दो घर्षणरहित आनत पथ, जिनमें से एक की ढाल अधिक है और दूसरे की ढाल कम है, बिंदु $A$ पर मिलते हैं। बिंदु $A$ से प्रत्येक पथ पर एक-एक पत्थर को विरामावस्था से नीचे सरकाया जाता है (चित्र)। क्या ये पत्थर एक ही समय पर नीचे पहुंचेंगे ? क्या वे वहां एक ही चाल से पहुंचेंगे? व्याख्या कीजिए। यदि $\theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}$ और $h=10\, m$ दिया है तो दोनों पत्थरों की चाल एवं उनके द्वारा नीचे पहुंचने में लिए गए समय क्या हैं ?

No; the stone moving down the steep plane will reach the bottom

first Yes; the stones will reach the bottom with the same speed $v_{ B }=$

$v_{ C }=14 m / s t_{1}=2.86 s ; t_{2}=1.65 s$

The given situation can be shown as in the following figure

Here, the initial height (AD) for both the stones is the same ( $h$ ). Hence, both will have the same potential energy at point $A.$

As per the law of conservation of energy, the kinetic energy of the stones at points $B$ and

C will also be the same, i. e., $\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{2}^{2}$

$v_{1}=v_{2}=v,$ say Where $, m=$ Mass of each

stone $v=$ Speed of each stone at points

$B$ and $C$

Hence, both stones will reach the bottom with the same speed, $v$

For stone I:

Net force acting on this stone is given by:

$F_{net}=m a_{1}=m g \sin \theta_{1}$

$a_{1}=g \sin \theta_{1}$

For stone II:

$a_{2}= g \sin \theta_{2}$

$\because \theta_{2}>\theta_{1}$

$\therefore \sin \theta_{2}>\sin \theta_{1}$

$\therefore a_{2}>a_{1}$

Using the first equation of motion, the time of slide can be obtained as:

$v=u+a t$

$\therefore t=\frac{v}{a} \quad(\because u=0)$

For stone I:

$t_{1}=\frac{v}{a_{1}}$

For stone II:

$t_{2}=\frac{v}{a_{2}}$

$\because a_{2}>a_{1}$

$\therefore t_{2}$

Hence, the stone moving down the steep plane will reach the bottom first.

The speed ( $v$ ) of each stone at points $B$ and $C$ is given by the relation obtained from the law of conservation of energy. $m g h=\frac{1}{2} m v^{2}$

$\therefore v=\sqrt{2 g h}$

$=\sqrt{2 \times 9.8 \times 10}$

$=\sqrt{196}=14 m / s$

The times are given as:

$t_{1}=\frac{v}{a_{1}}=\frac{v}{g \sin \theta_{1}}$$=\frac{14}{9.8 \times \sin 30}=\frac{14}{9.8 \times \frac{1}{2}}=2.86 s$

$t_{2}=\frac{v}{a_{2}}=\frac{v}{g \sin \theta_{2}}$$=\frac{14}{9.8 \times \sin 60}=\frac{14}{9.8 \times \frac{\sqrt{3}}{2}}=1.65 s$