Question 513 Marks
A particle is moving in a circular path of radius r with constant speed. Due to change in the direction of motion of the particle continuously, the velocity of the particle is changing. But thekinetic energy of the particle remains the same. Explain why.
AnswerKinetic energy is given by Since $\vec{\upsilon}.\vec{\upsilon}=\text{v}^2$, a scalar quantity, so it is the speed which is taken into account while calculating the kinetic energy of the particle. As the speed is constant, so kinetic energy of the particle will also remain constant.
View full question & answer→Question 523 Marks
A sphere of mass 'm' moving with a velocity u hits another stationary sphere of same mass at rest. If e is the coefficient of restitution. Find the ratio of the velocities of two spheres after the collision.
AnswerAccording to law of conservation of momentaum, $\text{mu}=\text{m}\upsilon_1+\text{m}\upsilon_2$ $\upsilon_1+\upsilon_2=\text{u}$ Also $\upsilon_1-\upsilon_2=-\text{eu}$ Solving eqn. (i) and (ii), we get $\upsilon_1=\frac{\text{u}(1-\text{e})}{2}$ and $\upsilon_2=\frac{\text{u}(1+\text{e})}{2}$ The ratio of velocities, $\frac{\upsilon_1}{\upsilon_2}=\Big(\frac{1-\text{e}}{1+\text{e}}\Big)$.
View full question & answer→Question 533 Marks
A block of mass 30.0kg is being brought down by a chain. If the block acquires a speed of 40.0cm/s in dropping down 2.00m, find the work done by the chain during the process.
AnswerGiven m = 30kg, v = 40cm/sec = 0.4m/sec, s = 2m From the free body diagram, the force given by the chain is,
F = (ma - mg) = m(a - g) [where a = acceleration of the block] $\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$ $=\frac{0.16}{0.4}=0.04\text{m}/\text{sec}^2$ So, work done $\text{W}=\text{Fs}\cos\theta=\text{m}(\text{a}-\text{g})\text{s}\cos\theta$ $\Rightarrow\text{W}=30(0.04-9.8)\times2$ $\Rightarrow\text{W}=-585.5$ $\Rightarrow\text{W}=-586\text{J}$ So, $\text{W}=-586\text{J}$
View full question & answer→Question 543 Marks
Suppose the average mass of raindrops is $3.0 \times 10^{-5} kg$ and their average terminal velocity $9 m s { }^{-1}$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
AnswerEnergy transferred by rain to surface of earth is kinetic energy The velocity of rain or water is $9 m / s$ For mass $m =$ volume $\times$ density $=$ Area of base $\times$ height $\times p =1 m^2 \times 1 m \times 1000=1000 kg$ So, energy transferred by 100 cm rainfall
View full question & answer→Question 553 Marks
A car of mass 1000kg travels up an incline of 1 in 25 at a constant velocity of 50km/ h. What power does the car engine have to develop if there is a resistive force of 300 N opposing the motion?
Answer$\sin\theta=\frac{1}{25}$. force along inclination due to gravitational Force $=\text{mg}\sin\theta$ Resistive force = 300 Net force to overcome $=300+1000\times10\times\sin\theta$ $=300+1000\times10\times\frac{1}{25}=700\text{ N}$ $\text{Power}=\vec{\text{F}}.\vec{\text{v}}=700\times50\times\frac{1000}{3600}=9722.22\text{ J/sec}$
View full question & answer→Question 563 Marks
Define kinetic energy. Prove that K.E. associated with a mass 'm' moving with velocity v $\frac{1}{2}\text{mv}^2$.
AnswerKinetic energy is defined as the energy associated with a body under motion. $\text{W}=\int\text{Fdx}=\int\text{m}\frac{\text{dv}}{\text{dt}}\text{dx}=\int\text{mv}\text{ dv}=\frac{1}{2}\text{mv}^2$ The work done is transformed as kinetic energy for anybody capable of moving.
View full question & answer→Question 573 Marks
One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (figure). Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.
Answer
$\theta=37^\circ,$ l = h = natural length
Let the velocity when the spring is vertical be ‘v’.
$\cos=37^\circ=\frac{\text{BC}}{\text{AC}}=0.8=\frac{4}{5}$
$\text{AC}=(\text{h}+\text{x})=\frac{5\text{h}}{4}$ (because BC = h)
So, $\text{x}=\frac{5\text{h}}{4}-\text{h}=\frac{\text{h}}{4}$
Applying work energy principle $=\frac{1}{2}\text{kx}^2+\frac{1}{2}\text{mv}^2$
$\Rightarrow\text{v}=\text{x}\sqrt{\Big(\frac{\text{k}}{\text{m}}\Big)}=\frac{\text{h}}{4}\sqrt{\frac{\text{k}}{\text{m}}}$ View full question & answer→Question 583 Marks
Give example of a situation in which an applied force does not result in a change in kinetic energy.
AnswerAssume a ball tied to a string and is moving in a vertical circle. Work done by tension force will be zero and hence tension force will not cause any change in KE of ball. Because at any instant of time the displacement is tangential and the force is central in nature, i.e., tension in the string and the small displacement at any instant are perpendicular to each other.
