Question

Plz do explain the procedure ! ASAP ​

Answer 1

Let $\displaystyle\small\bf {v_2}$ be the velocity of the shell with respect to ground and $\displaystyle\small\bf {v_1}$ be the recoil velocity of the tank, i.e., the velocity of the tank with respect to ground.

The tank moves horizontally leftwards after firing, so the recoil velocity will be,

• $\displaystyle\small\bf {v_1}=\sf{-v_1\,\hat i}$

Given that the velocity of the shell with respect to the tank is $\displaystyle\small\bf {V}$ and it makes an angle θ with horizontal, so,

$\displaystyle\small\longrightarrow\bf {V}=\sf{V\cos\theta\,\hat i+V\sin\theta\,\hat j\quad\quad\dots(1)}$

Also,

$\displaystyle\small\longrightarrow\bf{V=v_2-v_1}$

$\displaystyle\small\longrightarrow\bf{V=v_2}-\sf{(-v_1\,\hat i)}$

$\displaystyle\small\longrightarrow\bf{V=v_2}+\sf{v_1\,\hat i\quad\quad\dots(2)}$

From (1) and (2),

$\displaystyle\small\longrightarrow\bf{v_2}+\sf{v_1\,\hat i=V\cos\theta\,\hat i+V\sin\theta\,\hat j}$

$\displaystyle\small\longrightarrow\bf{v_2}=\sf{(V\cos\theta-v_1)\,\hat i+V\sin\theta\,\hat j}$

Since there is no external horizontal force acting on the system, the horizontal linear momentum of the system remains constant.

The initial horizontal linear momentum of the system is zero.

• $\displaystyle\small\bf {p_i}=\sf{0}$

The final horizontal linear momentum of the system is,

• $\displaystyle\small\bf {p_f}=\sf{m_1}\bf{(v_1)_x}+\sf{m_2}\bf{(v_2)_x}$

Here $\displaystyle\small\bf {(v_1)_x}$ and $\displaystyle\small\bf {(v_2)_x}$ are horizontal component of $\displaystyle\small\bf {v_1}$ and $\displaystyle\small\bf {v_2}$ respectively. So,

• $\displaystyle\small\bf {p_f}=\sf{m_1(-v_1)+m_2(V\cos\theta-v_1)}$

Then by linear momentum conservation,

$\displaystyle\small\longrightarrow\bf{p_i=p_f}$

$\displaystyle\small\longrightarrow\sf{m_1(-v_1)+m_2(V\cos\theta-v_1)=0}$

$\displaystyle\small\longrightarrow\sf{-m_1v_1+m_2(V\cos\theta-v_1)=0}$

$\displaystyle\small\longrightarrow\sf{-m_1v_1+m_2V\cos\theta-m_2v_1=0}$

$\displaystyle\small\longrightarrow\sf{m_2V\cos\theta-(m_1+m_2)v_1=0}$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{\sf{v_1=\dfrac{m_2V\cos\theta}{m_1+m_2}}}} }$

This is the recoil velocity of the tank.