Question

In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35 degree angle compressing the spring 18 cm cause the ball to hit the floor 1.6 m below the point at which it leaves the spring after traveling 5.2 m horizontally.
What is the spring constant?

Answer 1

Ball hits the floor at 1.6 m below the initial position

So here we can say that

$\Delta y = -1.6 m$

also it will hit at horizontal distance of 5.2 m

$\Delta x = 5.2 m$

let say its velocity in x and y directions are given as

$v_x ,v_y$

now we can say

$v_x* t = 5.2$

$-1.6 = v_y * t - \frac{1}{2}gt^2$

from above two equations

$-1.6 = v_y* \frac{5.2}{v_x} - 4.9t^2$

as we know that it is projected at an angle of 35 degree

so we know that

$v_y = v_x tan35$

$-1.6 = 5.2 tan35 - 4.9 t^2$

$4.9 t^2 = 3.64 + 1.6 = 5.24$

$t = 1.03 s$

now we have

$v_x * 1.03 = 5.2$

$v_x = 5.03 m/s$

also we can find vertical component of velocity as

$v_y = v_x tan35 = 3.52 m/s$

now we will find net velocity as

$v^2 = v_x^2 + v_y^2$

$v^2 = 5.03^2 + 3.52^2$

$v = 6.14 m/s$

now by energy conservation we can say

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

$k*(0.18)^2= 0.032*(6.14)^2$

$k = 37.2 N/m$

so spring constant is 37.2 N/m