How do I calculate the peak velocity of a 1kg object dropped 1m onto a spring?

[ehild]

The negative sign IS the direction. In my set up the coordinate system x increases up. So a negative velocity is going down. Note that in the initial conditions the initial velocity is negative. When speaking of velocities we must specify direction and magnitude, in this case negative means down. when looking for the min or max velocity it only makes sense to speak of the magnitude. Direction does not matter.

The question was peak velocity. Cannot it be positive as well? What do you think will happen after the body reaches that maximum negative velocity? I admit that "maximum peak velocity" means velocity of maximum magnitude, but this body moves first downward then upward so its peak velocities are both positive and negative, aren't they?

ehild

 

[HallsofIvy]

What makes you think the object will move faster going upward than downward?

 

[ehild]

What makes you think the object will move faster going upward than downward?

Oh my... My English must be very bad. I can not make myself understand. No, I did not say that the peak speed is higher upward than downward.
Well, Integral said that the maximum velocity was negative.

V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k

I wanted to say that the body reaches the same maximum SPEED both downward and upward, and I argued about the minus sign in Integral's post. It should be

V_{max}= \pm \sqrt { {V_i} ^2 + \frac {m g^2} k

if we speak about velocity, and just

V_{max}= \sqrt { {V_i} ^2 + \frac {m g^2} k

if it is speed.

ehild

 

[Integral]

Ideally each time the mass/spring combination oscillates past the point I called x_max it will have the same velocity. My solution is specific to a TIME at the TIME I found the ball is moving down. Notice that my solution is oscillatory in nature, so with a bit more effort I could find a sequence of times at which the ball passes through the point at which it has a maximum velocity.

Now, if it is not an ideal spring and losses are occurring which are not captured in our simple model, then when the spring/ball returns it will NOT be traveling at the same speed. So I could argue that the ONLY time the ball/spring system will have that maximum velocity is on the way down the first time.

Notice that the energy based solution the final result is found after taking the square root, by all rights you should take the negative root as the correct solution.