Question

A block of mass \[m\] moving with velocity $v$ collides with another stationary block of mass m as shown. The maximum compression in the spring is ?

Answer 1

Hint:To solve this question here we will apply energy conservation and also, we know about the spring force and using this law of conservation of energy of the system and formulas of kinetic energy and potential energy we will find the maximum compression.

Formula used:
${E_{{\text{spring}}}} = \dfrac{1}{2}k{x^2}$
Where, ${E_{{\text{spring}}}}$ is the energy of spring, $k$ is the spring constant and $x$ is the length of compression or expansion.

Complete step by step answer:
According to the question, the spring attached with a block of mass $M$ will compress until both the particles start to move together. So, external force acting on the system is $0$.Therefore, initial momentum is equal to final momentum
\[m{v_0} = (m + M)v \\
\Rightarrow v = \dfrac{{m{v_0}}}{{(m + M)}} \\ \]
Now, let’s assume maximum compression is $x$. Therefore, from law of conservation of energy,initial kinetic energy equals the sum of spring potential energy and kinetic energy of the system of blocks i.e.,
$\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}K{x^2} + \dfrac{1}{2}(m + M){\left( {\dfrac{{m{v_0}}}{{(m + M)}}} \right)^2} \\
\Rightarrow v_0^2\left( {m - \dfrac{{{m^2}}}{{(m + M)}}} \right) = K{x^2} \\
\Rightarrow {x^2} = \dfrac{{mM}}{{(m + M)K}}v_0^2 \\
\therefore x = \sqrt {\dfrac{{mM}}{{(m + M)K}}} v_0^2 \\ $
Hence, the maximum compression in the spring is $\sqrt {\dfrac{{mM}}{{(m + M)K}}} v_0^2$.

Note:Remember only \[v = \dfrac{{m{v_0}}}{{(m + M)}}\] because the external force acting on the system is zero if not then don’t proceed accordingly. Also, we know that the linear conservation of momentum says that the total energy of the blocks before and after the collision will be conserved.