Question

A ball of mass 2 kg is moving with speed
3m/s. It collides with another ball of mass 3 kg
which was at rest initially. Find the speed of the
3kg ball if it is known that the speed of the 2kg
block after collision becomes 1/3rd the original
speed in the same direction.​

Answer 1

Given :-

▪ A ball of mass, m = 2 kg is moving with speed u₁ = 3 m/s . It collides with another ball of mass, M = 3 kg which was at rest initially.

▪ After the collision, the velocity of the 2 kg ball becomes 1/3 of the original speed in the same direction.

To Find :-

▪ Velocity of 3 kg ball.

Solution :-

According to the law of conservation of momentum, Momentum of the system before collision is same as the momentum after the collision.

Momentum before Collision :

Ball 1 :

• mass, m = 2 kg
• Velocity, u = 3 m/s

Ball 2 :

• Mass, M = 3 kg
• Velocity, u' = 0 m/s

⇒ Momentum of system = Momentum of ball1 + Momentum of ball2

⇒ Momentum of system = mu + Mu'

⇒ Momentum of system = 2×3 + 3×0

⇒ Momentum of system = 6 + 0

⇒ Momentum of system = 6 kg.m/s

Momentum after Collision :

Ball 1 :

• Mass, m = 2 kg
• Velocity, v = u / 3 = 1 m/s

Ball 2 :

• Mass, M = 3 kg
• Velocity, v' = ?

⇒ Momentum of system = Momentum of Ball1 + Momentum of Ball2

⇒ Momentum of system = mv + Mv'

⇒ Momentum of system = 2×1 + 3×v'

⇒ Momentum of system = 2 + 3v'

As discussed above, momentum of the system is conserved. So,

⇒ Momentum before Collision = Momentum after Collision

⇒ 6 = 2 + 3v'

⇒ 4 = 3v'

⇒ v' = 4/3

⇒ v' = 1.3 m/s

Hence, The velocity of 3 kg ball after the collision will be 1.3 m/s in the direction of 2 kg ball.

Answer 2

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:

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

Explanation:

$\rule{300}{1.5}</p><p>$

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

$\sf \pink{★ \: v1=(m1+m2m1−m2)u1+(m1+m22m2)u2</p><p>}</p><p></p><p>$

$\begin{gathered}\mathfrak \red{Here}\begin{cases}\sf \orange{m_1}\text{ Denotes Mass of 1st Body}\\\sf \orange{m_2}\text{ Denotes Mass of 2nd Body}\\\sf \orange{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf \orange{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}\end{gathered} </p><p>$

Now,

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

Substituting the values,

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

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

$\rule{300}{1.5}$

$\rule{300}{1.5}</p><p>$

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

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

$\begin{gathered} \rm \red{Here}\begin{cases}\sf \pink {m_1}\text{ Denotes Mass of 1st Body}\\\sf \pink{m_2}\text{ Denotes Mass of 2nd Body}\\\sf \pink{u_1}\text{ Denotes Initial velocity of 1st body}\\\sf \pink{u_2}\text{ Denotes initial velocity of 2nd Body}\end{cases}\end{gathered} </p><p>$

Now,

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

Substituting the values,

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

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

$\rule{300}{1.5}$