CBSE Class 7 Science Advanced Physics Kinematics Long Answer Questions Set 01

Long Answer Questions

Teacher's Note:
1. Always label both axes with the quantity and its unit.
2. The slope of this line, 5 m ÷ 1 s, equals the speed of the bike.

Question 2. Describe the experimental method to determine the time period of a simple pendulum.
Answer:
Apparatus: a metal bob, thread, cork, iron stand with clamp, metre scale, wooden blocks and a stop watch.

Procedure
1. Find the radius R of the bob: place the bob between two wooden blocks, measure the gap between the blocks with a scale (this gives the diameter) and halve it.
2. Pass the thread through a vertically cut cork and hold the cork tightly between the clamps of the stand. For a pendulum of length 70 cm, the thread length should be (70 − R) cm, since the length is measured up to the centre of the bob.
3. Pull the bob to one side so that the thread makes about 5° with the vertical, and release it.
4. Start the stop watch when the bob is at the mean or extreme position and note the time taken for 20 oscillations.
5. Divide this time by 20 to get the time period T.
6. Repeat with other lengths such as 80 cm, 90 cm and 100 cm. The time period increases as the length increases.
In simple words: Time 20 swings of a pendulum of known length, divide by 20, and repeat for other lengths.

Teacher's Note:
1. Timing many oscillations reduces the error of the stop watch.
2. Keep the swing small (about 5°) so that the formula for the simple pendulum holds.

Question 3. When pendulum A completes 20 oscillations, pendulum B completes 30 oscillations. What is the ratio of their time periods?
Answer:
Let the time taken be t for both.
T_A = \( \frac{t}{20} \) and T_B = \( \frac{t}{30} \).
\( \frac{T_A}{T_B} = \frac{t/20}{t/30} = \frac{30}{20} = \frac{3}{2} \).
The ratio T_A : T_B = 3 : 2.
In simple words: The pendulum with fewer oscillations in the same time has the longer time period, so A : B = 3 : 2.

Teacher's Note:
1. More oscillations in the same time means a shorter time period.
2. The ratio of time periods is the inverse of the ratio of oscillations.

Question 4. Explain in detail about the different kinds of motion.
Answer:
The motion of bodies is classified as translatory, rotatory and oscillatory.
1. Translatory motion: the whole body moves from one place to another. It is of two types: (a) rectilinear motion along a straight path, for example soldiers marching on a straight road, a car on a straight road or a stone falling freely; (b) curvilinear motion along a curved path, for example a car taking a turn, a bus on a fly-over or a football kicked into the air.
2. Rotatory motion: the body turns about a fixed axis, for example the blades of a ceiling fan, a spinning top, a turbine and the earth around the sun. The wheels of a moving bicycle show rotatory and translatory motion together.
3. Oscillatory motion: the body moves to and fro about a fixed point, for example a child on a swing, a clock pendulum, a sewing-machine needle and an engine piston. A very fast oscillation, in which the body also changes slightly in shape or size, is called vibratory motion (a guitar string, a drum skin).
In simple words: Bodies may move along a line or curve, turn around an axis, or swing to and fro.

Teacher's Note:
1. Give at least two examples for every kind of motion.
2. Translatory = place to place; rotatory = about an axis; oscillatory = to and fro.

Question 5. Explain in detail about periodic motion and non-periodic motion.
Answer:
Periodic motion: a motion that repeats itself after equal intervals of time. All oscillatory motions are periodic, and some rotatory motions are periodic. Examples: the swing of a clock pendulum, the rotation of the earth about its axis (once in 24 h), the motion of the earth around the sun (once in a year) and the motion of a swing. The fixed time for one repeat is the time period.
Non-periodic motion: a motion that repeats itself, if at all, at unequal intervals of time. Examples: a ceiling fan slowing down after it is switched off, a drum rolling up or down an inclined plane, and the wheel of a bicycle when brakes are applied.
Every oscillatory motion is periodic, but a periodic motion need not be oscillatory (for example, the earth going around the sun).
In simple words: Periodic motion repeats in equal times; non-periodic motion does not.

Teacher's Note:
1. Periodic motion is the basis of clocks.
2. Remember the earth around the sun as the example of periodic but not oscillatory motion.

