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Answer 1

Answer :-

Option (3) $\sf{\dfrac{1}{2}u^2m_1\:\bigg(1\: + \: \dfrac{m_1}{m_2} \bigg)}$

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A shell at rest on a smooth horizontal surface explodes into two fragments.

The initial momentum of the shell is zero as it is in rest position.

But later, it explodes into two fragments of mass m1 and $\sf{m_2}$. After explosion $\sf{m_1}$ moves with speed u.

Let the -

• $\sf{m_2}$ moves with speed v after explosion.

So,

→ Initail momentum = 0

→ Final momentum = $\sf{m_1u\:+\:m_2v}$

According to law of conversation of linear momentum

The total momentum of the system can't be changed.

$\therefore$ Initial momentum = Final momentum

→ $\sf{0\:=\:m_1u\:+\:m_2v}$

→ $\sf{-\: m_2v\: =\: m_1u}$

→ $\sf{- v\: = \:\dfrac{m_1u}{m_2}}$

→ $\sf{v \:=\:\dfrac{ -\: m_1u}{m_2}}$ ...(1)

Now,

Work done = Kinetic energy of m1 and m2

→ $\sf{\dfrac{1}{2} m_1(u)^2\: +\:\dfrac{ 1}{2} m_2(v)^2}$

→ $\sf{\dfrac{1}{2} m_1u^2\: +\: \dfrac{1}{2} m_2v^2}$ ...(2)

Substitute of (1) in (2)

→ $\sf{\dfrac{1}{2} m_1u^2\: +\:\dfrac{ 1}{2} m_2 \bigg(\dfrac{-m _{1}u }{m_2} \bigg)^{2}}$

→ $\sf{\dfrac{1}{2} m_1u^2\: +\: \dfrac{1}{2} m_2\bigg( \dfrac{m _{1}^{2} {u}^{2} }{ m_{2} ^{2} } \bigg)}$

→ $\sf{\dfrac{1}{2}u^2\:\bigg(\dfrac{m_1m_2\:+\:m_1^2}{m_2}\bigg)}$

→ $\sf{\dfrac{1}{2}u^2m_1\:\bigg(\dfrac{m_2\:+\:m_1}{m_2}\bigg)}$

→ $\sf{\dfrac{1}{2}u^2m_1\:\bigg(\dfrac{m_2}{m_2} \: + \: \dfrac{m_1}{m_2} \bigg)}$

→ $\sf{\dfrac{1}{2}u^2m_1\:\bigg(1\: + \: \dfrac{m_1}{m_2} \bigg)}$

Answer 2

Let the fragment mass m2 move with speed v just after the explosion.

Apply the law of conservation of momentum.

m2 v + m1 u = (m1+m2)* 0

v = - m1 u/ m2

Work done = K E of m1 and m2.

= 1/2 m1 u^2 + 1/2 m2 v^2

= 1/2 m1 u^2 *( 1 + m1/m2)