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

An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000kg is moving with a speed of 36km 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 1.0m. 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 33 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 43 Marks

Suppose the average mass of raindrops is 3.0 × 10^-5kg 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 9m/s For mass m = volume × density

= Area of base × height × p

= 1m² × 1m × 1000

= 1000kg

So, energy transferred by 100cm rainfall Click and drag to move

Click and drag to move

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Question 53 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.

Click and drag to move

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 63 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

Answer

Efficiency of car engine = 0.5

$\therefore$ Energy given by car by 1 litre of petrol = 0.5 × 3 × 10⁷

= 1.5 × 10⁷

Work done by car in 15km = F.s

= f × 15000J {s = 15km = 15000m/ l}

This work done by car in only against force of friction as car is going horizontally only,

f × 15000 = 1.5 × 10⁷

$\therefore\ \text{f}=\frac{1.5\times10^7}{15000}=10^3\text{N}$

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Question 73 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 83 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 93 Marks

A ball of mass m, moving with a speed 2v₀, collides inelastically (e > 0) with an identical ball at rest. Show that:
For head-on collision, both the balls move forward.

Answer

Let the v₁, v₂ 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) $\text{v}_1=-\text{v}_2+2\text{v}_0$

$\text{v}_1=-\text{v}_1-2\text{ev}_0+2\text{v}_0$

$2\text{v}_1=2\text{v}_0-2\text{ev}_0$

$\text{v}_1=\text{v}_0(1-\text{e})$

$\because\ \text{e}<1$

v₀ is positive so the direction of v₁ is same as v₀ or v₁ is in forward direction. Hence proved.

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Question 103 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 113 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 123 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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Physics 3 Marks Question

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