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-dire

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 \dots (i)

m u = {m v}_{1} + {m v}_{2}

\Rightarrow v_{2} = u - v_{1} \dots (i i)

{m v}_{2} \frac{l}{2} - \frac{{m l}^{2}}{12} ω = 0

\Rightarrow w = \frac{6 v_{2}}{l} \dots (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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