A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What will be the compression if the same block collides with the spring at a speed of 2v?

Energy cannot be created or destroyed, according to the rule of conservation of energy, but it can be changed from one form to another.

The elastic potential energy of the spring is expressed as,

$U=\frac{1}{2}k{x}^2$

we have, $k$= spring constant and $x$= compression in the spring.

we right a expression for the kinetic energy of an object,

$K=\frac{1}{2}m{v}^2$

we have, $m$is the mass of the object and $v$is the velocity of the object.

The spring collides with the block as it glides along the horizontal surface, compressing it. As a result, the block's kinetic energy is transformed to the spring's potential energy. Thus,

$K_{\text{block}}=U_{\text{spring}}$

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

Velocity = $v_{\mathrm{i}}$and the compression is $x_{\mathrm{i}}$then,

$\frac{1}{2}m{v}_i^2=\frac{1}{2}k{x}_i^2$

velocity $v_{\mathrm{f}}$and the compression is $x_{\mathrm{f}}$then,

$\frac{1}{2}m{v}_{\mathrm{f}}^2=\frac{1}{2}k{x}_{\mathrm{f}}^2$

From above,

$\frac{\frac{1}{2}m{v}_{\mathrm{i}}^2}{\frac{1}{2}m{v}_{\mathrm{f}}^2}=\frac{\frac{1}{2}k{x}_{\mathrm{i}}^2}{\frac{1}{2}k{x}_{\mathrm{f}}^2}$

$\frac{v_{\mathrm{i}}^2}{v_{\mathrm{f}}^2}=\frac{x_{\mathrm{i}}^2}{x_{\mathrm{f}}^2}$

Rewrite the spring's final compression calculation in a different way.

$x_{\mathrm{f}}^2=\left(\frac{v_{\mathrm{f}}^2}{v_{\mathrm{i}}^2}\right)x_{\mathrm{i}}^2$

$x_{\mathrm{f}}=\left(\frac{v_{\mathrm{f}}}{v_{\mathrm{i}}}\right)x_{\mathrm{i}}$

The compression of the spring is,

$x_{\mathrm{f}}=\left(\frac{v_{\mathrm{f}}}{v_{\mathrm{i}}}\right)x_{\mathrm{i}}$

Putting $2v$for $v_{\mathrm{f}},v$for $v_{\mathrm{i}},2.0\mathrm{cm}$for $x_{\mathrm{i}}$.

$x_{\mathrm{f}}=\left(\frac{2v}{v}\right)(2.0\mathrm{cm})$

$=4.0\mathrm{cm}$

As a result, the spring's tension is reduced $4.0\mathrm{cm}$.