A ball with a mass of 25 grams needs to be fired at a 45 degree angle from a spring in order to go 2 meters, 4m and 6m. How much does the spring need to be compressed for each range?

Question

The Spring has a spring constant of 1.23 N/mm, Spring free length = 135mm, Spring minimum working length = 31mm. Load at minimum working length = 127.49 N.

Answer 1

The range of a projectile motion is R = v²sin2θ/g

==> v = √(Rg/sin2θ) for R =2m, and θ= 45°, v = 4.43m/s

The kinetic energy is equal to energy stored in the spring i.e

.5kx² = .5mv², but m = .025kg and k = 1.23N/mm = 1230N/m

Solving for the stretched lenght, x, you get

x = v√(m/k) = (4.43m/s)√(0.025kg/1230N/m) = 19.97mm [for R = 2m]

For R = 4m, v = 6.26m/s

==> x = v√(m/k) = 28.22mm

and for R = 6m, v = 7.67m/s

==> x = v√(m/k) = 34.57mm

Answer 2

I will explain how to work this type of problem. Use the range equation.

R = v²sin(2θ) / g

You have 3 values of R to work with ( 2, 4, and 6m) and the angle is 45°. For each of the 3 values of R, compute the required speed v. Then get the required spring compression distance by setting the potential energy of the compressed spring equal to the kinetic energy of the projectile.

kx²/2 = mv²/2 ⇒

x = v √(m/k)

m = 0.025 kg

k = 1.23 x 10³ N/m

Plug in the 3 values of v, and plug in m and k (all quantities in appropriate units) and get your 3 values of x in meters.