Question

A block of mass m and speed v collides with a spring, compressing it a distance Δx. What is the compression of the spring if the mass of the block is halved and its speed is doubled?

Answer 1

Given,

A block of mass m and speed v.

Collides with spring and compresses it by Δx.

To find,

Compression when mass is halved and speed doubled.

Solution,

Let us assume the spring constant be k.

Kinetic energy of the block = $\frac{1}{2} mv^{2}$

Now during compression all this energy is converted into spring potential energy, which is  $\frac{1}{2} kx^{2}$, where $x$ is the compression.

Having the relation,

                           $\frac{1}{2} mv^{2}$ =  $\frac{1}{2} k$Δ$x^{2}$  

                           Δx = $\sqrt{\frac{mv^{2} }{k} }$ ---------------eq(1)

Now, when mass is halved and velocity doubled the new Kinetic energy becomes,

                  $= \frac{1}{2}*\frac{m}{2}*(2v)^{2} \\\\ = mv^{2}$

Equating the energy equations, we have

              New kinetic energy =  Energy stored in spring

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

                                $x = \sqrt{\frac{2mv^{2}}{k}}$

                                x = $\sqrt{2}$Δx  [ using from eq(1) ]

Therefore, compression when mass of the block is halved and its speed is doubled is  $\sqrt{2}$Δx.