A spring with a constant of $1 \frac{k g}{s^{2}}$ $1 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $\frac{2}{5} k g$ $2/5 kg$ and speed of $\frac{1}{4} \frac{m}{s}$ $1/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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A spring with a constant of $1 \frac{k g}{s^{2}}$ $1 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $\frac{2}{5} k g$ $2/5 kg$ and speed of $\frac{1}{4} \frac{m}{s}$ $1/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-04-30T09:30:40

$0.16 m$ $0.16m$

Explanation:

According to the law of conservation of energy, the total energy before and after a reaction must be equal.

Kinetic energy before this reaction is $\frac{1}{2} m v^{2}$ $1/2mv^2$ , where $m$ $m$ is mass and $v$ $v$ is velocity. Potential energy stored in the spring is $\frac{1}{2} k x^{2}$ $1/2kx^2$ , where $k$ $k$ is the spring constant and $x$ $x$ is the distance it compresses.

Therefore, according to conservation,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$ $1/2mv^2=1/2kx^2$

$m v^{2} = k x^{2}$ $mv^2=kx^2$

Now we can put in what we know of the values and rearrange to find $x$ $x$ ,

$\frac{2}{5} \cdot {(\frac{1}{4})}^{2} = 1 \cdot x^{2}$ $2/5*(1/4)^2=1*x^2$

$\frac{1}{40} = x^{2}$ $1/40=x^2$

$x = \sqrt{\frac{1}{40}} \approx 0.16 m$ $x=sqrt(1/40)approx0.16m$

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