Question

A block of mass $m$ collides perfectly in elastically with another identical block attached to a spring of force constant $K$. What will be the amplitude of resulting SHM?

Answer 1

In this question, the concept of law of conservation of momentum will be used, that is the momentum of a system before the collision will be equal to the momentum after the collision.

Complete step by step answer:
In this question we have given the mass of the block is $m$ and the spring constant is $K$. The identical block of mass $m$ collides with the spring attached. The collision is perfectly elastic. In this question we need to calculate the amplitude of the simple harmonic motion after the collision.

As we know that the law of conservation of momentum states that the momentum of a system before the collision will be equal to the momentum after the collision.

Use the collision principle before and after collision as,
$\Rightarrow {m_1}{V_1} + {m_2}{V_2} = \left( {{m_1} + {m_2}} \right)V$
Here, $V$ is the final velocity of the combined system of blocks after collision and $v$ is the initial velocity of the block that is ${V_1}$.

In the above equation we substitute $0$ for ${V_2}$ and $m$ for ${m_1}$ and ${m_2}$.
$\Rightarrow mv + m\left( 0 \right) = \left( {m + m} \right)V$

Now, we simplify above equation as,
$\Rightarrow mv = \left( {2m} \right)V$
Now, we get the final velocity as,
$\Rightarrow V = \dfrac{v}{2}$
Now we apply the energy conservation after the collision. In this the kinetic energy of the blocks will be equal to the potential energy stored in the spring.
It can be expressed as,
$\Rightarrow \dfrac{1}{2}\left( {m + m} \right){V^2} = \dfrac{1}{2}K{A^2}$
Where, $A$ is the amplitude of the SHM.

After simplification we get,
$\Rightarrow 2m{V^2} = K{A^2}$
Now, we substitute $\dfrac{v}{2}$ for $V$in above equation as
$\Rightarrow 2m{\left( {\dfrac{v}{2}} \right)^2} = K{A^2}$
By simplification we get,
$\Rightarrow {A^2} = \dfrac{{m{v^2}}}{{2K}}$
Now, we take square root both the sides as,
$\Rightarrow \sqrt {{A^2}} = \sqrt {\dfrac{{m{v^2}}}{{2K}}}$
After simplification we get,
$\therefore A = v\sqrt {\dfrac{m}{{2K}}}$
Therefore, the amplitude of the resulting SHM is $A = v\sqrt {\dfrac{m}{{2K}}}$.

Note: As we know that the perfectly elastic collision means there is no loss of the kinetic energy of the system while in inelastic collision, the kinetic energy changes from one form energy to another form of the energy.