Sliding block hits and compresses a spring

[Antoha1]

Homework Statement

A 2 kg block slides along a smooth horizontal surface with a speed of 5 m/s. It collides with a horizontal spring (spring constant k = 400 N/m), compressing it until it comes to a complete stop. Find the maximum compression of the spring and the time it takes for the spring to compress to its maximum compression. We assume that there is no friction with the surface.

Relevant Equations

$\frac{1}{2}mv^2 ;~~~~~~~~~~\frac{1}{2}kx^{2};~~~~~~~~~a=\frac{v-v_{0}}{t}$

My solution is:
$\frac{1}{2}mv^2=\frac{1}{2}kx^{2} \Longrightarrow x=v\sqrt{\frac{m}{k}}$ x=0,354m
block gains negative acceleration from spring elastic force so:
$a=\frac{v-v_{0}}{t} ; x=\frac{at^{2}}{2} \Longrightarrow t=\frac{2v_{0}}{v-v_{0}}\sqrt{\frac{m}{k}}\approx 0,14~s$ where v0=5m/s and v=0m/s

The problem is, solution provided for second part is like this:

By compressing the spring and moving to the equilibrium position to the maximum and back, the block undergoes simple harmonic oscillation. The formula describing the spring-mass system is:
$\omega=\sqrt{\frac{k}{m}}$
In simple harmonic motion, the time to maximum compression (a quarter of a full period) is expressed as:
$t=\frac{T}{4}=\frac{\pi}{2\omega}\approx 0,111~s$

Am I in the wrong here?

 

[Steve4Physics]

My solution is:
$\frac{1}{2}mv^2=\frac{1}{2}kx^{2} \Longrightarrow x=v\sqrt{\frac{m}{k}}$ x=0,354m

That's OK.

But note that the data in the question are supplied to only one significant figure. Technically it would be best if your answer had only one significant figure, though two significant figures would be a reasonable compromise here. (Unless your teacher has advised otherwise.)

block gains negative acceleration from spring elastic force so:
$a=\frac{v-v_{0}}{t} ; x=\frac{at^{2}}{2} \Longrightarrow t=\frac{2v_{0}}{v-v_{0}}\sqrt{\frac{m}{k}}\approx 0,14~s$ where v0=5m/s and v=0m/s

You have used equations which apply forconstant acceleration. Is the acceleration constant while the spring is being compressed?

 

[kuruman]

As @Steve4Physics noted the calculation that finds $t = 0.14$ s is incorrect because it assumes contant acceleration. The calculation that finds $t=\frac{1}{4}T=0.11$ s is correct because the time from relaxed length to maximum compression is indeed one-quarter period.

Also note that things like $\frac{1}{2}mv^2$ and $\frac{1}{2}kx^2$ are not equations. Equations have two terms separated by an "equals" sign.

 

[haruspex]

By compressing the spring and moving to the equilibrium position

Have you quoted that correctly? When it contacts the spring, the spring is relaxed, so that is the equilibrium position. Does it perhaps say "moving from the equilibrium position "?

 

[Steve4Physics]

Have you quoted that correctly? When it contacts the spring, the spring is relaxed, so that is the equilibrium position. Does it perhaps say "moving from the equilibrium position "?

It might just be a not-that-great description (notably lacking commas!).

The OP wrote:

By compressing the spring and moving to the equilibrium position to the maximum and back, the block undergoes simple harmonic oscillation.

The intended meaning may be this (with block positions A-B-C):

"By compressing the spring" [B$\rightarrow$C]
"and moving to the equilibrium position" [C$\rightarrow$B]
"to the maximum" [B$\rightarrow$A]
"and back" [A$\rightarrow$B]
"the block undergoes [one full cycle of] simple harmonic oscillation."

 

[kuruman]

Have you quoted that correctly? When it contacts the spring, the spring is relaxed, so that is the equilibrium position. Does it perhaps say "moving from the equilibrium position "?

I, too, was puzzled by this. Then I noticed in the statement of the problem that the mass is said to come to a "complete" (as opposed to instantaneous) stop. I interpreted "complete" to mean that some mechanism (solenoid switch?) clicks in place and prevents the mass from moving any more. Of course, the mass undergoes only a quarter of cycle of harmonic motion but that does not affect the solution.