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#Long answer type

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15m \; at\; t = 2s.$ The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Let $u_{1}$ be the speed of the 1^st ball, $u_{1} = 2u\; m/s$

& $u_{2}$ be the speed of the second ball, $u_{2} = u \; m/s$

Let $h_{2}$ be the height of the two balls before coming to rest & $h_{1}$ be the height covered by 1 ball before coming to rest.

Now, we know that,

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

$H=\frac{v^{2}}{2g}$

Thus, $h_{1}=\frac{u{_{1}}^{2}}{2g}$

$=\frac{4u^{2}}{2g}$

and $h_{2}=\frac{u^{2}}{2g}$

Now, $h_{1}-h_{2}=15\; \; \; \; \; \; \; \; ........(given)$

Thus, $u=10\; m/s$

$h_{1}=20\; m\; and\; h_{2}=5\; m$

Calculating time for 1^st ball,

$V_{1} = u_{1} + gt$

$0 = 20 - 10t_{1}$

Thus, $t_{1} = 2 s$

Now, calculating time for second ball,

$V_{2} = u_{2} + gt_{2}$

$0 = 10 - 10t_{2}$

This, $t_{2}=1\; s$

Thus, time intervals between these two balls will be,

$=t_{1} - t_{2}$
$=(2-1)$

$= 1\; second.$

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#Long answer type

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by $v(t) = 2t (3 - t); 0<t<3$ and $v(t) = -(t - 3) ( 6 - t)$ for $3 < t < 6s$ on m/s. It repeats this cycle till it reaches the height of 20 m.

a) At what time is its velocity maximum?

b) At what time is its average velocity maximum?

c) At what times is its acceleration maximum in magnitude?

d) How many cycles are required to reach the top?

(a) for the velocity to be maximum,

$\frac{dv(t)}{dt}=0$

$d\frac{[2t(3-t)]}{dt}=0$

$d\frac{6t-2t^{2}}{dt}=0$

$6 - 4t = 0$

$4t = 6$

Thus, $t=\frac{3}{2}$

$= 1.5 s$

(b)Now, average velocity

$V=\frac{x}{t}$

$=6t-2t^{2}$

$\frac{ds(t)}{dt}=6t-2t^{2}$

$ds=(6t-2t^{2})dt$

Let us integrate both the sides from 0 to 3

$\int_{0}^{s}ds=\int_{0}^{3}(6t-2t^{2})dt$

$s=\left [ 6\frac{t^{2}}{2}-2\times\frac{t^{3}}{3} \right ]_{0}^{3}=\left [ 3t^{2}-\frac{2}{3}t^{3} \right ]_{0}^{3}$

$=\left [ 3\times9-\frac{2}{3}\times2 \right ]=27-18$

Thus, $s = 9m$

Now, average velocity $(V_{av})=\frac{9}{3}=\frac{3m}{s}$

$V(t) = 6t - 2t^{2} .......... since \; 0 < t < 3$

$3 = 6t - 2t^{2} .......... since V_{av}=3$

$2t^{2} - 6t + 3 = 0 \; \; \; \; \; \; .............. since ( a=2, b=-6 \; and\; c=3)$

$t=2.36s$

Thus, the average velocity is maximum at 2.36 seconds.

(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.

$Here, v=0$

$V(t) = 6t - 2t^{2}$

$0 = 6t - 2t^{2}$

$2t (3-t) = 0$

$t\neq 0$

Thus, the acceleration is maximum at t = 3 sec.

(d)Now, for 3 to 6 sec

$V(t) = -(t-3) (6-t)$

$\frac{ds}{dt}=(t-3)(6-t)$

$ds=(t^{2}-9t+18)dt$

Integrate from 3 to 6 s

$s_{2}=\int_{3}^{6}(t^{2}-9t+18)dt=\left [ \frac{t^{3}}{3}-\frac{9}{2}t^{2}+18t \right ]_{3}^{6}$

$=\frac{(6)^{3}}{3}-\frac{9}{2}(6)^{2}+18\times6-\left [ \frac{(3)^{3}}{3}-\frac{9}{2}(3)^{2}+18\times3 \right ]$

$=\frac{6\times6\times6}{3}-\frac{9\times6\times6}{2}+108-\frac{3\times3\times3}{2}-54$

$=180-162-63+40.5=18-22.5$

$S_{2}=-4.5\; m$ .............because distance is in downward direction

Thus, net distance $=4.5\; m$

Thus, in three cycle $=4.5(3)$

$=13.5\; m$

Remaining height will be

$20 -13.5 = 6.5 m$

The monkey can climb up to 9m without slipping but in the 4^thcycle it will slip and the height remaining to climb will be 6.5 m.

