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Matter Waves

A particle of...

Question

A particle of mass $m$ with an initial velocity $u \hat{i}$ collides perfectly elastically with a mass $3 m$ at rest. It moves with a velocity $v \hat{j}$ after collision, then $v$ is given by:

$v = \frac{1}{\sqrt{6}} u$
$v = \sqrt{\frac{2}{3} u}$
$v = \frac{u}{\sqrt{3}}$
$v = \frac{u}{\sqrt{2}}$

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Solution

The correct option is D
$v = \frac{u}{\sqrt{2}}$

Step 1: Given data:
Mass of $1 s t$ particle, $m_{1}$ $= m$
Mass of $2 n d$ particle, $m_{2}$ $= 3 m$
Initial velocity of $1 s t$ particle, $\vec{u_{1}}$ $= u \hat{i}$
Final velocity of $1 s t$ particle, $\vec{v_{1}}$ $= v \hat{j}$
Initial velocity of $2 n d$ particle, $\vec{u_{2}}$ $= 0$
Final velocity of $2 n d$ particle, $\vec{v_{2}}$ $= v_{x} \hat{i} + v_{y} \hat{j}$
Step 2: Evaluating the given conditions:
We know from the conservation of Kinetic Energy,
$\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2} = \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}$
Putting values in the above equation from given data, we get
$\frac{1}{2} m u^{2} + \frac{1}{2} 3 m {(0)}^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} 3 m {v_{2}}^{2}$
$u^{2} = v^{2} + {v_{2}}^{2}$ … $1$
Step 3. Find the value of $v_{x}$ :
Now, from the conservation of the Linear Momentum in x-direction, we have
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Putting values in above equation from given data, we get
$m u + 3 m (0) = m (0) + 3 m v_{x}$
$u = 3 v_{x}$
$v_{x} = \frac{u}{3}$ … $2$
Step 4. Find the value of $v_{y}$ :
Again, from the conservation of Linear Momentum in y-direction, we have
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Putting values in the above equation from given data, we get
$m (0) + 3 m (0) = m v + 3 m v_{y}$
$v_{y} = - \frac{v}{3}$ … $3$
Step 4. Find the final velocity $v$ of $1 s t$ particle
From Resolution of Vectors, we get
${v_{2}}^{2} = {v_{x}}^{2} + {v_{y}}^{2}$
From eq. $2$ and $3$ , we get
${v_{2}}^{2} = {(\frac{u}{3})}^{2} + {(- \frac{v}{3})}^{2}$
Putting above value of ${v_{2}}^{2}$ in eq. $1$ we get
$u^{2} = v^{2} + 3 ({(\frac{u}{3})}^{2} + {(\frac{v}{3})}^{2})$
⇒ $u^{2} = v^{2} + \frac{(u^{2} + v^{2})}{3}$
⇒ $3 u^{2} = 3 v^{2} + u^{2} + v^{2}$
⇒ $v^{2} = \frac{u^{2}}{2}$
⇒ $v = \frac{u}{\sqrt{2}}$
Hence, Option $(D)$ is the Correct Answer.

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A particle of mass m with an initial velocity ui^ collides perfectly elastically with a mass 3m at rest. It moves with a velocity vj^ after collision, then v is given by:v=16uv=23uv=u3v=u2

A particle of mass $m$ with an initial velocity $u \hat{i}$ collides perfectly elastically with a mass $3 m$ at rest. It moves with a velocity $v \hat{j}$ after collision, then $v$ is given by:

$v = \frac{1}{\sqrt{6}} u$
$v = \sqrt{\frac{2}{3} u}$
$v = \frac{u}{\sqrt{3}}$
$v = \frac{u}{\sqrt{2}}$