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

Short Answer Questions

Question 1. A car travels a certain distance with 42 km h⁻¹ for 20 minutes and the remaining distance at 60 km h⁻¹ for 30 minutes. What is its average speed of the whole journey?
Answer:
Distance in the first part = 42 × \( \frac{20}{60} \) = 14 km.
Distance in the second part = 60 × \( \frac{30}{60} \) = 30 km.
Total distance = 14 + 30 = 44 km; total time = 20 + 30 = 50 min = \( \frac{5}{6} \) h.
Average speed = \( \frac{44}{5/6} \) = 52.8 km h⁻¹.
In simple words: The car covers 44 km in 5/6 of an hour, so its average speed is 52.8 km h⁻¹.

Teacher's Note:
1. Change minutes to hours before multiplying by km h⁻¹.
2. Add the distances and the times, then divide; never average the two speeds.

Question 2. The distance-time graph of a car is as shown in the figure.
Is the car moving with constant speed? Find the speed of the car at the end of 2 hour and 6 hour, respectively.
Answer:
The graph is not a single straight line (the slope changes at 2 h), so the car is not moving with constant speed.
From 0 to 2 h: 100 km in 2 h, so the speed is \( \frac{100}{2} \) = 50 km h⁻¹ (the average speed up to the end of 2 h is also 50 km h⁻¹).
From 2 h to 6 h: (150 − 100) = 50 km in 4 h, so the speed is 12.5 km h⁻¹. The average speed over the whole 6 h = \( \frac{150}{6} \) = 25 km h⁻¹.
In simple words: The car goes quickly at first (50 km h⁻¹) and then slows down, so the speed is not constant.

Teacher's Note:
1. The speed on any part of a distance-time graph is the distance on that part divided by its time.
2. A bend in the graph shows that the speed has changed.

Question 3. A lorry driver sees an old man crossing the road. Suddenly he applies brakes and stops the lorry within 30 s. If the initial speed of the lorry is 108 km h⁻¹, find its deceleration.
Answer:
Initial velocity u = 108 km h⁻¹ = 108 × \( \frac{5}{18} \) = 30 m s⁻¹; final velocity v = 0; time t = 30 s.
a = \( \frac{v - u}{t} = \frac{0 - 30}{30} \) = −1 m s⁻².
So the deceleration is 1 m s⁻².
In simple words: The lorry loses 1 m s⁻¹ of speed every second until it stops.

Teacher's Note:
1. Convert km h⁻¹ into m s⁻¹ first.
2. A negative acceleration means deceleration; quote its magnitude.

Question 4. Give two examples for each of the motion along a straight line, circular motion and periodic motion.
Answer:
Motion along a straight line: (i) a car moving on a straight road, (ii) a stone dropped from a tower.
Circular motion: (i) the blades of a ceiling fan, (ii) a spinning top.
Periodic motion: (i) the pendulum of a clock, (ii) the earth going around the sun.
In simple words: Straight, circular and repeating motions each have everyday examples.

Teacher's Note:
1. Periodic motion repeats at equal intervals of time.
2. Circular motion goes round a fixed axis or centre.

Question 5. Define day and year. How do watches and clocks measure time? Explain.
Answer:
Day: the time the earth takes to turn once on its own axis (24 hours).
Year: the time the earth takes to go once around the sun (about 365¼ days).
Watches and clocks measure time with a periodic motion that repeats in equal intervals. Old wall clocks use the to-and-fro swing of a simple pendulum, and the clock counts the swings. Modern quartz clocks and watches use an electric circuit and are more accurate than pendulum and winding clocks.
In simple words: Day and year come from the earth's motions, and clocks count a regular repeating motion.

Teacher's Note:
1. Remember: day = earth's rotation, year = earth's revolution around the sun.
2. Any steady repeating motion can be used to keep time.