Net no. of cycle = 4.

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#Long answer type

A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump onto the truck. Human response time is 0.5 s.

Given: (for truck)

$U=72\; km/hr=2\times\frac{5}{18}m/s$

$=20\; m/s$

$V = 0, t = 5s$

$a =?$

We know that.

$V= u + at$

i.e., $0 = 20 + a (5)$

thus, $a = \frac{-20}{5} = \frac{-4m}{s}$

Given : (for car)

$U = 20 \; m/s$

$V =0, t = 3s$

$a = a_{c}$

Again, $v = u + at$

$0 = 20 + a{_{c}}^{3}$

$a_c = \frac{-20}{3} \; m/s^2$

A human takes at least 0.5 seconds to respond; thus, the time taken by the car driver to respond is $(t-0.5)$ sec …. (car takes t time to stop)

$Vc = u + a{_{c}}{t}$

$0 = 20-\frac{20}{3}.(t - 0.5) \; \; \; \; \; \; .......... (i)$

There is no responding time for the truck driver so he applies breaks with passing signal to car's back side; hence,

$V = u + at$

$0 = 20 - 4t \; \; \; \; \; \; .......... (ii)$

From (i) & (ii),

$20 - 4t = 20 - \frac{20}{3} ( t - 0.5)$

$-4t = - \frac{-20}{3} ( t - 0.5)$

$\\12t = 20t - 10\\-20 + 12t = -10\\$

Thus, $-8t = -10$

Thus, $t=\frac{10}{8}$

$=1.25\; sec$

Now the distance covered by the car & the truck in $\frac{5}{4}$ sec will be,

$S = 20 \left ( \frac{5}{4} \right ) + 0.5(-4) \left ( \frac{5}{4} \right )\left ( \frac{5}{4} \right )$

……. $( because \; s = ut + \frac{1}{2} at^{2})$

$= 25 - 3.125$

$=21.875 m$

In the first 0.5 seconds, the car moves at a uniform speed, but after responding, the brakes are applied for 0.5 seconds, and the retarding motion of the car starts.

$S_{c} = (20 \times 0.5) \times 20 (1.25 - 0.5) + \frac{1}{2} \left ( \frac{-20}{3} \right ) (1.25 - 0.5)^{2}$

$= 25 - 1.875$

$=21.875 m$

Thus, $s_{c} - s = 23.125 - 21.875$

$= 1.25 m$

Therefore, the car must be 1.25 metres behind the truck to avoid bumping into it.

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#Long answer type

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.

a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h

b) A typical raindrop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.

c) Estimate the time required to flatten the drop.

d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.

e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two raindrops is 5 cm.

$\\\; h=1\; km=1000\; m\\g=10\; m/s^{2}\\d=4\; mm\; and\; u=0\; m/s$

Thus, $\\r=\frac{4}{2}\; mm\\=2\times10^{-3}m$

(a) Let's find out the velocity of raindrop on the ground’

