Question

Block of mass m, moving on a smooth horizontal surface, collides with a
stationary block of mass m2. The coefficient of restitution for the
collision is e. The ratio for which my is at rest after collision is?​

Answer 1

Correct Question:-

Block of mass $\displaystyle\sf {m_1}$ moving on a smooth horizontal surface, collides with a stationary block of mass $\displaystyle\sf {m_2.}$ The coefficient of restitution for the collision is $\displaystyle\sf {e.}$ Find the ratio for which $\displaystyle\sf {m_1}$ is at rest after collision.

Solution:-

Before collision, let the block of mass $\displaystyle\sf {m_1}$ collide the other block, which was at rest initially, with velocity $\displaystyle\bf {u}.$

Let the block of mass $\displaystyle\sf {m_2}$ move with velocity $\displaystyle\bf {v}$ after collision, where the other block comes to rest.

Now coefficient of restitution is given by,

$\displaystyle\mathsf{\longrightarrow e=\dfrac {\mathbf{v-0}}{\mathbf{u-0}}}$

$\displaystyle\mathsf{\longrightarrow e=\dfrac {\mathbf{v}}{\mathbf{u}}\quad\quad\dots(1)}$

But by momentum conservation,

$\displaystyle\mathsf{\longrightarrow m_1}\mathbf{u}+\mathsf{m_2}(\mathbf{0})=\mathsf{m_1}(\mathbf{0})+\mathsf{m_2}\mathbf{v}$

$\displaystyle\mathsf{\longrightarrow m_1}\mathbf{u}=\mathsf{m_2}\mathbf{v}$

$\displaystyle\mathbf{\longrightarrow\dfrac {v}{u}}=\mathsf{\dfrac {m_1}{m_2}}$

Hence (1) becomes,

$\displaystyle\mathsf{\longrightarrow\underline {\underline {e=\dfrac {m_1}{m_2}}}}$

Answer 2

Correct Question:-

Block of mass $\displaystyle\sf {m_1}$ moving on a smooth horizontal surface, collides with a stationary block of mass $\displaystyle\sf {m_2.}$ The coefficient of restitution for the collision is $\displaystyle\sf {e.}$ Find the ratio for which $\displaystyle\sf {m_1}$ is at rest after collision.

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

Before collision, let the block of mass $\displaystyle\sf {m_1}$ collide the other block, which was at rest initially, with velocity $\displaystyle\bf {u}.$

Let the block of mass $\displaystyle\sf {m_2}$ move with velocity $\displaystyle\bf {v}$ after collision, where the other block comes to rest.

Now coefficient of restitution is given by,

$\displaystyle\mathsf{\longrightarrow e=\dfrac {\mathbf{v-0}}{\mathbf{u-0}}}$

$\displaystyle\mathsf{\longrightarrow e=\dfrac {\mathbf{v}}{\mathbf{u}}\quad\quad\dots(1)}$

But by momentum conservation,

$\displaystyle\mathsf{\longrightarrow m_1}\mathbf{u}+\mathsf{m_2}(\mathbf{0})=\mathsf{m_1}(\mathbf{0})+\mathsf{m_2}\mathbf{v}$

$\displaystyle\mathsf{\longrightarrow m_1}\mathbf{u}=\mathsf{m_2}\mathbf{v}$

$\displaystyle\mathbf{\longrightarrow\dfrac {v}{u}}=\mathsf{\dfrac {m_1}{m_2}}$

Hence (1) becomes,

$\displaystyle\mathsf{\longrightarrow\underline {\underline {e=\dfrac {m_1}{m_2}}}}$