Question 6. Can a particle have (i) zero speed and non-zero velocity? (ii) constant velocity but a varying speed? (iii) constant speed but a varying velocity?
Answer:
(i) No. If the speed is zero, the body is at rest, so its velocity is also zero.
(ii) No. The magnitude of velocity is the speed. If the velocity is constant, its magnitude (the speed) is also constant.
(iii) Yes. When a body moves along a circular path with constant speed, its direction keeps changing, so its velocity is varying even though the speed is constant.
In simple words: Zero speed or constant velocity forces the other quantity to be fixed, but changing direction changes only the velocity.

Teacher's Note:
1. Velocity has two parts, magnitude (speed) and direction.
2. Uniform circular motion is the standard example for part (iii).

Question 7. A physics student observed a train 100 m long crossing a bridge of length 245 m. If the train is moving with a velocity of 54 km h⁻¹, he determines the time taken by it to cross the bridge. Find his answer.
Answer:
Velocity of the train = 54 km h⁻¹ = 54 × \( \frac{5}{18} \) = 15 m s⁻¹.
To cross the bridge, the train must cover a distance equal to the length of the bridge plus its own length = 245 + 100 = 345 m.
Time = \( \frac{345}{15} \) = 23 s.
In simple words: The whole train must pass the bridge, so it travels 345 m at 15 m s⁻¹, which takes 23 s.

Teacher's Note:
1. Crossing starts when the engine enters and ends when the last coach leaves the bridge.
2. Add the length of the train to the length of the bridge.

Question 8. A body travels along a straight line for the first half of the total time with a constant speed of 20 m s⁻¹ and for the second half of the total time with a constant speed of 30 m s⁻¹. Find the average speed of the body.
Answer:
Let the total time be t.
Distance in the first half of the time = 20 × \( \frac{t}{2} \) = 10t.
Distance in the second half of the time = 30 × \( \frac{t}{2} \) = 15t.
Average speed = \( \frac{10t + 15t}{t} \) = 25 m s⁻¹.
In simple words: When the two speeds last for equal times, the average speed is simply the mean of the two speeds.

Teacher's Note:
1. Equal times give the simple mean; equal distances do not.
2. Keep t as a symbol; it cancels.

Question 9. A body starts moving from a point A along the boundaries of a square field as shown in the figure and reaches the point C after 30 s. If its speed is 1 m s⁻¹, what is the shortest distance between A and C?

Answer:
Distance travelled along the boundary (A → B → C) = speed × time = 1 × 30 = 30 m.
This is two sides of the square, so each side x = \( \frac{30}{2} \) = 15 m.
The shortest distance AC is the diagonal = \( \sqrt{15^2 + 15^2} = \sqrt{450} \) = 15√2 ≈ 21.2 m.
In simple words: The body walks 30 m along two sides, so each side is 15 m and the straight line AC is about 21.2 m.

Teacher's Note:
1. The shortest distance between two points is the straight line joining them (the displacement).
2. The diagonal of a square of side x is x√2.

Question 10. Ram and Raj travel along a straight line from X to Y with different uniform speeds. Ram started from X and reaches the place Y in t s. If Raj travels half of the total distance between X and Y in 2t s, determine the ratio of speeds of Ram and Raj.
Answer:
Let the distance XY = d.
Speed of Ram = \( \frac{d}{t} \).
Speed of Raj = \( \frac{d/2}{2t} = \frac{d}{4t} \).
Ratio of speeds Ram : Raj = \( \frac{d}{t} : \frac{d}{4t} \) = 4 : 1.
In simple words: Ram covers the full distance in t seconds, while Raj would need 4t seconds, so Ram is four times as fast.

Teacher's Note:
1. Write both speeds in terms of d and t, then divide.
2. Half the distance in 2t means the whole distance in 4t.

Question 11. A car moving along a straight road with a speed of 72 km h⁻¹ is brought to rest within 3 s after the application of brakes. Calculate the deceleration produced by the brakes.
Answer:
Initial velocity u = 72 km h⁻¹ = 72 × \( \frac{5}{18} \) = 20 m s⁻¹; final velocity v = 0; time t = 3 s.
a = \( \frac{v - u}{t} = \frac{0 - 20}{3} \) = −6.67 m s⁻².
The deceleration produced is 6.67 m s⁻².
In simple words: The brakes reduce the car's velocity by about 6.67 m s⁻¹ every second.

Teacher's Note:
1. The negative sign shows deceleration.
2. Convert the speed to m s⁻¹ first.