$\\v^{2}=u^{2}-2as\\=u^{2}-2g(-h)\\=u^{2}+2gh\\=0^{2}+2(10)(1000)$

Thus,

$\\v=100\sqrt{2}\; m/s\\v=100\sqrt{2}\left ( \frac{18}{5} \right )km/hr$

$\\=360\sqrt{2}\; km/hr\\=510\; km/hr$

(b) Momentum of the raindrop when it touches the ground

mass of drop(m) = Vol. × density

$=\frac{4}{3}\pi \; r^{3}\rho$ $(\rho = density\; of\; water)$

Now, density of water $=10^{3}kg/m^{3}$

Thus, $M=\frac{4}{3}\pi (2\times10^{-3})^{3}\times1000$

$=\frac{4}{3}\times\frac{22}{7}\times2\times2\times2\times10^{-9}\times10^{3}$

$=\frac{704}{21}\times10^{-6}kg$

$=3.35\times10^{-5}kg$

Now, we know that Momentum (p) = mv

$p=3.35\times10^{-5}\times100\sqrt{2}$

$=4.7\times10^{-3}kg\; ms^{-1}$

(c) Time required for a drop to be flattened-

$Time=\frac{distance}{speed}$

$=\frac{4\times10^{-3}}{100\sqrt{2}}m\; \; \; \; \; \; .....(distance=4mm=4\times10^{-3})$

$=4\frac{(\sqrt{2})}{100(2)}\times10^{-3}$

$=\frac{2(1.414)}{100}\times10^{-3}$

$=2.8\times10^{-5}\; sec$

(d) Now, we know that,

$Force=\frac{dp}{dt}$

$=\frac{mv-0}{t-0}$

$Force=\frac{4.7\times10^{-3}}{2.8\times10^{-5}}$

$=1.68(10^{2})$

$=168\; N$

(e) Here,

Radius of umbrella $=\frac{1}{2}m\; \; \; \; \; \; .......(since\; its\; \; diameter=1m)$

Thus, Area of umbrella $=\pi R^{2}$

$=\frac{22}{7}.\frac{1}{2}.\frac{1}{2}m^{2}$

$=\frac{11}{14}m^{2}$

Now, the square area covered by one drop

$= (5 \times 10^{-2})^{2}$

$= 25 \times 10^{-4} m^{2}$

Therefore, no. of drops falling on the umbrella $= \frac{\pi R^{2}}{25} \times 10^{-4}$

$=\frac{11(10)^{4}}{14(25)}$

$=0.0314\times10^{4}$

Therefore 314 drops fell on the umbrella.

Thus, the net force on the umbrella $= 314 \times 168N = 52752 N.$

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#Short Answer type

The velocity-displacement graph of a particle is shown in the figure.

a) Write the relation between v and x.

b) Obtain the relation between acceleration and displacement and plot it.

a) Consider the point P(x,v) at any time t on the graph such that angle ABO is $\theta$ such that

$\tan \theta = \frac{AQ}{QP} = \frac{(v_{0}-v)}{x} = \frac{v_{0}}{x_{0}}$

When the velocity decreases from $v_{0}$ to zero during the displacement, the acceleration becomes negative.

$v_{0}-v=\left ( \frac{v_{0}}{x_{0}} \right )x$

$v=v_{0}(1-\frac{x}{x_{0}})$

is the relation between v and x.

$b) a = \frac{dv}{dt} = (\frac{dv}{dt})(\frac{dx}{dx}) = \left ( \frac{dv}{dx} \right )\left ( \frac{dx}{dt} \right )$

$a=\frac{-v_{0}}{x_{0}}v$

$a=\left ( \frac{v{_{0}}^{2}}{x{_{0}}^{2}} \right )x-\left ( \frac{v{_{0}}^{2}}{x_{0}}\right )$

$At \; x = 0$

$a=\frac{-v{_{0}}^{2}}{x^{0}}$

$At \; a = 0$

x=x₀

The points are

$(0, \frac{-v{_{0}}^{2}}{x_{0}})$ and $B(x_{0},0)$

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#Short Answer type

A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

For the first ball-

$U = 0, h = 45m$

$a = g, t=t \; and\; V = v_{1} =?,$

we know that,

$v = u + at$

$v1 = 0 + gt = gt\; downward$

(for second ball)

$V = v_{2}$

$u = 40 m/s, a = -g,$

$t =t$

now, $V = u + at$

$v_{2} = (40 - gt) \; upward$

Now relative velocity of the 1st ball w.r.t. 2^nd

$v_{12} = v_{1} - v_{2} = -gt -(40 - gt) = - gt - 40 + gt = - 40\; m/s$

The speed of one ball increases and the speed of the other decreases with the same rate due to acceleration.

Hence, relative speed = 40 m/s.

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#Short Answer type

A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building?