Question 6. What are the different points one must keep in mind while choosing a suitable scale for drawing a graph? Discuss.
Answer:
While choosing a scale we keep these points in mind:
1. The difference between the highest and the lowest values of the quantities, so that the whole range fits on the paper.
2. The intermediate values, so that they can be plotted easily with the scale chosen.
3. The scale should use as much of the graph paper as possible.
Example: for a bus journey of 80 km in 2 hours on a 20 cm × 25 cm sheet, 1 km = 1 cm needs an 80 cm axis (too long), and 10 km = 1 cm gives a small graph. The suitable scale is 5 km = 1 cm for distance and 6 min = 1 cm for time.
In simple words: Choose a scale that fits the paper, is easy to plot and fills the page well.

Teacher's Note:
1. Try one or two scales and check the length of the axis for each.
2. Easy scales such as 1, 2, 5 or 10 units per cm are the best.

Question 7. Define velocity and explain how it is different from speed.
Answer:
Velocity is speed in a specified direction.
Differences: (1) Speed has only magnitude, so it is a scalar; velocity has magnitude and direction, so it is a vector. (2) Speed is distance ÷ time; velocity is displacement ÷ time. (3) Speed can never be negative or zero for a moving body, whereas velocity changes when the direction changes even if the speed is constant.
Example: 10 m s⁻¹ is a speed, but 10 m s⁻¹ towards the east is a velocity.
In simple words: Velocity is speed together with the direction of travel.

Teacher's Note:
1. A body moving in a circle at constant speed has a changing velocity.
2. State the direction whenever you quote a velocity.

Question 8. Derive the relation between acceleration, change in velocity and time.
Answer:
Let a body have initial velocity u, and let its velocity become v after a time t.
Change in velocity = v − u.
Change in velocity per unit time = \( \frac{v - u}{t} \).
By definition this is the acceleration a, so \( a = \frac{v - u}{t} \), that is, v − u = at and v = u + at.
In simple words: Acceleration is the change in velocity divided by the time taken for that change.

Teacher's Note:
1. Write the initial velocity as u and the final velocity as v.
2. Check that the unit is m s⁻¹ ÷ s = m s⁻².

Question 9. What are uniform speed and non-uniform speed? Explain, with suitable examples.
Answer:
Uniform speed: a body covers equal distances in equal intervals of time, however small the intervals are. Example: a car covering 10 m in every second.
Non-uniform speed: a body covers unequal distances in equal intervals of time, or equal distances in unequal intervals. Example: a bus after the brakes are applied, or a car in city traffic.
In simple words: Same distance every second is uniform speed; different distances every second is non-uniform speed.

Teacher's Note:
1. Both words 'equal' matter in the definition of uniform speed.
2. The graph of uniform speed is a straight line.

Question 10. Slow and fast are relative. Explain it with examples.
Answer:
A body is called fast or slow only by comparing it with another body.
Example 1: a bicycle at 15 km h⁻¹ is fast compared with a walking man (5 km h⁻¹) but slow compared with a car (60 km h⁻¹).
Example 2: a car is slow compared with an aeroplane but fast compared with a bullock cart.
So the same body can be fast in one comparison and slow in another.
In simple words: Whether something is fast or slow depends on what we compare it with.

Teacher's Note:
1. Use numbers to make the comparison clear.
2. The same idea is true for rest and motion.

Question 11. Rest and motion are relative. Explain.
Answer:
A body is at rest or in motion with respect to the chosen surroundings (the observer).
Example: for a person sitting in a moving bus, the fellow passengers do not change their position, so they are at rest. For a person standing on the ground, the same passengers change their position and are in motion.
Hence one body can be at rest for one observer and in motion for another; rest and motion are relative.
In simple words: Whether a body is moving depends on who is watching.

Teacher's Note:
1. Always name the reference point.
2. A passenger is at rest with respect to the bus, in motion with respect to the road.

Question 12. Define different kinds of velocity. Give one example of each kind.
Answer:
Uniform velocity: the speed and direction both stay the same. Example: an aeroplane flying east at 500 km h⁻¹.
Variable velocity: the speed, or the direction, or both change. Example: a car taking a turn.
Average velocity: total displacement ÷ total time taken, used when the velocity is variable. Example: a bus whose journey of 60 km displacement takes 2 h has an average velocity of 30 km h⁻¹ in the direction of the displacement.
In simple words: Velocity can be steady, changing or averaged over a trip.

Teacher's Note:
1. Name the three kinds: uniform, variable and average.
2. Average velocity uses displacement, not distance.

