CBSE Class 8 Science Advanced Physics Force Short Answer Questions Set 01

Short Answer Type

Question 1. Force can change the direction of a moving body. Explain with the help of some examples.
Answer:
A force can turn a moving body and make it go in a new direction. Examples:
1. A cricketer hits the ball with a bat. The ball changes its speed and its direction.
2. A football player changes the direction of a moving ball by angling his foot.
3. We change the direction of a moving bicycle by applying force on the handle.
In simple words: A force on a moving object can turn it, as when a bat hits a ball.

Teacher's Note:
1. Direction changes only when a force acts sideways or against the motion.
2. Pick two everyday examples when you write the answer.

Question 2. What are Balanced and Unbalanced Forces ?
Answer:
Balanced forces: forces acting on an object that do not change its state of rest, its uniform motion or its direction of motion. Examples: a book resting on a table, a person holding a briefcase, squeezing a lemon. In a tug of war with 300 N on each side the rope does not move.
Unbalanced forces: forces acting on an object that change its state of rest or motion or its direction. Examples: a stone dropped from the hand, a briefcase released from the hand. In a tug of war with 400 N against 300 N the rope moves towards the stronger side.
In simple words: Balanced forces cancel and nothing changes; unbalanced forces leave a net force and the motion changes.

Teacher's Note:
1. Net force zero means balanced.
2. Net force not zero means unbalanced and acceleration follows.

Question 3. A force can change the shape of object. Explain.
Answer:
When a force acts on a body it can squeeze, stretch or bend it.
1. A spring fixed to a wall is stretched when we pull it, so its shape changes.
2. A sheet of paper is crumpled by pressing it, which changes its shape.
3. Squeezing a rubber ball changes its shape.
In simple words: Pulling, pressing or squeezing an object can change its shape.

Teacher's Note:
1. The shape of soft or elastic things changes easily.
2. Springs and rubber return to their shape when the force is removed.

Teacher's Note:
1. Use a two-column table in the exam.
2. Give one example in each column.

Question 5. When a force of 40 N is applied on a body, it moves with an acceleration of 5 ms⁻². Calculate the mass of the body.
Answer:
F = ma, so m = F ÷ a.
m = 40 ÷ 5 = 8 kg.
In simple words: Mass equals force divided by acceleration.

Teacher's Note:
1. Write the formula first, then substitute.
2. Check: 8 × 5 = 40 N.

Question 6. What is the difference between rest & equilibrium ?
Answer:
Rest: a body is at rest when its velocity is zero and its position does not change with time.
Equilibrium: a body is in equilibrium when the net force on it is zero. Then it is either at rest or moving with constant velocity in a straight line.
A body can be at rest for a moment without being in equilibrium. For example, a ball thrown up is momentarily at rest at the top, but gravity still acts on it, so the net force is not zero.
In simple words: Rest is about not moving; equilibrium is about zero net force.

Teacher's Note:
1. Equilibrium includes uniform motion as well as rest.
2. The ball at its highest point is a good example for comparison.

Question 7. Equal forces F act on the isolated bodies A & B as shown. The mass of B is three times that of A. Find the ratio of magnitude of the acceleration.

Answer:
F ÷ m is the acceleration, so a_A = F/m and a_B = F/(3m).
a_A : a_B = 1/m : 1/(3m) = 3 : 1.
So A accelerates three times as much as B.
In simple words: With the same force, the lighter body gets three times the acceleration.

Teacher's Note:
1. Same force means acceleration is inversely proportional to mass.
2. Write the ratio as A : B.

Question 8. How can two forces of 3 N & 4 N combine to give forces of (a) 7 N (b) 1 N
Answer:
(a) 7 N: when both forces act in the same direction, they add: 3 + 4 = 7 N.
(b) 1 N: when they act in opposite directions, they subtract: 4 − 3 = 1 N, in the direction of the 4 N force.
In simple words: Same direction gives the sum, opposite directions give the difference.

Teacher's Note:
1. Only these two cases are possible on a straight line.
2. The 1 N resultant points the way of the larger force.

Question 9. A force acting on a particle of mass 200 g. The velocity of the particle changes from 15 m/s to 25 m/s in 2.5 s. Assuming the force to be constant, find its magnitude.
Answer:
m = 200 g = 0.2 kg.
a = (v − u) ÷ t = (25 − 15) ÷ 2.5 = 4 m/s².
F = ma = 0.2 × 4 = 0.8 N.
In simple words: Find the acceleration first, then multiply by the mass in kg.

Teacher's Note:
1. Convert grams into kilograms before using F = ma.
2. The constant force means constant acceleration.

Question 10. Two cart A & B of mass 10 kg each are placed on a horizontal track. They are joined tightly by a light but strong rope C. A man hold a cart B & pulls it towards the right with a force of 70 N. The total force of friction by the track and the air on each cart is 15 N acting towards the left. Find the acceleration of the carts

Answer:
Forward force = 70 N. Friction on the two carts = 15 + 15 = 30 N backwards.
Net force = 70 − 30 = 40 N.
Total mass = 10 + 10 = 20 kg (the rope is light, so the carts move together).
a = 40 ÷ 20 = 2 m/s² towards the right.
In simple words: The net force 40 N moves both carts of total mass 20 kg, so a = 2 m/s².

