3 Marks Question

Question 13 Marks

A raindrop of mass $1.00g$ falling from a height of $1km$ hits the ground with a speed of $50m s^{-1}$. Calculate:

The loss of $P.E.$ of the drop.
The gain in $K.E.$ of the drop.
Is the gain in $K.E.$ equal to loss of $P.E.$? If not why.

Take $g = 10m s^{-2}$

Answer

Drop $m = 0.001kg, h = 1km = 1000m$ Speed of $v = 50m/ s u = 0$
$PE$ at highest point of drop $= mgh = 0.001 × 10 × 1000 = 10J$ So loss $pf \ PE = 10J$

Gain in $\text{KE}=\frac{1}{2}\text{mv}^2$
$=\frac{1}{2}\times0.001\times50\times50=1.250$
Gain in $= 1.250J$

Gain in $KE$ is not equal to the loss in $PE$. It is due to the loss of $PE$ or $KE$ against resistance or dragging force of air.

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Question 23 Marks

In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls $($i.e., when they are in contact$)$.

Kinetic energy.
Total linear momentum?

Give reason for your answer in each case.

Answer

When two billiard balls collide each other then their linear momentum and kinetic energy remains conserved. Because here it is considered that there is not any non conservative force $($like air resistance$/$ friction on surface etc.$)$ and speed of ball is not so high so that they deformed on collision.

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Question 33 Marks

Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Answer

When elevator is descending then it is not its free fall under gravity it descends with uniform speed. Power is required to decrease the velocity due to free fall.
Power of motor or system of an elevator is constant and a limited or specified power can stop the speed of freely falling of passenger along with elevator.

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Question 43 Marks

A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.

Answer

Work done by a body moving along closed loop can be zero if only conservative force acting on the body during motion. Work done by a body moving along a loop is not zero if any non-conservative force, i.e., frictional, electrostatic, magnetic force are acting on body.

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Question 53 Marks

A ball of mass m, moving with a speed $2v_0,$ collides inelastically $(e > 0)$ with an identical ball at rest. Show that:
For head-on collision, both the balls move forward.

Answer

Let the $v_1, v_2$ are the velocities of the two balls after the collision. Now by the principle of law of conservation of momentum.$\text{mv}_0=\text{mv}_1+\text{mv}_2$
$2\text{v}_0=\text{v}_1+\text{v}_2\ ...(\text{i})$
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{v}_0+\text{v}_0}$
$\text{v}_2-\text{v}_1=2\text{ev}_0$
From (i) $\text{v}_1=-\text{v}_2+2\text{v}_0$$\text{v}_1=-\text{v}_1-2\text{ev}_0+2\text{v}_0$
$2\text{v}_1=2\text{v}_0-2\text{ev}_0$
$\text{v}_1=\text{v}_0(1-\text{e})$
$\because\ \text{e}<1$
$v_0$ is positive so the direction of $v_1$ is same as $v_0$ or $v_1$ is in forward direction. Hence proved.

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Question 63 Marks

A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify

Answer

For a body falling freely under gravity. The mechanical energy is not conserved because some part of mechanical energy utilized against force of friction of air molecules which is non conservative force. But if a body is falling freely under gravity in vacuum, the total mechanical energy remain conserved.

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Question 73 Marks

An adult weighing 600N raises the centre of gravity of his body by 0.25m while taking each step of 1m length in jogging. If he jogs for 6km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.

Answer

Energy used up by raising the centre of gravity by 0.25m by jogger in one step = mgh mg = 600N h = 0.25m$\therefore$ Number of steps in $6\text{Km}=\frac{600\text{m}}{1\text{m}}=600\text{ steps}$
Energy utilised in $6000\text{m}=6000\times600\times0.25\text{J}$ Since 10% of energy utilised in jogging.$\therefore$ Energy utilised in jogging $=\frac{10}{100}\times6000\times600\times0.25$
$=360000\times0.25$
$=90000\text{J}=9\times10^4\text{J}$

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Question 83 Marks

Suppose the average mass of raindrops is $3.0 \times 10^{-5}$ kg and their average terminal velocity $9m s^{-1}$. 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.

Answer

Energy 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

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Question 93 Marks

Give example of a situation in which an applied force does not result in a change in kinetic energy.

Answer

Assume 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.

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Question 103 Marks

An engine is attached to a wagon through a shock absorber of length $1.5 m$ . The system with a total mass of $50,000 kg$ is moving with a speed of $36 km h ^{-1}$ 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
$\frac{1}{2}\text{kx}^2=\frac{10}{100}\times2500000$
$\text{x}=1\text{m}$
$\frac{1}{2}\text{k}\times1\times1=250000$
$\text{K}=500000=5\times10^5\text{N/m}$

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Question 113 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?

Answer

As block is at rest on inclined plane as shown in figure.

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.

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Question 123 Marks

On complete combustion a litre of petrol gives off heat equivalent to $3 \times 107J$. 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 $0.5$

Answer

Efficiency 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 $15 km= F . s = f \times 15000 J\{ s =15 km=15000 m / l \}$
This work done by car in only against force of friction as car is going horizontally only, $f \times 15000=1.5 \times 10^7$
$\therefore\ \text{f}=\frac{1.5\times10^7}{15000}=10^3\text{N}$

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