Two massless string of length 5, m hang from | vedclass

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Question

$m_{B}=0.5 \mathrm{Kg}$

$m_{A}=0.25 K g$

$h=0.45 m$

velocity of ball A just before it collides with $\mathrm{B}$ is $v_{A i}$

$m_{A} g h=\

Vedclass · Accepted Answer

$1\, ms^{-1}$ to the left

Vedclass · Answer

$m_{B}=0.5 \mathrm{Kg}$

$m_{A}=0.25 K g$

$h=0.45 m$

velocity of ball A just before it collides with $\mathrm{B}$ is $v_{A i}$

$m_{A} g h=\frac{1}{2} m_{A}\left(v_{A i}ight)^{2}$

Hence $v_{A i}=3$

$E_{B}=1 J$

Now, after collision velocity of ball B can be obtained from

$E_{B}=\frac{1}{2} m_{B}\left(v_{B}ight)^{2}$

$\Rightarrow\left(v_{B}ight)^{2}=\frac{2}{m_{B}}$

$	herefore\left(v_{B}ight)^{2}=\frac{2}{0.5}$

$	herefore\left(v_{B}ight)^{2}=4$

$	herefore v_{B}=2$

since, in collision elastic momentum is conserved hence,

$m_{A} v_{A i}=m_{A} v_{A}+m_{B} v_{B}$

$.25 	imes 3=.25 	imes v_{A}+.5 	imes 2$

$.75=.25 v_{A}+1$

$v_{A}=-1 m s^{-1}$

Hence, the velocity of ball A just after the collision is $1 \mathrm{ms}^{-1}$, to the left.

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