Question

A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed of 5v. What distance was the spring compressed in the second case?

Answer 1

Answer:

$x' = 10 x$

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

so we will have

$v = \sqrt{\frac{k}{m}}(x)$

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

$5v = \sqrt{\frac{k}{4m}}(x')$

now from above two equations we have

$\frac{5v}{v} = \frac{x'}{2x}$

so we have

$x' = 10 x$