A body of mass m 1 elastically collide with another stationary body of mass m 2 begin array - WhineItext b After the collision- the momentum - q-1 c After the collision- K-E- of second - r -1 D For elastic collision- coefficient of - s -0 restitution is - lendarray A- a-r b-s c-s d-rB- a-r b-r c-s d-rC- a - p b - q c - r d - r D- a-p b-p c-s d-r

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Partially Inelastic Collision

A body of mas...

Question

A body of mass $m_{1}$ elastically collide with another stationary body of mass $m_{2}$
$\begin{matrix} Column - I & Column -II (a) After the collision, velocity of second & (p) > 1 body is maximum when \frac{m_{1}}{m_{2}} is (b) After the collision, the momentum & (q) < 1 of second body is maximum when \frac{m_{1}}{m_{2}} is (c) After the collision, K.E. of second & (r) = 1 body is maximum when \frac{m_{1}}{m_{2}} is (D) For elastic collision, coefficient of & (s) = 0 restitution is \end{matrix}$

a - (r) b - (s) c - (s) d - (r)

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a - (r) b - (r) c - (s) d - (r)

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a - (p) b - (q) c - (r) d - (r)

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a - (p) b - (p) c - (s) d - (r)

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Solution

The correct option is C $a - (p) b - (q) c - (r) d - (r)$
From conservation of linear momentum
$m_{1} u + m_{2} \times 0 = m_{1} v_{1} + m_{2} v_{2} . . . . . . . . . . (i)$
$1 = \frac{v_{2} - v_{1}}{u} . . . . . . . . (i i)$
From equal (i) and (ii)
$v_{2} = \frac{2 m_{1} u}{m_{1} + m_{2}}, v_{2} = \frac{2 u}{1 + \frac{m_{2}}{m_{1}}}$
$v_{2}$ is maximum
when $\frac{m_{2}}{m_{1}} = 0, i . e m_{1} >> m_{2}$
(or) $\frac{m_{1}}{m_{2}} > 1$
So $(a) \to (p)$
Momentum of second body $P_{2} = m_{2} v_{2} = \frac{2 m_{1} m_{2} u}{m_{1} + m_{2}}$
$P_{2}$ is maximum, when $\frac{m_{1}}{m_{2}} \to 0$
i.e., $m_{1} << m_{2} o r \frac{m_{1}}{m_{2}} < 1$
So $(b) \to (q)$
After the collision kinetic energy of second body
$K = \frac{1}{2} m_{2} \frac{4 m_{1}^{2} u^{2}}{(m_{1} + m_{2})^{2}}$
$\frac{d K}{d m_{1}} = 0$
For maximum or minima
$m_{1} = m_{2}$ i.e. $\frac{m_{1}}{m_{2}} = 1$
So $(c) \to (r)$
For elastic collision coefficient of resitution $e = 1$
So $(d) \to (r)$

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