Centre Of Mass Ques 3
- A body of mass $m_1$ moving with an unknown velocity of $v_1 \hat{\mathbf{i}}$, undergoes a collinear collision with a body of mass $m_2$ moving with a velocity $v_2 \hat{\mathbf{i}}$. After collision, $m_1$ and $m_2$ move with velocities of $v_3 \hat{\mathbf{i}}$ and $v_4 \hat{\mathbf{i}}$, respectively.
If $m_2=0.5 m_1$ and $v_3=0.5 v_1$, then $v_1$ is
(2019 Main, 8 April II)
(a) $v_4+v_2$
(b) $v_4-\frac{v_2}{4}$
(c) $v_4-\frac{v_2}{2}$
(d) $v_4-v_2$
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Answer:
Correct Answer: 3.( d )
Solution:
- Key Idea Total linear momentum is conserved in all collisions, i.e. the initial momentum of the system is equal to final momentum of the system.
Given,
$ m_2=0.5 m_1 \quad \Rightarrow \quad m_1=2 m_2 $
Let $m_2=m$, then, $m_1=2 m$
Also, $v_3=0.5 v_1$
Given situation of collinear collision is as shown below
$\therefore$ According to the conservation of linear momentum, Initial momentum $=$ Final momentum
$ \begin{aligned} & m_1 v_1 \hat{\mathbf{i}}+m_2 v_2 \hat{\mathbf{i}}=m_1 v_3 \hat{\mathbf{i}}+m_2 v_4 \hat{\mathbf{i}} \\ & \Rightarrow 2 m v_1 \hat{\mathbf{i}}+m v_2 \hat{\mathbf{i}}=2 m\left(0.5 v_1\right) \hat{\mathbf{i}}+m v_4 \hat{\mathbf{i}} \\ & \Rightarrow v_4=v_1+v_2 \Rightarrow v_1=v_4-v_2 \end{aligned} $
BETA