View full question & answer→Question 593 Marks
A body of mass 'M' at rest is struck by a body of mass 'm'. Show that the fraction of K.E. of mass m transferred to the struck particle is $\frac{4\text{mM}}{(\text{m}+\text{M})^2}.$
Answer$\mathrm{m}_1=\mathrm{m} \mathrm{u}_1=u \mathrm{~m}_2=\mathrm{M} \mathrm{u}_2=0, \mathrm{v}_2$= ? $\text{v}_2=\frac{2\text{m}_1\text{u}_1}{\text{m}_1+\text{m}_2}+\frac{\text{m}_2+\text{u}_2}{\text{m}_1+\text{m}_2}$
$=\frac{2\text{mu}}{\text{m}+\text{M}}+0=\frac{2\text{m}\text{u}}{\text{m}+\text{M}}$ K.E of body stuck after collision, $\text{E}_2=\frac{1}{2}\text{m}_2\text{v}_2^2=\frac{1}{2}\text{M}\Big(\frac{2\text{m}\text{u}}{\text{m}+\text{M}}\Big)^2$
$=\frac{2\text{M}\text{m}^2\text{u}^2}{(\text{m}+\text{M})^2}$ Initial K.E., $\text{E}_1=\frac{1}{2}\text{m}_1\text{u}_1^2=\frac{1}{2}\text{m}\text{u}^2$
$\therefore$ fraction of initial K.E. transferred $\frac{\text{E}_2}{\text{E}_1}=\frac{2\text{M}\text{m}^2\text{u}^2}{(\text{m}+\text{M})^2\Big(\frac{1}{2}\text{m}\text{u}^2\Big)}$
$=\frac{4\text{m}\text{M}}{(\text{m}+\text{M})^2}$
View full question & answer→Question 603 Marks
Figure shows a light rod of length l rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth h below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the block moves in a complete circle about the ring?
Answer
The minimum velocity required to cross the height point $\text{c}=\sqrt{2\text{gl}}$
Let the rod released from a height h.
Total energy at A = total energy at B
$\text{mgh}=\frac{1}{2}\text{mv}^2$
$\text{mgh}=\frac{1}{2}\text{m}(2\text{gl})$
[Because v = required velocity at B such that the block makes a complete circle.]
So, h = l View full question & answer→Question 613 Marks
Consider the decay of a free neutron at rest: $n → p + e^-$ Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus. 
[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like $e^-$, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + $e^-$ + v] AnswerThe decay process of free neutron at rest is given as, $n → p + e^-$ From Einstein’s mass-energy relation, we have the energy of electron as $\Delta\text{mc}^2$ Where, $\Delta\text{m}$ = Mass defect = Mass of neutron - (Mass of proton + Mass of electron) c = Speed of light $\Delta\text{m}$ and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the $\beta$-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.
View full question & answer→Question 623 Marks
Work done by a force is given by $\text{W}=\vec{\text{F}.}\vec{\text{S}}\text{ Where}\vec{\text{ F}}$ is the force and $\vec{\text{S}}$is the displacement. Show that:
- Work done is also equal to change in K.E.
- Work done is also equal to change in potential energy using this expression.
Answer
-
m = mass of body at rest
$\vec{\text{ds}}$ = small displacement in the direction of force Small amount of work done by force
$\text{dW}=\vec{\text{F}}.\vec{\text{ds}}$
$=\text{FdS}\cos0^\circ=\text{FdS}$
If is acceleration Produced in the body, then
$\vec{\text{F}}=\text{m}\vec{\text{a}}=\text{m}\frac{\text{dv}}{\text{dt}}$
$\text{dW}=\bigg(\text{m}\frac{\text{dv}}{\text{dt}}\bigg).\text{dS}=\text{m}\bigg(\frac{\text{dS}}{\text{dt}}\bigg)\text{dv}$
$=\text{mv}\text{ dv}$
Total work done by the force
$\text{W}=\int_\limits{0}^{\text{v}}\text{vdv}=\text{m}\bigg[\frac{\text{v}^2}{2}\bigg]_0^\text{v}$
$\text{W}=\frac{1}{2}\text{mv}^2$
- F = mg
As the distance is moved in the direction of force applied
Work done = Force distance
W = F × h = mgh View full question & answer→Question 633 Marks
Water falling from a 50m high fall is to be used for generating electric energy. If $1.8 x 10^5kg$ of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100W lamps can be lit?
Answer$h =50 m, m =1.8 \times 10^5 kg / hr , P =100$ watt, P.E. $= mgh =1.8 \times 10^5 \times 9.8 \times 50=882 \times 10^5 J / hr$ Because, half the energy is converted into electricity. Electrical energy $=\frac{1}{2}\text{P.E.}$ $= 441 \times 10^5 J/hr$
So, power in watt (J/sec) is given by $=\frac{441\times10^5}{3600}$
$\therefore$ number of $100W$ lamps,
that can be lit $=\frac{441\times10^5}{3600\times100}=122.5\approx122$
View full question & answer→Question 643 Marks
Two masses 10 kg and 20 kg are connected by a massless spring. A force of 200 N acts on 20 kg mass. At the instant when the 10 kg mass has an acceleration $12 \mathrm{~m} / \mathrm{s}^2$, what will be the energy stored in the spring? (Given $\mathrm{k}=2400 \mathrm{~N} /$ $\mathrm{m})$.
