Let, contraction of a horizontal spring, $x$, is $4$ cm and mass of a bullet, $m$, is $10 g$. Spring constant, $k$, is $200 N/m$.
When bullet leaves the gun, what will be the velocity of it?
Can I solve it in the following way?
$0.5mv^2=0.5kx^2$ Or, $mv^2=kx^2$ Or, $v=5.6569 m/s$.
Please tell me whether this method of solution is correct or not.
But if I want to solve it in an alternative way, I may write:
$$F=kx\\ \implies ma=kx\\ \implies 0.01a=(200)(0.04)\\ \implies a=800 m/s^2$$
Again we know $v^2=u^2+2as$ Or, $v=8 m/s$.
Which is totally different from the first method. Why is this occuring? I knew that, $v^2=u^2+2as$ can be used when accelaration, a, is constant over the distance, $s$. Since in case of spring, accelaration created due to expansion or contraction of spring is not constant with the displacement of edge from equilibrium position, we cannot use $v^2=u^2+2as$. But I want to solve it with alternatives way except the first one.
Please tell me accelaration of which point $a=800 m/s^2$ is indicating to, i.e at to which point accelaration will be $800 m/s^2$.
If I want to solve this mathematical problem in the second way, which steps I should follow?