9th Standard CBSE Science Model Question paper

Class 9th Science - Force and Laws of Motion Case Study Questions and Answers 2022 - 2023

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Force and Laws of Motion Case Study Questions With Answer Key

9th Standard CBSE

Science

Time : 00:15:00 Hrs

Total Marks : 10

Case Study

We take a glass tumbler and place a thick square card on its mouth as shown in Figure (a). A coin is then placed above this card in the middle. Let us flick the card hard with our fingers. On flicking, the card moves away but the coin drops into the glass tumbler [see Figure (b)].

(i) Give reason for the above observation.
(a) The coin possesses inertia of rest, it resists the change and hence falls in the glass.
(b) The coin possesses inertia of motion; it resists the change and hence falls in the glass.
(c) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.
(d) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.
(ii) Name the law involved in this case.

(a) Newton's second law of motion	(b) Newton's first law of motion.
(c) Newton's third law of motion.	(d) Law of conservation of energy

(iii) If the above coin is replaced by a heavy five rupee coin, what will be your observation. Give reason.
(a) Heavy coin will possess more inertia so it will not fall in tumbler.
(b) Heavy coin will possess less inertia so it will fall in tumbler.
(c) Heavy coin will possess more inertia so it will fall in tumbler.
(d) Heavy coin will possess less inertia so it will not fall in tumbler.
(iv) Name the law which provides the definition of force.

(a) Law of conservation of mass	(b) Newton's third law.
(c) Newton's first law	(d) Newton's second law.

(v) State Newton's first law of motion.
(a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant.
(b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force.
(c) For every action in nature there is an equal and opposite reaction.
(d) The acceleration in an object is directly related to the net force and inversely related to its mass.

The car A of mass 1500 kg, travelling at 25 m/s collides with another car B of mass 1000 kg travelling at 15 m/s in the same direction. After collision the velocity of car A becomes 20 m/s. Calculate the velocity of car B after the collision.

(i) What is the momentum of the car A before collision?

(a) 30000 kg. m/s	(b) 37500 kg. m/s
(c) 15000 kg. m/s	(d) 52500 kg. m/s

(ii) What is the total momentum of car A and car B before collision?

(a) 3750 kg. m/s	(b) 37500 kg. m/s
(c) 15000 kg. m/s	(d) 52500 kg. m/s

(iii) What is the momentum of the car A after collision?

(a) 30000 kg. m/s	(b) 37500 kg. m/s
(c) 15000 kg. m/s	(d) 52500 kg. m/s

(iv) What is the total momentum of car A and car B after collision?

(a) 30000 kg. m/s	(b) 37500 kg. m/s
(c) 15000 kg. m/s	(d) 52500 kg. m/s

(v) What is the velocity of car B after the collision?

(a) 22 m/s	(b) 25 m/s
(c) 20 m/s	(d) None of these

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Answers

Force and Laws of Motion Case Study Questions With Answer Key Answer Keys

Case Study

(i) (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass.
(ii) (b) Newton's first law of motion.
(iii) (c) Heavy coin will possess more inertia so it will fall in tumbler.
(iv) (d) Newton's second law.
(v) (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force.

(i) (b) 37500 kg. m/s
Momentum of car A (before collision) = Mass of car A × Velocity of car A
= 1500 × 25
= 37500 kg.m/s
(ii) (c) 15000 kg. m/s
Momentum of car B (before collision) = Mass of car B × Velocity of car B
= 1000 × 15
= 15000 kg.m/s
Total momentum of car A and car B (before collision) = 37500 + 15000
= 52500 kg.m/s
(iii) (a) 30000 kg. m/s
After collision, the velocity of car A of mass 1500 kg becomes 20 m/s.
So, Momentum of car A (after collision) = 1500 × 20
= 30000 kg.m/s
(iv) (d) 52500 kg. m/s
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
∴ Total momentum after collision = 52500 kg. m/s
(v) (a) 22 m/s
Momentum of car B (after collision) = 1000 × v
= 1000 v kg.m/s
Total momentum of car A and car B (after collision) = 30000 + 1000 v
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision = 52500 kg. m/s
∴ 52500 = 30000 + 1000 v
⇒1000 v = 52500 – 30000
⇒ 1000 v = 22500
⇒ v = 22500/1000
⇒ v = 22.5 m/s

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