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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

  • Option 1)

    \frac{1}{4}V\hat{i}+\frac{3}{2}\: V\hat{j}

  • Option 2)

    \frac{1}{3}\: V\hat{i}+\frac{2}{3}V\: \hat{j}

  • Option 3)

    \frac{2}{3}V\hat{i}+\frac{1}{3}V\: \hat{j}

  • Option 4)

    \frac{3}{2}V\hat{i}+\frac{1}{4}V\hat{j}

 

Answers (1)

best_answer

As discussed in

Perfectly Inelastic Collision -

v= frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}

- wherein

Two bodies stick together after the collision ,so there will be a final common velocity (v)

m_{1},m_{2}= masses

v_{1}=: initial : velocity : mass: m_{1}

v_{2}=: initial : velocity : mass: m_{2}

 

 \therefore mv\hat{i}+3m(2v)\hat{i} = 4 m\vec{v}

\vec{v} = \frac{v}{4}\hat{i}+\frac{6}{4}\hat{j}\Rightarrow \frac{v}{4}\hat{i}+\frac{3}{2}\hat{j}


Option 1)

\frac{1}{4}V\hat{i}+\frac{3}{2}\: V\hat{j}

This solution is correct

Option 2)

\frac{1}{3}\: V\hat{i}+\frac{2}{3}V\: \hat{j}

This solution is incorrect

Option 3)

\frac{2}{3}V\hat{i}+\frac{1}{3}V\: \hat{j}

This solution is incorrect

Option 4)

\frac{3}{2}V\hat{i}+\frac{1}{4}V\hat{j}

This solution is incorrect

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