Q&A - Ask Doubts and Get Answers

#Long answer type

A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m × 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he walk only at a speed of v m/s. What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Since PQ is the diagonal, we can use Pythagoras theorem to find out,

$PQ=\sqrt{50\times 50+50\times 50}=50\sqrt{2}\; meters$

$AC=\sqrt{100\times 100+100\times 100}=100\sqrt{2}\; meters$

Now let us calculate the time taken by the man to travel the path A-P-Q-C,

$T1=\frac{AC-PQ}{1}+\frac{PQ}{v}=\frac{100\sqrt{2}-50\sqrt{2}}{1}+\frac{50\sqrt{2}}{v}$

$T1=50\sqrt{2}[1+\frac{1}{v}]$

Now let us calculate the time taken by the man to travel the path A-R-C,

$T2=2\; AR$

$AR=\left ( \frac{100\sqrt{2}}{2} \right )^{2}+\left ( \frac{50\sqrt{2}}{2} \right )^{2}$

$AR=25\sqrt{10}$

$T2=50\sqrt{10}$

The case when ,

$T1>T2$

We get,

$50\sqrt{2}\left [ 1+\frac{1}{v} \right ]<50\sqrt{10}$

$\frac{1}{v}<\sqrt{5}-1$

$v<0.82\; m/s$

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infoexpert23

#Long answer type

A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle $\theta$ to the horizontal, find

a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator

b) what will be time of flight?

c) what is the distance from the point of projection at which the ball will land?

d) find $\theta$ at which he should throw the ball that would maximise the horizontal range as found c)

e) how does $\theta$ for maximum range change if $u>v_{o}=v_{o}$ and $u<v_{o}$

f) how does $\theta$ in e) compare with that for u = 0?

a) u is the horizontal velocity with which the cricketer runs. The ball is thrown by him while running and hence the speed of ball also contains a component of the cricketer’s speed.

$U_{x}=u+v\; \cos \; \theta$

Vertical component,

$U_{y}=v\; \sin \; \theta$

$\tan \theta =\frac{v\; \sin \theta }{u+v\cos \theta }$

$\theta =\tan ^{-1}[\frac{v\; \sin\; \theta }{u+v\; \cos \; \theta }]$

b) Time of flight

$S_{y}=U_{y}t+\frac{1}{2}a_{y}t^{2}$

Since the ball returns back to the same position, Sy = 0

$U_{y}=v\; \sin \theta$

So, $0=v\; \sin \theta (T)-\frac{1}{2}g\; T^{2}$

$T[v\; \sin \theta -\frac{1}{2}g\; T]=0$

Since T cannot be zero, we have

$T=2v\; \sin \frac{\theta }{g}$

c) Maximum range

for the max range, the condition is

$\frac{dR}{d\theta} =0$

$d\frac{\left \{ \frac{v}{g}[2u]\sin \; \theta +\frac{v}{g}\; \sin 2\theta \right \}}{d\; \theta }=0$

$\theta =\cos ^{-1}\left [ \frac{-u\pm \sqrt{u^{2}+8\; v_{0}^{2}}}{4v_{0}} \right ]$

$\cos \; \theta =\left [ \frac{-u\pm \sqrt{u^{2}+8\; v_{0}^{2}}}{4v_{0}} \right ]$

e) In the case when $u=v,\cos \theta =$

$\frac{-v_{0}\pm \sqrt{v_{0}^{2}+8v_{0}^{2}}}{4v_{0}}$

$=\frac{-v_{0}+3v_{0}}{4v_{0}}$

$\cos\; \theta =-1+\frac{(-3)}{4}$

$\cos\; \theta =\frac{1}{2}$ (as $\theta$ is taken as an acute angle here)

hence, $\theta =60^{o}$

for the case of u << v

$\cos \theta =-u+\frac{(-2\sqrt{2}v)}{4v}$

Since $\theta$ is an acute angle here,

as u << v here, we can neglect the last term

$\cos \theta =\frac{1}{\sqrt{2}}$

$\theta =\frac{\pi }{4}$

For u >> v,

$\cos\; \theta =\frac{-u+(-u)}{4v}$

$\cos\; \theta =0=\cos 90$

$\theta =\frac{\pi }{2}$

f) when $u=0$

$\cos\; \theta =\frac{-u\pm \sqrt{u^{2}+8v_{0}^{2}}}{4v_{0}}$

$\cos\; \theta =\frac{2\sqrt{2}v}{4v}=\frac{1}{\sqrt{2}}$

$\cos\; \theta =45$

$\cos\; \theta =\frac{\pi }{4}$

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#Long answer type

A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s (Fig. 4.8).

