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A smooth slide $AB$ is fixed so that its highest point $A$ is $3{\text{ }}m$ above horizontal ground. $B$ is $h{\text{ }}m$ above the ground. A particle $P$ of mass $0.2{\text{ }}kg$ is released from rest at a point on the slide. The particle moves down the slide and, after passing $B$, continues moving until it hits the ground (see diagram).

The speed of $P$ at $B$ is ${v_B}$ and the speed at which $P$ hits the ground is ${v_G}$.

a) In the case that $P$ is released at $A$, it is given that the kinetic energy of $P$ at $B$ is $1.6{\text{ }}J$. Find

(i) the value of $h$,

(ii) the kinetic energy of the particle immediately before it reaches the ground,

(iii) the ratio ${v_G}{\text{ }}:{\text{ }}{v_B}$.

b) In the case that $P$ is released at the point $X$ of the slide, which is $H{\text{ }}m$ above the ground (see diagram), it is given that ${v_G}{\text{ }}:{\text{ }}{v_B} = 2.55$. Find the value of $H$ correct to 2 significant figures.

پاسخ تشریحی :
نمایش پاسخ

a)(i) $PE$ loss $ = 0.2g\left( {3 - h} \right)$

$\left[ {0.2g\left( {3 - h} \right) = 1.6} \right]$

$h = 2.2$

(ii) $KE$ is $6{\text{ }}J$

(iii) $[{v_G}/{v_B} = {\left( {3/\left( {3 - 2.2} \right)} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$

or ${v_G}/{v_B} = \sqrt {6/1.16} ]$

Ratio is $1.94$

b) $H/\left( {H - 2.2} \right) = {2.55^2}$

$H = 2.6$

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