In vertical motion,

$u_{y} = 0$

$a = 10 m/s^{2}$

$s = 9m$ & $t=t$

now, $s = u_{y}t+\frac{1}{2}at^{2}$

thus, $9=0(t)+\frac{1}{2}(10)(t^{2})$

Therefore,

$t=\sqrt{\frac{9}{5}}$

$=\frac{3}{\sqrt{5}}sec$

Now, the horizontal distance covered by the person will be,

$u_{x}\times\; t=9\left ( \frac{3}{\sqrt{5}} \right )$

$=\frac{27}{\sqrt{5}}=\frac{27}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}$

$=27\times\frac{2.236}{5}$

$=12.07\; m$

Therefore, the person will reach the building which s next farther the first edge by $12.7 - 10 = 2.07\; m$ .

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#Short Answer type

A bird is tossing between two cars moving towards each other on a straight road. One car has a speed of 18 m/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?

We can find out the relative speed of the cars by adding the speed of the two cars

$= 27 + 18$

$= 45 km/hr$

$Time \; taken \: to \: meet = \frac{Distance \; between \; cars}{relative \; speed \; of \; the \; cars}$

$=\frac{36}{45}=\frac{4}{5}$ hours

Hence, the distance that the bird will cover in $\frac{4}{5}$ hours $=36(\frac{4}{5})=28.8\; km.$

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#Short Answer type

A particle executes the motion described by $x(t) = x_{0} (1 - e^{-\gamma t})$ where $t \geq 0, x0 > 0$

a) Where does the particles start and with what velocity?

b) Find maximum and minimum values of $x(t), v(t), a(t)$ . Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.

Here, $x(t)=x_{0}[1-e^{-\gamma t}]$

So, $v(t)=\frac{dx(t)}{dt}$

$=\frac{d}{dt}[x_{0}(1-e^{-\gamma t})]$

$=+x_{0}\gamma e^{-\gamma t}\; \; \; \; \; \; ..........(i)$

&

+ $a(t)=\frac{dv}{dt}$

$=-x_{0}\gamma ^{2}e^{-\gamma t}\; \; \; \; \; \; ..........(ii)$

$(i) x(0) = x_{0} [1 - e^{0}]$

$= x_{0} (1 - 1) =0$

$v(0) = x_{0}\gamma e^{0}$

$=x_{0}\gamma$

Thus, $x=0$ is the starting point of the particle and its velocity is $v_{0}=x_{0}\gamma$

(b) $x (t)\ is,$

Maximum at $t = \infty$ since $t = \infty$ $[x(t)]_{max} = \infty$

Minimum at $t = 0$ since at $t = 0$ , $[x(t)]min = 0$

v(t) is,

maximum at $t = 0$ since $t=0, v(0) = x_{0}\gamma$

minimum at $t = \infty$ since, $t = \infty$ , $v(\infty ) = 0$

a(t) is,

maximum at $t = \infty$ since at $t = \infty$ , $a(\infty )=0$

minimum at $t = 0$ since at $t = 0$ , $a(0)=-x_{0}\gamma ^{2}$

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#Short Answer type

A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards.

a) Plot qualitatively velocity vs time graph

b) Plot qualitatively acceleration vs time graph

If we observe the graph we know that the displacement (x) is always positive. The velocity of the body keeps on increasing till the displacement becomes zero, after that the velocity decreases to zero in the opposite direction till the maximum value of x is reached, viz., smaller than earlier. When the body reaches towards x=0, the velocity increases and acceleration is in the downward direction. And when the body’s displacement is $x>0$ , i.e., the body moves upwards, the direction will be downwards and velocity will decrease, i.e., $a=-g$ .

(a) Velocity time graph

(b) Acceleration time graph.

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JEE Main high-scoring chapters and topics

The velocity-displacement graph of a particle is shown in the figure. a) Write the relation between v and x. b) Obtain the relation between acceleration and displacement and plot it.

Posted by

infoexpert22

A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

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The velocity-displacement graph of a particle is shown in the figure.

a) Write the relation between v and x.

b) Obtain the relation between acceleration and displacement and plot it.

A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards.

a) Plot qualitatively velocity vs time graph

b) Plot qualitatively acceleration vs time graph