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on south bank and reach opposite point B on north bank,

(a) which direction should he swim?

(b) what will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

a) If swimmer starts swimming due north, what will be his resultant velocity

$V_{s}=4\; m/s$ due north

$V_{r}=4\; m/s$ due east

Now since both the directions are perpendicular,

$\left | V_{r} \right |^{2}=4^{2}+3^{2}=5\; m/s$

$\tan \; \theta =\frac{V_{r}}{V_{s}}=0.75=36^{o}54'$ in the North direction

b) The swimmer wants to start from point A on the south bank and reaches the opposite point B on the north bank

The swimmer makes an angle $\theta$ with the north.

From the figure we have the relation,

$V^{2}=v{_{s}}^{2}-v{_{r}}^{2}=16-9=7$

Henece $v=\sqrt{7}\; m/s$

Now we calculate the value of θ through the below formula,

$\tan \theta =\frac{v_{r}}{v}=\frac{3\sqrt{7}}{7}=1.13$

So, $\theta =48^{o}29'30''$ in th edirection from North to West

c) we need to find from the above two scenarios that for the swimmer to reach the opposite bank in the shorter time

we know that the velocity component perpendicular to the river is 4m/s

let us assume the width of the river to be ‘w’

Time taken - North

$\frac{w}{4}=t1$

time taken in part b) when $v=\sqrt{7}\; m/s$

$\frac{w}{\sqrt{7}}=t2$

taking ratio,

$\frac{t1}{t2}=\frac{(\frac{w}{4})}{(\frac{w}{\sqrt{7}})}$

$4\; t1=\sqrt{7}\; t2$

Now as, $4>\sqrt{7}$

$t1<t2$

So, the swimmer will take a shorter time in case a)

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#Long answer type

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at $45^{o}$ to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

Let the north direction be i and south direction be j

The velocity of rain is $v=a\hat{i}+b\hat{J}$

Case 1 (v=5i)

The velocity of rain with respect to the girl is:

$v_{r}-v_{g}=(a\hat{i}+b\hat{j})-5\hat{i}=(a-5)\hat{i}+b\hat{j}$

Since the horizontal component is zero, $a-5=0 \; or\; a=5$

Case 2 (v=10i)

$v_{r}-v_{g}=(a\hat{i}+b\hat{j})-10\hat{i}=(a-10)\hat{i}+b\hat{j}$

The angle of rain appears to be 45 degrees.

$\tan 45=\frac{b}{a}=\frac{b}{-5}$

$b=-5$

So,

$\left | v_{r} \right |=\sqrt{5^{2}+(-5)^{2}}=\sqrt{50}=5\sqrt{2}\; m/s$

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#Long answer type

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle $\theta$ with speed $v_{o}$ and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.

The particle rebounds from P. when it strikes plane inclined at v0 speed. Hence the speed of particle after it rebounds from P will be v0 We assume the new axis X’OX and YOY’ axis at P as origin ‘O’. The components of g and v0 in the new OX and OY axis are:

Focusing on the motion of the particle from O to A,

$s=ut+\frac{1}{2}gt^{2}$

Here, t=T which is the time of flight

$0=T[v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T]$

So, either $T=0\; or \left [ v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T \right ]=0$

$S_{x}=u_{x}t+\frac{1}{2}a_{x}t^{2}$

$L=[\frac{2v}{g}]v\; \sin \theta +\frac{1}{2}g\; \sin\; \theta [\frac{2v}{g}]^{2}$

$L=\frac{4v^{2}}{g}.\sin \theta$

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#Long answer type

A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal.

