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Sagot :
Solution :
Given :
M = 0.35 kg
[tex]$m=\frac{M}{2}=0.175 \ kg$[/tex]
Total mechanical energy = constant
or [tex]$K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$[/tex]
But [tex]$K.E._{top} = 0$[/tex] and [tex]$P.E._{bottom} = 0$[/tex]
Therefore, potential energy at the top = kinetic energy at the bottom
[tex]$\Rightarrow mgh = \frac{1}{2}mv^2$[/tex]
[tex]$\Rightarrow v = \sqrt{2gh}$[/tex]
[tex]$=\sqrt{2 \times 9.8 \times 0.35}$[/tex] (h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :
[tex]$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$[/tex]
here, [tex]$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$[/tex]
∴ [tex]$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}[/tex]
[tex]$=\frac{0.175}{0.525}+0$[/tex]
= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
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