Question

A `0.473` `kg` object is pushed down `1.087` `m` on a spring with spring constant `K=312.4` `Nm^(-1)`. When the spring is released it is shot out at an angle of `49.8^o` to the horizontal. How far does it fly before striking the ground?

Answer 1

This problem has two steps: find the initial velocity and then use that (by dividing into horizontal and vertical components) to find the range.

The range is `78.61` `m`.

Explanation:

The first step in this problem uses energy to calculate the initial velocity of the projectile.

Summarise the data we have:

`k = 312.4` `Nm^(−1)`
`d=1.087` `m` (this distance will be called '`x`' in the equation)
`m=0.473` `kg`
`theta=49.8^o`

The spring potential energy when the spring is compressed is given by:

`E_(sp)=1/2kx^2=1/2xx312.4xx1.087^2=184.6` `J`

As the dodgeball leaves the mouth of the catapult all of this spring potential energy will have been converted into kinetic energy:

`E_(sp)=E_k=1/2mv^2`

Rearranging to make `v` the subject:

`v=sqrt((2E_k)/m)=sqrt((2xx184.6)/0.473)=27.94` `ms^-1`

This is close to `100` `kmh^-1` or `60` `mph`, so it's an impressive catapult!

The second step requires us to divide the velocity into its horizontal and vertical components and solve a projectile motion problem.

`v_x=vcostheta=27.94xxcos49.8=18.03` `ms^-1`
`v_y=vsintheta=27.94xxsin49.8=21.34` `ms^-1`

The horizontal velocity, `v_x`, will remain constant, and we'll use it again at the end of our calculation.

The vertical velocity, `v_y`, will undergo acceleration due to gravity, which will cause it to slow down as it travels upward, until its vertical velocity is `0` at the top of its motion. That will occur at the halfway mark of its trajectory.

Take the equation `v=u+at` and rearrange to make `t` the subject:

`t = (v-u)/a = (0-21.34)/-9.8 = 2.18` `s`

(the acceleration due to gravity is negative because it is on the opposite direction to the initial velocity)

This is the time for half the flight, so we need to double it for the full flight. The total time of flight is `4.36` `s`.

The range is simply the horizontal distance covered in this time:

`v=d/t`

Rearrange to make `d` the subject:

`d=v_xt=18.03xx4.36=78.61` `m`