Answer
Since F = ma
⇒ F = 10 × 12 = 120N
$\therefore\text{x}=\frac{1}{20}$
Energy stored in the spring, $\text{E}=\frac{1}{20}\text{kx}^2$
$\Rightarrow\text{E}=\frac{1}{2}\times2400\times\big(\frac{1}{20}\big)^2$
$\text{E}=\frac{1}{2}\times2400\times\frac{1}{400}=3\text{J}$. View full question & answer→Question 653 Marks
A body of mass 3kg makes an elastic collision with anthor body at rest and continues to move in the original direction with a speed equal to one-third of its original speed. Find the mass of the second body.
AnswerHere, $m _1=3 kg$ Let $u _1= x ms ^{-1}$ and $m _2= m _1 kg_{ u _2=0 \text {, }}$
$\text{v}_1=\frac{\text{x}}{3}\text{ms}^{-1}$
Since collision is elastic, so both momentaum and K.E. remain conserved.
According to law of conservation of linear momentau, $\text{m}_1\text{u}_1+\text{m}_1\text{v}_1+\text{m}_2\text{v}_2$
$3\text{x}+0=\frac{3\text{x}}{\text{x}_3}+\text{m}\text{v}_2$
$\text{m}\text{v}_2=2\text{x}\cdots(1)$ According to the law of conservation of KE.,
$\frac{1}{2}\text{m}_1\text{u}_1^2+\frac{1}{2}\text{m}_1\text{v}_1^2+\frac{1}{2}\text{m}_2\text{v}_2^2$
$\frac{1}{3}\text{x}^2+0=\frac{1}{2}\times3\frac{\text{x}^2}{9}+\frac{1}{2}\text{m}\text{v}_2^2$
$\text{m}\text{v}_2^2=\frac{8\text{x}^2}{3}\cdots(2)$ Dividing (2) by (1), $\text{v}_2=\frac{4\text{x}}{3}$
Put this value in eqn. (1). we get $\text{m}\times\frac{4\text{x}}{3}=2\text{x}$
$\text{m}=\frac{3}{2}=1.5\text{kg}$. Thus, mass of second body is 1.5kg.
View full question & answer→Question 663 Marks
A $60kg$ man skating with a speed of $10m/s$ collides with a $40kg$ skater at rest and they cling to each other. Find the loss of kinetic energy during the collision.
Answer$\text { Mass of the man }=m_1=60 \mathrm{~kg} \text { Speed of the man }=v_1$=$10 \mathrm{~m} / \mathrm{s} \text { Mass of the skater }=m_2=40 \mathrm{~kg} \text { let its velocity }=\mathrm{v}^{\prime} \therefore$
$60 \times 10+0=100 \times \mathrm{v}^{\prime} \Rightarrow \mathrm{v}^{\prime}=6 \mathrm{~m} / \mathrm{s} \text { loss in } \Delta \mathrm{KE}=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2-\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{v}^2$
$\text { K.E. }=\left(\frac{1}{2}\right) 60 \times(10)^2-\left(\frac{1}{2}\right) \times 100 \times 36=1200 \mathrm{~J}$
View full question & answer→Question 673 Marks
Two bodies make an elastic head-on collision on a smooth horizontal table kept in a car. Do you expect a change in the result if the car is accelerated on a horizontal road because of the noninertial character of the frame? Does the equation "Velocity of separation = Velocity of approach" remain valid in an accelerating car? Does the equation "Final momentum = Initial momentum" remain valid in the accelerating car?
AnswerIn case car is accelerated it would affect velocity of both bodies,
- In this case the change in velocity would affect velocity of both bodies. (body moving in direction of car would slow down and other one moving in opposite direction would speed up in case car is accelerated)
- Velocity of separation would be equal to velocity of approach. As only change would be in velocity but everything would remain same.
- Yes final momentum would be still equal to initial momentum as with increase in velocity of one body the velocity of other body does decrease.
View full question & answer→Question 683 Marks
A molecule in a gas container hits a horizontal wall with speed $200ms^{-1}$ and angle $30°$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
AnswerThe momentum of the gas molecule remains conserved whether the collision is elastic or inelastic. The gas molecule moves with a velocity of 200m/s and strikes the stationary wall of the container, rebounding with the same speed. It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.
View full question & answer→Question 693 Marks
Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.
