- #1
Jazz
- 103
- 5
Homework Statement
A toy gun uses a spring with a force constant of ##\small{300 N/m}## to propel a ##\small{10.0\!-\!g}## steel ball. If the spring is compressed ##\small{7.00\ cm}## and friction is negligible:
At what angles above the horizontal may a child aim to hit a target ##\small{3.00\ m}## away at the same height as the gun?
Data:
##m = 10.0 \times 10^{-3}\ kg##
##k = 300\ N/m##
##x = 7.00 \times 10^{-2}\ m##
##Range = 3.00\ m##
Homework Equations
##PE_{spring} = \frac{1}{2}kx^2##
##KE = \frac{1}{2}mv^2##
##v^2 = v_0^2 - 2gh##
##\bar{v}t = d##
The Attempt at a Solution
It's assumed conservation of energy in order to find the initial velocity:
##\Delta PE_{spring} = \Delta KE##
##\sqrt{\frac{kx^2}{m}} = v##
##v= 12.1\ m/s##
Then I broke the velocity and the displacement into their components (I skipped the units for simplicity). I used just half of the range, hence I took ##\small{d = 1.50\ m}##:
##\cos(\theta) = \frac{1.50}{x} \Rightarrow x = \frac{1.50}{\cos(\theta)}##
##y = 1.50\tan(\theta) \hspace{35pt}(1)##
I wanted to know the height (y) the ball reached:
##v^2 = v_0^2 - 2gy##
##y = \frac{(12.12\sin(\theta))^2}{(2)(9.8)} = 7.50\sin^2(\theta) \hspace{35pt}(2)##
Combining equations ##(1)## and ##(2)##:
##1.50\tan(\theta) = 7.50\sin^2(\theta)##
##\cos(\theta)\sin(\theta) = \frac{1}{5}##
##2\cos(\theta)\sin(\theta) = \frac{2}{5}##
##\sin(2\theta) = \frac{2}{5}##
##2\theta = \arcsin(\frac{2}{5})##
##\theta = 11.79º\ or\ 78.21º##
Then by finding the time I wanted to comprobe whether the results I got are corrects. The time it took the ball to reach its maximum height, its horizontal displacement should have been ##\small{d = 1.50\ m}##:
##t= \frac{v\sin(\theta)}{g}= \frac{(12.12\ m/s)\sin(11.79º)}{9.8\ m/s^2} = 0.253\ s##
##v\cos(\theta)t = d##
##(12.12\ m/s)\cos(11.79º)(0.2527\ s) = 3.00\ m##
I don't know why I get this horizontal displacement. I expected to get ##\small{d = 1.50\ m}## and not the whole range.
Thanks!
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