Question

A small ball of mass 10 kg is moving with speed 4m/s collide elastically with another ball of mass 6kg at rest.If collision is head on then the respective speed of ball after collision will be

Answer 1

Answer:

• The Final velocity (v₁) of 10 Kg block is 1 m/s.
• The final velocity (v₂) of 6 Kg block is 5 m/s.

Given:

$\boxed{\begin{minipage}{10 em} \textbf{\underline{10 kg Block}}\colon \\\\ \bullet \rm u_1 = 4 \;m/s \\\bullet \rm v_1 = ? \\\bullet \rm m_1 = 10 \;Kg\end{minipage}}\quad\boxed{\begin{minipage}{10 em}\textbf{\underline{6 Kg Block}}\colon \\\\ \bullet \rm u_2 = 0 \; m/s \\\bullet \rm v_2 = \; ? \\\bullet \rm m_2 = 6 \;Kg \end{minipage}}$

Explanation:

$\rule{300}{1.5}$

From the final velocity of one - dimensional collision of first body (m₁ = 10 kg),

$\large \bigstar\; \boxed{\tt v_1 = \Bigg(\dfrac{m_1 - m_2}{m_1 + m_2}\Bigg)u_1+\Bigg(\dfrac{2\;m_2}{m_1+m_2}\Bigg)u_2}$

$\mathfrak{Here}\begin{cases}\sf{m_1}\text{ Denotes Mass of 1st Body}\\\sf{m_2}\text{ Denotes Mass of 2nd Body}\\\sf{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}$

Now,

$\large \boxed{\tt v_1 = \Bigg(\dfrac{m_1 - m_2}{m_1 + m_2}\Bigg)u_1+\Bigg(\dfrac{2\;m_2}{m_1+m_2}\Bigg)u_2}$

Substituting the values,

$\displaystyle\dashrightarrow \tt v_1=\Bigg(\dfrac{10 - 6}{10 + 6}\Bigg)\times4+\Bigg(\dfrac{2\times6}{10+6}\Bigg)\times0\\\\\\\dashrightarrow \tt v_1=\Bigg(\dfrac{10 - 6}{10 + 6}\Bigg)\times4+0\\\\\\\dashrightarrow\tt v_1=\Bigg(\dfrac{10 - 6}{10 + 6}\Bigg)\times4\\\\\\\dashrightarrow\tt v_1=\dfrac{4}{16}\times4\\\\\\\dashrightarrow\tt v_1 = \dfrac{16}{16}\\\\\\\dashrightarrow\tt v_1 = \cancel{\dfrac{16}{16}}\\\\\\\dashrightarrow\tt \large\underline{\boxed{\red{\tt v_1 = 1\;m/s}}}$

∴ The Final velocity of 10 Kg Block is 1 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the final velocity of one - dimensional collision of second body (m₂ = 6 kg),

$\large \bigstar\; \boxed{\tt v_2 = \Bigg(\dfrac{m_2 - m_1}{m_1 + m_2}\Bigg)u_2+\Bigg(\dfrac{2\;m_1}{m_1+m_2}\Bigg)u_1}$

$\mathfrak{Here}\begin{cases}\sf{m_1}\text{ Denotes Mass of 1st Body}\\\sf{m_2}\text{ Denotes Mass of 2nd Body}\\\sf{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}$

Now,

$\large \boxed{\tt v_2 = \Bigg(\dfrac{m_2 - m_1}{m_1 + m_2}\Bigg)u_2+\Bigg(\dfrac{2\;m_1}{m_1+m_2}\Bigg)u_1}$

Substituting the values,

$\displaystyle\dashrightarrow \tt v_2=\Bigg(\dfrac{6 - 10}{10 + 6}\Bigg)\times0+\Bigg(\dfrac{2\times10}{10+6}\Bigg)\times4\\\\\\\dashrightarrow \tt v_2=\Bigg(\dfrac{2\times10}{10 + 6}\Bigg)\times4+0\\\\\\\dashrightarrow\tt v_2=\Bigg(\dfrac{2\times10}{10 + 6}\Bigg)\times4\\\\\\\dashrightarrow\tt v_2=\dfrac{20}{16}\times4\\\\\\\dashrightarrow\tt v_2=\dfrac{80}{16}\\\\\\\dashrightarrow\tt v_2=\cancel{\dfrac{80}{16}}\\\\\\\dashrightarrow\tt \large\underline{\boxed{\red{\tt v_2 = 5\;m/s}}}$

∴ The Final velocity of 6 Kg Block is 5 m/s.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

1. The Final velocity (v₁) of 10 Kg block is 1 m/s.
2. The Final velocity (v₁) of 10 Kg block is 1 m/s.The final velocity (v₂) of 6 Kg block is 5 m/s.