a) find an expression of range on the plane surface

b) time of flight

c) $\beta$ at which range will be maximum

a) expression of range on the plane surface

Now, $a_{y}=-g\; \cos \alpha$ and $a_{x}= g\; \sin \alpha$

$Y=0$ at O and P,

So

$U_{y}=v\; \sin \beta$ where t = T

We calculate Time of Flight Part (b) before Part (a)

b) motion of a particle along Y axis

$s=ut+\frac{1}{2}gt^{2}$

$s=0,u=v\; \sin \beta , g=a=-g\; \cos \alpha ,t=T$

$0=v\; \sin \beta .T+\frac{1}{2}(-g\; \cos \alpha )T^{2}$

$T[v\; \sin \beta -T.\frac{g}{2}\cos \alpha ]=0$

$\frac{Tg}{2}.\cos \alpha =v\sin \beta$

So, $T=2v\; \sin \beta /g.\cos \alpha$

a) Now we continue the part a

$T=2v\; \sin \beta /g.\cos \alpha$

$s=ut+\frac{1}{2}gt^{2}$

$L=v\; \cos \beta (T)+\frac{1}{2}(-g\; \sin \alpha )T^{2}$

$L=2\; v^{2}\; \sin \beta[\cos \beta \cos \alpha -\sin \beta \sin \alpha ]/g \cos ^{2}\alpha$

$L=2\; v^{2}\; \sin \beta \times \cos [\alpha +\beta ]/g \cos ^{2}\alpha$

c) on the axis X, L will be maximum when $\sin \beta \cos [\alpha +\beta ]$ will be maximum

let $z=\sin \beta \cos [\alpha +\beta ]$

$=\sin \beta [\cos \beta \cos \alpha -\sin \beta \sin \alpha ]$

$=\frac{1}{2}[\cos \alpha\; 2\sin\; \beta +\sin \alpha \cos \; 2\; \beta-\sin \alpha ]$

$=\frac{1}{2}[\sin(2\beta +\alpha )-\sin\; \alpha ]$

In order to make Z maximum, we put $[\sin(2\beta +\alpha )-\sin\; \alpha ]=1$

Opening the brackets, we get $(2\beta +\alpha )=90$

$2\beta =90-\alpha$

$\beta =45-\frac{\alpha }{2}$ radian

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#Long answer type

A gun can fire shells with maximum speed $v_{0}$ and the maximum horizontal range that can be achieved is $R=v{_{0}}^{2}/g$

If a target farther away by $\Delta x$ beyond R has to be hit with the same gun as shown in the figure. Show that it could be achieved by rating the gun to a height at least

$h=\Delta x[1+\frac{\Delta x}{R}]$

The solution to this problem can be given in two ways:

i) the target is present at the horizontal distance of $(R+\Delta x)$ and is h meters below the projection point. $Y=-h$

ii) the motion of projectile starting from point P till reaching point T. vertical height covered is -h and horizontal range is $\Delta x$

max range of a projectile is given by $R=\frac{v^{2}}{g}$

here, $\theta =45$

we assume that the gun is raised to a height h to hit the target T

total range $=(R+\Delta x)$

the horizontal component of velocity is, $v\; \cos \theta$

horizontal velocity at $P=V_{x}=-V\; \cos \theta$

vertical velocity at $P=V_{y}=V\sin \theta$

now, $h=ut+\frac{1}{2}at^{2}$

So, $h=-V\; \sin \theta t+\frac{1}{2}gt^{2}\; \; \; \; \; \; ---------------(1)$

$V\; \cos \theta .t=(R+\Delta x)$

$t=\frac{(R+\Delta x)}{V\; \cos \theta }$

Substituting the value of t in (1) we get,

$h=-V\; \sin \theta \left ( \frac{(R+\Delta x)}{V\; \cos \theta } \right )+\frac{1}{2}g\left ( \frac{(R+\Delta x)}{V\; \cos \theta } \right )^{2}$

$h=-(R+\Delta x)+\frac{1}{R}(R^{2}+\Delta x^{2}+2R\; \Delta x)=-R-\Delta x+R+\frac{\Delta x^{2}}{R}+2\Delta x$

$h=\Delta x[1+\frac{\Delta x}{R}]$

hence proved.

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#Long answer type

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g = 10 m/s².

Packet speed = 125 m/s, height of hill = 500m

In order to cross the hill, the vertical component of the packet should be reduced to make the height of 500m attainable. The distance between canon and hill should also be half of that of the packet’s range.

$V^{2}-u^{2}=2gh$