=√(37.24 m^2/s^2)`

`v_i=6.1 m/s`

`v_i=V*sin(31)`

`V=(v_i)/(sin(31))=(6.1m/s)/(sin(31))`

`V=11.84 m/s`

b).

`K_k=(1)/(2)*K*x^2`

No friction on the ball so:

`x^2=(2*K_k)/(K)`

`x=\sqrt{(2*0.053kg*9.8m/s^2*1.9m)/(730N/m)}`

`x=\sqrt{2.7x10^(-3)m^2}=0.052m`