Answer
Given, N = mg
As shown in the figure, $\frac{\text{mv}^2}{\text{R}}=\text{mg}$
$\Rightarrow\text{v}^2=\text{gR}\ \dots(1)$
Total energy at point A = energy at P
$\frac{1}{2}\text{kx}^2=\frac{\text{mgR}+2\text{mgR}}{2}$ [because $v^2$ = gR]
$\Rightarrow\text{x}^2=\frac{{3\text{mgR}}}{\text{k}}$
$\Rightarrow\text{x}=\sqrt{\frac{{3\text{mgR}}}{\text{k}}}$ View full question & answer→Question 703 Marks
An engine is attached to a wagon through a shock absorber of length 1.5 m . The system with a total mass of $50,000 \mathrm{~kg}$ is moving with a speed of $36 \mathrm{~km} \mathrm{~h}^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m . If $90 \%$ of energy of the wagon is lost due to friction, calculate the spring constant.
Answer$\text{KE}=\frac{1}{2}\text{mv}^2$ $\text{m}=50,000\text{kg}$ $\text{v}=36\times\frac{5}{18}\text{m/ s}=10\text{m/ s}$ $\text{KE}=\frac{1}{2}\times50,000\times10\times10$ $\text{KE}=2500000\text{J}$ 90% of KE of wagon lost due to friction by breaks only 10% are passed to spring. KE of spring = 10% of KE wagon $\frac{1}{2}\text{kx}^2=\frac{10}{100}\times2500000$ $\text{x}=1\text{m}$ $\frac{1}{2}\text{k}\times1\times1=250000$ $\text{K}=500000=5\times10^5\text{N/m}$
View full question & answer→Question 713 Marks
A small block of mass 100g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0cm (figure). The spring constant is 100N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2m below the spring?
Answer
m = 100g = 0.1kg, x = 5cm = 0.05m, k = 100N/m
When the body leaves the spring, let the velocity be v,
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{k}\text{x}^2$
$\Rightarrow\text{v}=\text{x}\sqrt{\frac{\text{k}}{\text{m}}}$
$\Rightarrow0.05\times\sqrt{\frac{100}{0.1}}=1.58\text{m}/\text{sec}$
For the projectile motion, $\theta=0^\circ,\text{Y}=-2$
Now, $\text{Y}=(\text{u}\sin\theta)\text{t}-\frac{1}{2}\text{gt}^2$
$\Rightarrow-2=\Big(\frac{-1}{2}\Big)\times9.8\times\text{t}^2$
$\Rightarrow\text{t}=0.63\text{sec}$
So, $\text{x}=(\text{u}\cos\theta)\text{t}$
$\Rightarrow1.58\times0.63=1\text{m}$ View full question & answer→Question 723 Marks
A body of mass $1.0kg$ initially at rest is moved by a horizontal force of $0.5N$ on a smooth frictionless table. Calculate the work done by the force in $10s$ and show that this is equal to the change in kinetic energy of the body.
AnswerAcceleration of a body, $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{0.5}{1.0}=0.5 \mathrm{~ms}^{-2}$ Distance travelled, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$
$\mathrm{s}=0+\frac{1}{2} \times 0.5(10)^2=25 \mathrm{~m}$ Work done $=\mathrm{F} \times \mathrm{s}=0.5 \times 25=12.5 \mathrm{~J} \mathrm{v}=\mathrm{u}+$ at $0+0.5 \times 10=5 \mathrm{~ms}^{-1}$ Change in, $\mathrm{KE}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^2-\mathrm{u}^2\right)=\frac{1}{2} \times 1.0\left(5^2-0\right)=12.51$
View full question & answer→Question 733 Marks
The potential energy function for a particle executing linear simple harmonic motion is given by $\text{V(x)} = \frac{\text{kx}^2}{2},$ where k is the force constant of the oscillator. For $k = 0.5Nm^{-1}$, the graph of V(x) versus x is shown in Show that a particle of total energy $1J$ moving under this potential must ‘turn back’ when it reaches $\text{x} =\pm 2\text{m}.$
AnswerGiven, Potential energy for a particle executing linear simple harmonic motion is, $\text{V(x)}=\frac{1}{2}\text{kx}^2$ Where, $\text{k}=\frac{1}{2}\text{N/m}$
Total energy of particle E = 1J
Since at extreme position total energy is potential energy, $\therefore\text{V}=\frac{1}{2}\text{kx}^2=1\text{J}$ $\Rightarrow\text{x}=\pm\sqrt{\frac{2}{\text{k}}}=\pm\sqrt{4}$ $\Rightarrow\text{x}=\pm2\text{m}$ Hence the results. View full question & answer→Question 743 Marks
In one-dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a onedimensional collision be interchanged if the masses are not equal?
AnswerNo, it’s not possible as total momentum is to be constant thus if heavy body is at rest and light body is moving the light body will move back with same speed in opposite direction. If a heavy body strikes a lighter body at rest the light body would start moving with double as velocity of heavy body but the heavy body would retain its velocity.
View full question & answer→Question 753 Marks
A vertical spring with constant $200N/ m$ has a light platform on its top. When a $500g$ mass is kept on the platform spring compresses $2.5cm$. Mass is now pushed down $7.50cm$ further and released. How far above later position will the mass fly? $(g = 10ms^{-2})$.
AnswerWhen the external force is removed after the push, the mass gets detached when spring obtain its natural length and say, mass m rises h height from the pushed position.
