Two particles A and B, move with constant velocities \mathop {\upsilon _1 }\limits^ \to and \mathop {\upsilon _2 }\limits^ \to.  At the initial moment their position vectors are \mathop {\text{r}_1 }\limits^ \to and \mathop {\text{r}_2 }\limits^ \to respectively. The condition for particles A and B for their collision is :

 

  • Option 1)

    \mathop {\text{r}_1 }\limits^ \to \cdot \mathop {\upsilon _1 }\limits^ \to = \mathop {\text{r}_2 }\limits^ \to \cdot \mathop {\upsilon _2 }\limits^ \to

  • Option 2)

    \mathop {\text{r}_1 }\limits^ \to \times \mathop {\upsilon _1 }\limits^ \to = \mathop {\text{r}_2 }\limits^ \to \times \mathop {\upsilon _2 }\limits^ \to

  • Option 3)

    \mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to = \mathop {\upsilon _1 }\limits^ \to - \mathop {\upsilon _2 }\limits^ \to

  • Option 4)

    \frac{{\mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to }} {{\left| {\mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to } \right|}} = \frac{{\mathop {\upsilon _2 }\limits^ \to - \mathop {\upsilon _1 }\limits^ \to }} {{\left| {\mathop {\upsilon _2 }\limits^ \to - \mathop {\upsilon _1 }\limits^ \to } \right|}}

 

Answers (1)
P Plabita

As we discussed in

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

frac{1}{2}m_{1}u_{1}^{2}+frac{1}{2}m_{2}u_{2}^{2}= frac{1}{2}m_{1}v_{1}^{2}+frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

m_{1},m_{2}:masses

u_{1},v_{1}:initial : and: final : velocity: of : the: mass m_{1}

u_{2},v_{2}:initial : and: final : velocity: of : the: mass m_{2}

 

 Let the farticle A and B collide aat time t their Collision, the position vecters both particled should be same at time t.

r\vec{_{1}}+v\vec{_{1}}t= r\vec{_{2}}+v\vec{_{2}}t

r\vec{_{1}}-r\vec{_{2}}= v\vec{_{2}}t-v\vec{_{1}}t=\left | r\vec{_{1}}-r\vec{_{2}} \right |=\left | v\vec{_{2}}-v\vec{_{1}} \right |t

t= \frac{\left | r\vec{_{1}}-r\vec{_{2}} \right |}{\left | v\vec{_{2}}-v\vec{_{1}} \right |}

\left | r\vec{_{1}}-r\vec{_{2}} \right |=\left | v\vec{_{2}}-v\vec{_{1}} \right |= \frac{\left | r\vec{_{1}}-r\vec{_{2}} \right |}{\left | v\vec{_{2}}-v\vec{_{1}} \right |}

= \frac{\vec{r}_{1}-\vec{r}_{2}}{\left | \vec{r}_{1}-\vec{r}_{2} \right |}\:=\:\frac{\vec{v}_{2}-\vec{v}_{2}}{\left | \vec{v}_{2}-\vec{v}_{1} \right |}


Option 1)

\mathop {\text{r}_1 }\limits^ \to \cdot \mathop {\upsilon _1 }\limits^ \to = \mathop {\text{r}_2 }\limits^ \to \cdot \mathop {\upsilon _2 }\limits^ \to

Incorrect Option

Option 2)

\mathop {\text{r}_1 }\limits^ \to \times \mathop {\upsilon _1 }\limits^ \to = \mathop {\text{r}_2 }\limits^ \to \times \mathop {\upsilon _2 }\limits^ \to

Incorrect Option

Option 3)

\mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to = \mathop {\upsilon _1 }\limits^ \to - \mathop {\upsilon _2 }\limits^ \to

Incorrect Option

Option 4)

\frac{{\mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to }} {{\left| {\mathop {\text{r}_1 }\limits^ \to - \mathop {\text{r}_2 }\limits^ \to } \right|}} = \frac{{\mathop {\upsilon _2 }\limits^ \to - \mathop {\upsilon _1 }\limits^ \to }} {{\left| {\mathop {\upsilon _2 }\limits^ \to - \mathop {\upsilon _1 }\limits^ \to } \right|}}

Correct Option

Exams
Articles
Questions