Teacher's Note:
1. Treat the two carts joined by a rope as one body.
2. Do not forget to subtract friction on both carts.

Question 11. A force acting on a particle of mass 200 g. The velocity of the particle changes from 15 m/s to 25 m/s in 2.5 s. Assuming the force to be constant, find its magnitude.
Answer:
m = 0.2 kg.
a = (25 − 15) ÷ 2.5 = 4 m/s².
F = 0.2 × 4 = 0.8 N.
In simple words: A change of 10 m/s in 2.5 s is 4 m/s², and 0.2 kg × 4 = 0.8 N.

Teacher's Note:
1. This question appears twice in the book (questions 7 and 11).
2. Change 200 g to 0.2 kg.

Question 12. How can two forces of 3 N & 4 N combine to give forces of (a) 7 N (b) 1 N
Answer:
(a) Act in the same direction: 3 + 4 = 7 N.
(b) Act in opposite directions: 4 − 3 = 1 N.
In simple words: Add them for 7 N, subtract them for 1 N.

Teacher's Note:
1. This question appears twice in the book (questions 6 and 12).
2. Direction decides whether forces add or subtract.

Question 13. A wooden block of mass m₁ accelerates at 10 ms⁻² when a force of 5 N acts on it. Another block of mass m₂ accelerates at 20 ms⁻² when same force acts on it. Find the acceleration if both the blocks are tied together and same force acts on their combination.
Answer:
m₁ = F ÷ a = 5 ÷ 10 = 0.5 kg.
m₂ = 5 ÷ 20 = 0.25 kg.
Combined mass = 0.5 + 0.25 = 0.75 kg.
a = 5 ÷ 0.75 = 6.67 m/s².
In simple words: Find each mass from F = ma, add them, and divide the force by the total.

Teacher's Note:
1. Masses add when blocks are tied together.
2. The answer is smaller than both separate accelerations.

Question 14. A car of mass 1000 kg is moving with a certain speed when a constant braking force 1000 N acts on it for 5 s and the speed of the car reduces to half its original speed. Find the further time required to stop the car, if the same constant force acts.
Answer:
Retardation a = F ÷ m = 1000 ÷ 1000 = 1 m/s².
In 5 s the speed falls by 1 × 5 = 5 m/s, and this is half the original speed u. So u = 10 m/s and the speed after 5 s is 5 m/s.
Further time to stop = 5 ÷ 1 = 5 s.
In simple words: The car loses 5 m/s each 5 s, and 5 m/s is left, so it needs 5 s more.

Teacher's Note:
1. A constant force gives a constant retardation.
2. The speed drop is the same in equal times.

Question 15. Two balls of identical shape and size, but of different materials are acted upon by same force, such that the acceleration of first ball is only one fourth that of the second ball. Calculate the ratio of the mass of the first ball with respect to that of the second ball.
Answer:
For the same force, m ∝ 1/a.
m₁ : m₂ = a₂ : a₁ = 4 : 1.
The ratio is 4 : 1.
In simple words: The ball with one fourth of the acceleration has four times the mass.

Teacher's Note:
1. Same force means mass and acceleration are inversely related.
2. Write the ratio of the first ball to the second.

Question 16. A car having a mass of 1000 kg is moving at a speed of 30 m/s. Brakes are applied to bring the car to rest. If the frictional force between the tyre and the road surface is 5000 N, calculate the time in which the car will come to rest.
Answer:
Retardation a = F ÷ m = 5000 ÷ 1000 = 5 m/s².
Time t = (u − v) ÷ a = (30 − 0) ÷ 5 = 6 s.
In simple words: A retardation of 5 m/s² takes away 30 m/s in 6 s.

Teacher's Note:
1. First find acceleration, then time.
2. Final speed is 0 when the car stops.

Question 17. A ball of mass 10 g is moving with a velocity of 50 m/s on applying a constant force on ball for 2 sec. It acquires a velocity of 70 m/s. Calculate (i) The acceleration of ball (ii) The magnitude of force applied.
Answer:
(i) a = (70 − 50) ÷ 2 = 10 m/s².
(ii) m = 10 g = 0.01 kg, so F = ma = 0.01 × 10 = 0.1 N.
In simple words: The velocity rises 20 m/s in 2 s, giving 10 m/s², and force = 0.01 × 10 = 0.1 N.

Teacher's Note:
1. Change 10 g to 0.01 kg.
2. Part (i) is needed for part (ii).

Question 18. A force produces an acceleration of 5.0 cm/s² in a body of mass 20 g. Find the force acting on the body in Newton.
Answer:
a = 5 cm/s² = 0.05 m/s² and m = 20 g = 0.02 kg.
F = ma = 0.05 × 0.02 = 0.001 N = 10⁻³ N.
(Check in CGS: 20 × 5 = 100 dyne = 100 × 10⁻⁵ N = 10⁻³ N.)
In simple words: Change to SI units first; the force is 0.001 N.

Teacher's Note:
1. Both the CGS and the SI method must give the same answer.
2. This is Illustration 1.1 of the chapter, repeated as an exercise.