Loss in Potential energy of spring = Gain in gravitational potential energy. $\frac{1}{2}\text{k}[\text{x}^2]=\text{mgh}$ $\frac{1}{2}\times200[0.1]^2$ = 0.5 × 10 × h 1 = 5h ⇒ h = 0.2m View full question & answer→Question 763 Marks
The spring constant of the spring shown in figure is $250N/m.$ Find the maximum compression of the spring. 
AnswerGiven: K = spring constant = 250N/ m Considering a non-elastic collision between blocks. The combined mass moves with the velocity v. $m_1v_1 + m_2v_2 = (m_1 + m_2 )v = ( 5 + 1 ) v 5 \times 15 + 0 = 6v$
$\Rightarrow \text{v}=\frac{75}{6}\text{m/ s}$ Applying conservation of energy, $\frac{1}{2}(\text{m}_1\text{m}_2)\text{v}^2=\frac{1}{2}\text{k}\text{x}^2$
where, x = maximum compression of the spring.
$\Rightarrow6\times\frac{75}{6}\times\frac{75}{6}=250\times\text{x}^2$
$\Rightarrow\text{x}^2=\frac{5625}{250\times6}=\frac{5625}{1500}=3.75$
$\Rightarrow\text{x}=1.93\text{m}$
View full question & answer→Question 773 Marks
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
AnswerWhen two bodies of same mass undergo elastic collisions their velocities get interchanged. In the given situation, bob A is moving with certain speed and bob B is at rest. Therefore, after collision, bob A comes to rest and the bob B starts moving with the speed of bob A. The whole momentum of bob A will get transfer to bob B and so bob A, will not rise at all after the second collision. Therefore bob A will come to rest and bob B will keep on moving.
View full question & answer→Question 783 Marks
A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
AnswerAs block is at rest on inclined plane as shown in figure.
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Force of friction on body is due to the tendency of block M to slide Mg sinq down over the inclined plane. As there is no displacement in block so work done
f by and block is zero. As there is no work, so no dissipation of energy takes place.
View full question & answer→Question 793 Marks
Two bodies A and B having masses me and my respectively have equal kinetic energies. If $p_A$ and $P_B$ are their respective momentaa, then prove that the ratio of momenta is equal to the square root of ratio of respective masses.
AnswerLet $v_A$ and $v_B$_ be the velocities of A and B respectively. Since their kinetic energies are equal, $\therefore\frac{1}{2}\text{m}_\text{A}\upsilon_\text{A}^2=\frac{1}{2}\text{m}_\text{B}\upsilon_\text{B}^2$
$\text{m}_\text{A}\upsilon_\text{A}^2=\text{m}_\text{B}\upsilon_\text{B}^2$
$(\text{m}_\text{A}\upsilon_\text{A})\upsilon_\text{A}=(\text{m}_\text{B}\upsilon_\text{B})\upsilon_\text{B}$
$\text{p}_\text{A}\upsilon_\text{A}=\text{p}_\text{B}\upsilon_\text{B}\Rightarrow\frac{\text{p}_\text{A}}{\text{p}_\text{B}}=\frac{\upsilon_\text{B}}{\upsilon_\text{A}}$ From equation (i), $\frac{\upsilon_\text{A}^2}{\upsilon_\text{B}^2}=\frac{\text{m}_\text{B}}{\text{m}_\text{A}}$
$\frac{\upsilon_\text{A}}{\upsilon_\text{B}}=\sqrt{\frac{\text{m}_\text{B}}{\text{m}_\text{A}}}$ From equation (ii), $\frac{\text{p}_\text{A}}{\text{p}_\text{B}}=\sqrt{\frac{\text{m}_\text{A}}{\text{m}_\text{B}}}$
View full question & answer→Question 803 Marks
Figure shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
Answer
H = 1m, h = 0.5mApplying law of conservation of Energy for point A & B
$\text{mgH}=\frac{1}{2}\text{mv}^2+\text{mgh}$
$\Rightarrow\text{g}=\frac{1}{2}\text{v}^2+0.5\text{g}$
$\Rightarrow\text{v}^22(\text{g}-0.59)=\text{g}$
$\Rightarrow\text{v}=\sqrt{\text{g}}=3.1\text{m}/\text{s}$
After point B the body exhibits projectile motion for which
$\theta=0^\circ,\text{v}=-0.5$
So, $-0.5=(\text{u}\sin\theta)\text{t}-\Big(\frac{1}{2}\Big)\text{gt}^2$
$\Rightarrow0.5=4.9\ \text{t}^2$
$\Rightarrow\text{t}=0.31\text{sec}$
So, $\text{x}=(\text{v}\cos\theta)\text{t}$
$=3.1\times3.1=1\text{m}$
So, the particle will hit the ground at a horizontal distance in from B. View full question & answer→Question 813 Marks
Figure shows a spring fixed at the bottom end of an incline of inclination $37^\circ$. A small block of mass 2kg starts slipping down the incline from a point 4.8m away from the spring. The block compresses the spring by 20cm, stops momentarily and then rebounds through a distance of 1m up the incline. Find.
- The friction coefficient between the plane and the block.