Question 13. Explain how will you change the time period of a simple pendulum.
Answer:
The time period T = 2π√(ℓ/g) depends only on the length ℓ of the pendulum (at a given place).
To increase the time period, make the pendulum longer; to decrease it, make the pendulum shorter. Since T ∝ √ℓ, the length must be made four times to double the time period.
Changing the mass, size or shape of the bob does not change the time period.
In simple words: Change the length of the thread; that is the only practical way to change the time period.

Teacher's Note:
1. Remember T ∝ √ℓ, not ℓ.
2. The mass of the bob does not matter.

Question 14. In a race, two athletes A and B reach the finishing point in 20 second and 22 second, respectively. (a) What is the ratio of their speeds? (b) If both are allowed to run with their respective speeds for a given time, what is the ratio of the distance covered by them?
Answer:
Let the race distance be d. Speed of A = \( \frac{d}{20} \), speed of B = \( \frac{d}{22} \).
(a) Ratio of speeds = \( \frac{d/20}{d/22} = \frac{22}{20} \) = 11 : 10.
(b) In the same time, distance = speed × time, so the distances are in the same ratio as the speeds, 11 : 10.
In simple words: A is faster in the ratio 11 to 10, so A also covers 11 parts for every 10 parts covered by B.

Teacher's Note:
1. For equal distances, speed is inversely proportional to time.
2. For equal times, distance is directly proportional to speed.

Question 15. The odometer of a motorcycle shows a reading of 10,237 km at 8.30 a.m. At 9.00 a.m., it shows a reading 10,267 km. What is the speed of the motorcycle in km h⁻¹?
Answer:
Distance = 10,267 − 10,237 = 30 km.
Time = 8.30 a.m. to 9.00 a.m. = 30 min = 0.5 h.
Speed = \( \frac{30}{0.5} \) = 60 km h⁻¹.
In simple words: The motorcycle covers 30 km in half an hour, so it moves at 60 km h⁻¹.

Teacher's Note:
1. Subtract the two odometer readings for the distance.
2. Half an hour is 0.5 h.

Question 16. A simple pendulum of length ℓ and time period T on the earth is taken onto the surface of the moon. How should the length of the simple pendulum be changed on the moon such that the time period is constant? (Take g_e = 6 g_m)
Answer:
On the earth: T = 2π√(ℓ/g_e). On the moon with a new length ℓ′: T = 2π√(ℓ′/g_m).
For the same T: \( \frac{\ell'}{g_m} = \frac{\ell}{g_e} \), so ℓ′ = ℓ × \( \frac{g_m}{g_e} \) = \( \frac{\ell}{6} \).
The length must be reduced to one-sixth of its value on the earth.
In simple words: Gravity on the moon is weaker, so the pendulum must be six times shorter to keep the same time period.

Teacher's Note:
1. Equate the two expressions for T and cancel 2π.
2. Weaker gravity makes a pendulum slower, so shorten it to compensate.

Question 17. The length of a seconds pendulum is doubled. What happens to the time period of the pendulum?
Answer:
The time period of a seconds pendulum is T = 2 s, and T ∝ √ℓ.
If the length is doubled, T′ = T × √2 = 2 × 1.414 = 2.83 s (about 2.8 s).
The time period increases by a factor of √2 (about 1.41 times).
In simple words: A pendulum that is twice as long swings about 1.4 times slower.

Teacher's Note:
1. The time period depends on the square root of the length.
2. Do not double the time period; it goes up only by √2.

Question 18. What is a simple pendulum? Explain the terms time period, mean position and extreme positions of a simple pendulum.
Answer:
Simple pendulum: a small metallic ball (the bob) hung from a rigid support by a light thread.
Mean position: the position of the bob when it hangs at rest (point O).
Extreme positions: the two end points (A and B) up to which the bob swings on either side of the mean position.
Time period: the time taken by the bob for one complete oscillation, for example O → B → O → A → O.
In simple words: A pendulum is a ball on a string; its middle rest position, its two end positions and the time for one swing are the key terms.

Teacher's Note:
1. Draw the pendulum with O, A and B to explain the terms.
2. The time period is measured for a full to-and-fro swing.