A block of mass m is pushed against a spring of spring constant k. The spring is compressed by a distance d, the block is then released. It is launched by the spring along a horizontal frictionless surface with a final speed v. A second block, this one having mass 4m is pushed against the same spring and released, gaining a final speed 2v. By what distance was the spring compressed in the second case?
A) d
B) 2d
C) 4d
D) 16d
E) 25d

Question

Quizplus · Accepted Answer

Firstly, we can calculate the initial potential energy and final kinetic energy of the first block using the equation:
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
where x = d (distance compressed by the spring)

Then, we can find the spring constant for the second block by rearranging the equation above:
$k = \frac{2mv^2}{d^2}$

Next, we can use conservation of energy to find the distance compressed by the spring for the second block:
$\frac{1}{2}kx^2 = \frac{1}{2}(4m)(2v)^2$
$\frac{1}{2}\left(\frac{2mv^2}{d^2}\right)x^2 = 2mv^2$
$x^2 = 8d^2$
$x = \sqrt{8}d$
$x = 2\sqrt{2}d$

Therefore, the spring is compressed by 4d in the second case.

A Block of Mass M Is Pushed Against a Spring