- The spring constant of the spring. Take $g = 10m/s^2.$
Answer
$m = 2kg, s_1 = 4.8m, R = 20cm = 0.2m, s_2 = 1m,$
$\sin37^\circ=0.60=\frac{3}{5},\theta=37^\circ,$
$ \cos37^\circ=.79=0.8=\frac{4}{5},\text{g}=10\text{m}/\text{sec}^2$
Applying work – Energy principle for downward motion of the body
$0-0=\text{mg}\sin37^\circ\times5-\mu\text{R}\times5-\frac{1}{2}\text{kx}^2$
$\Rightarrow20\times(0.60)\times1-\mu\times20\times(0.80)\times1+\frac{1}{2}\text{k}(0.2)^2=0$
$\Rightarrow60-80\mu-0.02\text{k}=0$
$\Rightarrow80\mu+0.02\text{k}=60\ \dots(1)$
Similarly, for the upward motion of the body the equation is,
$0-0=(-\text{mg}\sin37^\circ)\times1-\mu\text{R}\times1+\frac{1}{2}\text{k}(0.2)^2$
$\Rightarrow-20\times(0.60)\times1-\mu\times20\times(0.80)\times1+\frac{1}{2}\text{k}(0.2)^2=0$
$\Rightarrow-12-16\mu+0.02\text{K}=0\ \dots(2)$
Adding equation (i) & equation (ii), we get $96\mu=48$
$\Rightarrow\mu=0.5$
Now putting the value of $\mu$ in equation (1), K = 1000N/m View full question & answer→Question 823 Marks
On complete combustion a litre of petrol gives off heat equivalent to $3 × 107J$. In a test drive a car weighing $1200kg$. including the mass of driver, runs $15km$ per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were $0.5$
AnswerEfficiency of car engine $=0.5 \therefore$ Energy given by car by 1 litre of petrol $=0.5 \times 3 \times 10^7=1.5 \times 10^7$ Work done by car in $15 \mathrm{~km}=\mathrm{F} . \mathrm{s}=\mathrm{f} \times 15000 \mathrm{~J}\{\mathrm{~s}=15 \mathrm{~km}=15000 \mathrm{~m} / \mathrm{l}\}$ This work done by car in only against force of friction as car is going horizontally only, $\mathrm{f} \times 15000=1.5 \times 10^7 \therefore \mathrm{f}=\frac{1.5 \times 10^7}{15000}=10^3 \mathrm{~N}$
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A mass m displaces in a straight line, guided by a machine delivering constant power. Prove that the displacement (x) is proportional to $\text{t}^\frac{3}{2}$
AnswerPower = Constant Fv = Constant $\text{m}\frac{\text{dv}}{\text{dt}}\text{v}=\text{Constant}\text({K})$ $\int\text{vdv}=\frac{\text{k}}{\text{m}}\int\text{dt}$ $\frac{\text{v}^2}{2}=\frac{\text{k}}{\text{m}}\text{t}$ $\text{v}^2=\frac{2\text{k}}{\text{m}}\text{t}\Rightarrow\text{v}=\sqrt{\frac{2\text{k}}{\text{m}}}\text{t}$ $\frac{\text{dx}}{\text{dt}}=\sqrt{\frac{2\text{k}}{\text{m}}}\int\text{t}^\frac{1}{2}\text{dt}$ $\text{x}=\sqrt{\frac{2\text{k}}{\text{m}}}\frac{\text{t}^\frac{3}{2}}{\frac{3}{2}}$ $=\frac{2}{3}\sqrt{\frac{2\text{k}}{\text{m}}}\text{t}^\frac{3}{2}$
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A body of mass M at rest is struck by a moving body of mass m. Show that the fraction of the initial kinetic energy of moving mass m transferred to the struck body is $4Mm /(m + M)^2.$
AnswerHere, $m_1 = m, m_2 = Mu_1 = u$ (say) and $u_2 = 0.$
View full question & answer→Question 853 Marks
A constant force of 2.50N accelerates a stationary particle of mass 15g through a displacement of 2.50m. Find the work done and the average power delivered.
Answer
F = 2.50N, S = 2.5m, m = 15g = 0.015kg.
So, $\text{w} =\text{F} \times \text{S}$
$\Rightarrow\text{a}=\frac{\text{F}}{\text{m}}=\frac{2.5}{0.015}$
$=\frac{500}{3}\text{m/s}^2$
$=\text{F}\times\text{S}\cos0^\circ$ (acting along the same line)
$=2.5\times2.5=6.25\text{J}$
Let the velocity of the body at b = U. Applying work-energy principle $\frac{1}{2}\text{mv}^2-0=6.25$
$\Rightarrow\text{V}=\sqrt{\frac{6.25\times2}{0.015}}=28.86\text{m/sec}$
So, time taken to travel from A to B.
$\Rightarrow\text{t}=\frac{\text{v-u}}{\text{a}}=\frac{28.86\times3}{500}$
$\therefore$ Average power $=\frac{\text{W}}{\text{t}}=\frac{6.25\times500}{(28.86)\times3}=36.1$ View full question & answer→Question 863 Marks
What is the amount of work done by
- A weightlifter in holding a weight of 120 kg on his shoulder for 30 s, and.
