A particle moving horizonatally collides perpendiculrly at one end of a rod having equal mass and placed on a horizontal surface. The book says that particle will continue to move along the same direction regardless of value of e (coefficient of restitution). I did not understand the logic. please help.


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Question Number 24477 by sushmitak last updated on 18/Nov/17

$A particle moving horizonatally$ $collides perpendiculrly at one end$ $of a rod having equal mass and$ $placed on a horizontal surface .$ $The book says that particle will$ $continue to move along the same$ $direction regardless of value of$ $e (coefficient of restitution) .$ $I did not understand the logic .$ $please help .$
Commented by sushmitak last updated on 18/Nov/17

Commented by sushmitak last updated on 18/Nov/17

$My work so far$ $initial velocity of ball u$ $final velocity of ball v_{1}$ $velocity of center of mass of rod v_{2}$ $angular velocity of rod at CM = ω$ $e = 1$ $v = v_{1} + v_{2}$ $a n g u l a r m o m e n t u m a t C M$ $m u \frac{l}{2} = {m v}_{1} \frac{l}{2} + I w = {m v}_{1} \frac{l}{2} + m \frac{l^{2}}{12} ω$ $\frac{1}{2} {m u}^{2} = \frac{1}{2} {m v}_{1}^{2} + \frac{1}{2} {m v}_{2}^{2} + \frac{1}{2} I ω^{2}$ $Are the equation correct ?$ $thnls for your help$
	Answered by mrW1 last updated on 19/Nov/17

	Commented by mrW1 last updated on 19/Nov/17

	$v e l o c i t y o f b u t t o m o f r o d$ $v_{2}^{'} = v_{2} + \frac{l ω}{2}$ $v_{2}^{'} - v_{1} = e u$ $\Rightarrow v_{2} + \frac{l ω}{2} - v_{1} = e u . . . (i)$ $m u = {m v}_{1} + {m v}_{2}$ $\Rightarrow v_{2} = u - v_{1} . . . (i i)$ ${m v}_{2} \frac{l}{2} - \frac{{m l}^{2}}{12} ω = 0$ $\Rightarrow w = \frac{6 v_{2}}{l} . . . (i i i)$ $(i i) a n d (i i i) i n t o (i) :$ $u - v_{1} + \frac{l}{2} \times \frac{6}{l} (u - v_{1}) - v_{1} = e u$ $4 u - 5 v_{1} = e u$ $\Rightarrow v_{1} = \frac{4 - e}{5} \times u$ $s i n c e 0 ⩽∣ e ∣⩽ 1, \Rightarrow v_{1} > 0, t h a t m e a n s$ $a f t e r t h e h i t t h e b a l l m o v e s a l w a y s$ $i n t h e s a m e d i r e c t i o n a s b e f o r e t h e h i t .$
	Commented by mrW1 last updated on 19/Nov/17

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