Name: Vertical Circular Motion
Uploaded: 2019-07-19
Duration: 5 min 40 s
Description: Have you seen vertical circular motion? The answer is going to be yes because you have seen the giant wheel's rotation. In this video, let us analyse this mo...

Question

Consider two bodies of masses

m_{1}

and

m_{2}

moving with velocities

u_{1}

and

u_{2}

along the same straight line with

u_{1} > u_{2} .

Find their velocities after the collision, assuming an elastic collision.

Accepted Answer

Let the velocities of $A$ and $B$ After the collision be $u_{1}$ and $u_{2}$ Respectively.

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} . . . (i)$ (Applying law of conservation of linear momentum)

$m_{2} (v_{2} - u_{2}) = m_{1} (u_{1} - v_{1}) . . . (ii)$

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ (Applying law of conservation of $K E$ .)

$u_{1}$

$m_{2} (v_{2}^{2} - u_{2}^{2}) = m_{1} (u_{1}^{2} - v_{1}^{2}) . . . (iii)$

Divide equation (iii) by equation (ii)

$\frac{m_{2} (v_{2}^{2} - u_{2}^{2})}{m_{2} (v_{2} - u_{2})} = \frac{m_{1} (u_{1}^{2} - v_{1}^{2})}{m_{1} (u_{1} - v_{1})}$ .

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

$v_{2} - v_{1} = u_{1} - u_{2} . . . (iv)$

In one-dimensional elastic collision.

Relative velocity of separation after collision = Relative velocity of approach before collision

$\frac{v_{2} - v_{1}}{u_{1} - u_{2}} = 1 = e (coefficient of restiution)$

From (iii),

$v_{2} = u_{1} - u_{2} + v_{1} . . . (v)$

Put this value in (i)

$m_{1} v_{1} + m_{2} (u_{1} - u_{2} + v_{1}) = m_{1} u_{1} + m_{2} u_{2}$

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

$v_{1} (m_{1} + m_{2}) = (m_{1} - m_{2}) u_{1} + 2 m_{2} u_{2}$

$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}} . . . (vi)$

Put this value in (v)

$v_{2} = u_{1} - u_{2} + \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$

$= u_{1} (1 + \frac{m_{1} - m_{2}}{m_{1} + m_{2}}) + u_{2} (\frac{2 m_{2}}{m_{1} + m_{2}} - 1)$

$= u_{1} (\frac{m_{1} + m_{2} + m_{1} - m_{2}}{m_{1} + m_{2}}) + u_{2} (\frac{2 m_{2} - m_{1} - m_{2}}{m_{1} + m_{2}})$

$v_{2} = \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} + \frac{(m_{2} - m_{1}) u_{2}}{m_{1} + m_{2}} . . . (vii)$

Hence, mass of first body is $v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$ And that of second is $v_{2} = \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} + \frac{(m_{2} - m_{1}) u_{2}}{m_{1} + m_{2}}$ .

Consider two bodies of masses $m_{1}$ and $m_{2}$ moving with velocities $u_{1}$ and $u_{2}$ along the same straight line with $u_{1} > u_{2} .$ Find their velocities after the collision, assuming an elastic collision.

Important Questions on Work, Energy and Power

Learn and Prepare for any exam you want !

Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along the same straight line with u1>u2. Find their velocities after the collision, assuming an elastic collision.

Important Questions on Work, Energy and Power

Learn and Prepare for any exam you want !

Consider two bodies of masses $m_{1}$ and $m_{2}$ moving with velocities $u_{1}$ and $u_{2}$ along the same straight line with $u_{1} > u_{2} .$ Find their velocities after the collision, assuming an elastic collision.