A body of mass m1 moving with an unknown velocity of v1 - undergoes a collinear collision with a body of mass m2 moving with a velocity v2 - After collision- m1 and m2 move with velocities of v3 - and v4 - respectively-If m2 - 0-5 m1 and v3 - 0-5 v1 then v1 is equal to-

According to conservation of momentum principle, m_1v_1 + m_2v_2=m_1v_3 + m_2v_4 ⇒ m_1v_1+(0.5 m_1)v_2=m_1 (0.5 v_1) + (0.5 m_1)v_4 ⇒ v_1+0.5 v_2=0.5 v_1 + 0 ...

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Standard IX

Physics

Third Law of Motion

A body of mas...

Question

A body of mass $m_{1}$ moving with an unknown velocity of $v_{1}^i$ , undergoes a collinear collision with a body of mass $m_{2}$ moving with a velocity $v_{2}^i$ . After collision, $m_{1}$ and $m_{2}$ move with velocities of $v_{3}^i$ and $v_{4}^i$ , respectively.
If $m_{2} = 0.5 m_{1}$ and $v_{3} = 0.5 v_{1}$ then $v_{1}$ is equal to,

v_{4} - v_{2}

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v_{4} - \frac{v_{2}}{2}

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v_{4} + v_{2}

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v_{4} - \frac{v_{2}}{4}

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Solution

The correct option is A $v_{4} - v_{2}$
According to conservation of momentum principle,

$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{3} + m_{2} v_{4}$

$\Rightarrow m_{1} v_{1} + (0.5 m_{1}) v_{2} = m_{1} (0.5 v_{1}) + (0.5 m_{1}) v_{4}$

$\Rightarrow v_{1} + 0.5 v_{2} = 0.5 v_{1} + 0.5 v_{4}$

$∴ v_{1} = v_{4} - v_{2}$

Hence, $(B)$ is the correct answer.

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A body of mass m1 moving with an unknown velocity of v1^i, undergoes a collinear collision with a body of mass m2 moving with a velocity v2^i. After collision, m1 and m2 move with velocities of v3^i and v4^i, respectively. If m2=0.5 m1 and v3=0.5v1 then v1 is equal to,

The correct option is A v4−v2According to conservation of momentum principle, m1v1+m2v2=m1v3+m2v4 ⇒m1v1+(0.5m1)v2=m1(0.5v1)+(0.5m1)v4 ⇒v1+0.5v2=0.5v1+0.5v4 ∴v1=v4−v2 Hence, (B) is the correct answer.