- A locomotive against gravity, if it is travelling on a level plane?
Question 873 Marks
Calculate the power of a motor which is capable of raising of water in 5 min from a well 120m deep.
AnswerHere, the volume of water raised V = 2000)L Density of water $\rho=1\text{kg/ L} $ $\therefore$ Mass of water raised $\text{m}=\text{V}\rho=2000\times1=2000\text{kg}$ Power$\text{P}=\frac{\text{W}}{\text{t}}$ $=\frac{\text{mgh}}{\text{t}}$ $=\frac{2000\times9.8\times120}{5\times60}$ $=7840\text{W}$ $=7.840\text{kW}$ $[1\text{kW}=1000\text{W}]$
View full question & answer→Question 883 Marks
A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.
Answer
Let ‘dx’ be the length of an element at a distance × from the table
mass of ‘dx’ length $=\Big(\frac{\text{m}}{\ell}\Big)\text{dx}$
Work done to put dx part back on the table,
$\text{W}=\Big(\frac{\text{m}}{\ell}\Big)\text{dx }\text{g(x)}$
So, total work done to put $\frac{\ell}{3}$ part back on the table,
$\text{W}=\int\limits_0^{\frac{1}{3}}\Big(\frac{\text{m}}{\ell}\Big)\text{gx dx}$
$\Rightarrow\text{W}=\Big(\frac{\text{m}}{\ell}\Big)\text{g}\Big[\frac{\text{x}^2}{2}\Big]_0^\frac{\ell}{3}$
$\Rightarrow\frac{\text{mg}\ell^2}{18\ell}=\frac{\text{mg}\ell}{18}$ View full question & answer→Question 893 Marks
When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground?
AnswerWhen an apple falls from a tree, its gravitational potential energy decreases as it reaches the ground. After it strikes the ground, its potential energy will remain unchanged.
View full question & answer→Question 903 Marks
A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation $\text{v}=\text{a}\sqrt{\text{x}},$ where a is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = d.
AnswerGiven, $\text{v}=\text{a}\sqrt{\text{x}},$ (uniformly accelerated motion) Displacement $\text{s} =\text{d} – 0 =\text{d}$ Putting $\text{x} = 0, \text{v}_1= 0$ Puttin $\text{x}=\text{d},\text{v}_2=\text{a}\sqrt{\text{d}}$ $\text{a}=\frac{\text{v}_2^2-\text{u}_2^2}{2\text{s}}$ $=\frac{\text{a}^2\text{d}}{2\text{d}}=\frac{\text{a}^2}{2}$ Force $\text{F}=\text{ma}=\frac{\text{ma}^2}{2}$ Work done $\text{W}=\text{Fs}\cos\theta$ $\text{W}=\frac{\text{ma}^2}{2}\times\text{d}=\frac{\text{ma}^2\text{d}}{2}$
View full question & answer→Question 913 Marks
The pulley shown in figure has a radius of 20cm and moment of inertia $0.2kg-m^2$. The string going over it is attached at one end to a vertical spring of spring constant 50N/m fixed from below, and supports a 1kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10cm. Take g $= 10m/s_2$. 
Answer
$l = 0.2kg-m^2, r = 0.2m, K = 50N/m,$
$m = 1kg, g = 10ms^2, h = 0.1m$
Therefore applying laws of conservation of energy
$\text{mgh}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{kx}^2$
$\Rightarrow1=\frac{1}{2}\times1\times\text{v}^2+\frac{1}{2}\times0.2\times\frac{\text{v}^2}{0.04}\\+\Big(\frac{1}{2}\Big)\times50\times0.01$ $(\text{x}=\text{h})$
$\Rightarrow1=0.5\text{v}^2+2.5\text{v}^2+\frac{1}{4}$
$\Rightarrow\text{3v}^2=\frac{3}{4}$
$\Rightarrow\text{v}=\frac{1}{2}=0.5\text{m/s}.$ View full question & answer→Question 923 Marks
A block of mass 5.0kg is suspended from the end of a vertical spring which is stretched by 10cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2.0m/s. How high will it rise? Take g = $10m/s^2.$
Answer$m =5 kg, x =10 cm=0.1 m, v =2 m / sec , h =? G =10 m / sec ^2$
So, $\text{k}=\frac{\text{mg}}{\text{x}}=\frac{50}{0.1}=500\text{N}/\text{m}$
Total energy just after the blow $\text{E}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{kx}^2\ \dots(1)$
Total energy a a height h $=\frac{1}{2}\text{k}(\text{h}-\text{x})^2+\text{mgh}\ \dots(2)$
$=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{kx}^2$
$=\frac{1}{2}\text{k}(\text{h}-\text{x})^2+\text{mgh}$ On, solving we can get, H = 0.2m = 20cm
View full question & answer→Question 933 Marks
An electric fan of mass $2kg$ falls from the ceiling of a lift moving down with uniform speed of $2ms^{-1}$. It hits the floor of the lift (length of the lift = $2m$) and does not rebound. How much heat will be produced by the impact?
