Two incitned frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given $\theta_{1}=30^{\circ}, \theta_{2}=60^{\circ},$ and $h=10\; m ,$ what are the speeds and times taken by the two stones?

887-38

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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$

887-s38

Similar Questions

Write the equation of total mechanical energy of a body having mass $m$ and stationary at height $H$.

A body of mass $50\, kg$ is projected vertically upwards with velocity of $100 \,m/sec$. $5 \,seconds$ after this body breaks into $20\, kg$ and $30 \,kg$. If $20\, kg $ piece travels upwards with $150 \,m/sec$, then the velocity of other block will be

A uniform chain of length $3\, meter$ and mass $3\, {kg}$ overhangs a smooth table with $2\, meter$ laying on the table. If $k$ is the kinetic energy of the chain in joule as it completely slips off the table, then the value of ${k}$ is (Take $\left.g=10\, {m} / {s}^{2}\right)$

  • [JEE MAIN 2021]

A bomb of mass $12\,\,kg$  at rest explodes into two fragments of masses in the ratio $1 : 3.$  The $K.E.$  of the smaller fragment is $216\,\,J.$  The momentulm of heavier fragment is (in $kg-m/sec$ )

A particle of mass $m$ moving horizontally with $v_0$ strikes $a$ smooth wedge of mass $M$, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to $a$ height $h$. Choose the correct statement related to the wedge $M$