AnswerMass of the fan, $\mathrm{m}=2 \mathrm{~kg}$ Length of the elevator, $\mathrm{h}=2 \mathrm{~m}$. The kinetic energy acquired by the fan during its fall is: $\mathrm{E}=$ $\mathrm{mgh}=2 \times 9.8 \times 2=39.2 \mathrm{~J}$ The amount of heat produced will also be 39.2 J , as in accordance with the law of conservation of energy, whole of the kinetic energy will be converted into heat.
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A particle of mass 4m which is at rest explodes into three fragments. Two of these fragments each of mass m, are found to move with a speed of v, each in mutually perpendicular direction. What is the total energy released in this process?
Answer
$\sqrt{2}\text{ mv}=(2\text{m})\text{V}$ $\text{V}=\frac{\text{v}}{\sqrt{2}}$ Total K.E. $=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{mv}^2+\frac{1}{2}(2\text{m})\frac{\text{v}^2}{2}$ $=\frac{3}{2}\text{mv}^2$ View full question & answer→Question 953 Marks
Can normal force do a nonzero work on an object. If yes, give an example. If no, give reason.
AnswerYes. Let us consider an elevator accelerating upward with a body placed in it. In this case, the normal reaction offered by the floor of the elevator on the body is greater than the weight of the body acting in the downward direction. If a person is observing this from the ground, then, for him, the normal reaction is doing a positive work, as the elevator is moving upward.
View full question & answer→Question 963 Marks
A simple pendulum of length lm has a wooden bob of mass $1kg$. It is struck by a bullet of mass $10^{-2}kg$ moving with a speed of $2 \times 10^2m/ sec$. The bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back.
AnswerMomentum of the bullet = $10^{-2} \times 2 \times 10^2kg\ ms^{-1}$ Momentum with bob after collision = $(10^{-2} + 1)$. v Since momentum is conserved $\text{v}=\frac{2}{(1.01)}\text{ ms}^{-1}$ According to conservation of energy $\frac{1}{2}(\text{M}+\text{m})\text{v}^2=\text{M}+\text{m})\text{ gh}$
$\therefore\text{h}=\frac{\text{v}^2}{2\text{g}}=\bigg(\frac{2}{1.01}\bigg)^2\times\frac{1}{20}$ $=0.196\text{ m}=0.2\text{ m}$
View full question & answer→Question 973 Marks
A body of mass $2\ kg$ is initially at rest. A constant force of $5\ N$ acts on it for $10\ s.$ Calculate the average power of the force.
AnswerHere, $m = 2\ kg, u = 0, F = 5\ N, t = 10\ s,$
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A block of mass M is pulled along a horizontal surface by applying a force at an angle with horizontal. Coefficient of friction between block and surface is u. If the block travels with uniform velocity, find the work done by this applied force during ú displacement d of the block.
AnswerThe forces acting on the block are shown in Figure. As the block moves with uniform velocity the forces add up to zero.
$\therefore\text{F}\cos\theta=\mu\text{N}\cdots\text{(i)}$ $\text{F}\sin\theta+\text{N}=\text{mg}\cdots\text{(ii)}$ Eliminating N from equation (i) and (ii), $\text{F}\cos\theta=\mu(\text{Mg}-\text{F}\sin\theta)$ $\text{F}=\frac{\mu\text{Mg}}{\cos\theta+\mu\sin\theta}$ Work done by this force during a displacement d $\text{W}=\text{F}.\text{d}\cos\theta=\frac{\mu\text{Mgd}\cos\theta}{\cos\theta+\mu\sin\theta}$ View full question & answer→Question 993 Marks
- What is the scale factor of human relative to monkey in relation to heartbeats?
- What is the monkey's heart rate?
Answer
- Human heart rate is about 70 beats/ min and scale factor is 2.5.
- Monkey's heart rate = 70 ~ 2.5 = 175.
View full question & answer→Question 1003 Marks
If stretch in a spring of force constant k is doubled, calculate:
- Ratio of final to initial force in the spring.
- Ratio of elastic energies stored in the two cases.
- Work done in changing to the state of double stretch.
Answer
- For a given spring F = kx
$\therefore\frac{\text{F}_2}{\text{F}_1}=\frac{\text{kx}_2}{\text{kx}_1}=\frac{2\text{x}}{\text{x}}=2$
- For a given spring, $\text{U}=\frac{1}{2}\text{kx}^2$
$\frac{\text{U}_2}{\text{U}_1}=\frac{\frac{1}{2}\text{kx}_2^2}{\frac{1}{2}\text{kx}_1^2}=\frac{(2\text{x})^2}{\text{x}^2}=4$
- Since work done in stretching the spring is stored in the spring in the form of elastic potential energy of the spring, therefore,
$\text{W}=\text{U}_2-\text{U}_1=\frac{1}{2}\text{kx}_2^2-\frac{1}{2}\text{kx}_1^2$
$\Rightarrow\text{W}=\frac{1}{2}\text{k}\Big[(2\text{x})^2-\text{x}^2\Big]=\frac{3}{2}\text{kx}^2$ View full question & answer→