## Force and Laws of Motion Class 9 Science Most Important Questions

Hello everyone, in this article, you will all the important questions of chapter 9 Force and Laws of Motion (Class 9 Science). Students must prepare these important questions from Force and Laws of motion to score maximum in their exams. Let's start!

**Q. No. 1) If the net force acting on an object is zero, will the body necessarily be in the rest position?**

**Ans.** No, according to Newton’s First law of motion the body may move uniformly along a straight line.

**Q. No. 2) A goalkeeper in a game of football pulls his hands backward after holding the ball shot at the goal. This enables the goalkeeper to**

**a. Exert larger force on the ball**

**b. Reduce the force exerted by the ball on the hands**

**c. Increase the rate of change of momentum**

**d. Decrease the rate of change of momentum**

**Ans.** Option (d).

**Q. No. 3) Give the difference between balanced and unbalanced forces.**

**Ans.**

Balanced force |
Unbalanced force |

1. Forces acting on a body from opposite directions are equal. | 1. Forces acting on a body from opposite directions are not equal. |

2. It does not change the state of rest or motion of an object. | 2. It does change the state of rest or motion of an object. |

**Q. No. 4) Heavier bodies require greater force to put them in motion as compared to lighter bodies. Why?**

**Ans.** According to Newton’s Second law of motion, F = ma

The more the mass, the more will be inertia hence greater force will be required to move the object.

**Q. No. 5) An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is**

**a. 32 N**

**b. 0 N**

**c. 2 N**

**d. 8 N**

**Ans.** 0 N

(Since no unbalanced force is required to change the motion of an object moving with uniform velocity.)

**Q. No. 6) Two trucks of mass equal to 5000 kg are moving on a road. One is moving with a velocity of 50 m/s while the other is moving with a velocity of 2 m/s. Calculate the force required to stop them in 20 seconds. Which one of them requires more force and why?**

**Ans.** Here, final velocity = 0 (for both trucks)

F_{1} = ma = m(v-u)/t = 5000 x (0 – 50)/20 = - 12500 N

F_{2} = ma = m(v-u)/t = 5000 x (0 – 2)/20 = - 500 N

Therefore, F_{1} > F_{2}, due to greater momentum.

Also see: Class 9 Science Most Important Questions

**Q. No. 7) Why is it difficult to achieve a zero unbalanced force in practical situations? In practice what happens to a rolling marble? How can we reduce the effect of friction on a marble?**

**Ans.** It is difficult to achieve a zero unbalanced force because of the presence of the frictional force acting opposite to the direction of motion.

In practice, the rolling marble stops after traveling some distance.

The effect of frictional force may be reduced by using a smoother marble and a smoother plane and providing a lubricant on top of the planes.

**Q. No. 8) There are three solids made up of aluminium, steel, and wood, of the same shape and same volume. Which of them would have the highest inertia?**

**Ans.** Steel. As mass is the measure of inertia, steel due to its high mass, will have the highest inertia among the three.

**Q. No. 9) Describe our walking in terms of Newton’s Third law of motion.**

**Ans.** When we walk on the ground, our feet push the ground in the backward direction (action) and then the ground pushes our foot in the forward direction (reaction).

**Q. No. 10) i. State the law of conservation of momentum.**

**ii. A man throws a ball of mass 0.4 kg vertically upwards with a velocity of 15 m/s. What will be its momentum at max height?**

**Ans.** i. Law of conservation of momentum: The sum of momenta of the two objects before the collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum.

ii. At max height, velocity = 0 m/s.

Therefore, momentum(p) = mv = 0.4 x 0 = 0 kg m/s.

**Q. No. 11) Water sprinkler used for grass lawns begins to rotate as soon as the water is supplied. Explain the principle on which it works.**

**Ans.** The working of the rotation of the sprinkler is based on the third law of motion. As the water comes out of the nozzle of the sprinkler, an equal and opposite reaction force comes into play. So the sprinkler starts rotating.

**Q. No. 12) The action and reaction forces do not balance each other. Why?**

**Ans.** This is because they act on different bodies.

**Q. No. 13) What will you observe in the below case? Why? Name the law involved in this case.**

**Ans.** When the card is flicked with the finger the coin placed over it falls in the tumbler. This is because the inertia of the coin (inertia of rest) tries to maintain its state of rest even when the card flows off.

The law involved is Newton’s First law of motion. It states that an object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.

**Q. No. 14) A target of mass 400 g moving with a horizontal speed of 10 m/s is hit by a bullet of mass 0.01 kg moving in the opposite direction. If both the bullet and the target come to rest after the collision, calculate the velocity of the bullet at the time of striking the target.**

**Ans.** Mass of target (m_{t}) = 400 g = 400/1000 kg = 0.4 kg

Initial velocity of target (u_{t}) = 10 m/s

Final velocity of target (v_{t}) = 0 m/s

Mass of bullet (m_{b}) = 0.01 kg

Initial velocity of bullet = u_{b} (let)

Final velocity of bullet (v_{b}) = 0 m/s

According to the law of conservation of momentum,

Total momentum before collision = Total momentum after the collision

⇒ m_{t}u_{t} + m_{b}u_{b} = m_{t}v_{t} + m_{b}v_{b}

⇒ 0.4 x 10 + 0.01 x u_{b }= 0.4 x 0 + 0.01 x 0

⇒ 4 + 0.01u_{b}_{ }= 0

⇒ u_{b }= - 4/0.01 = -400 m/s

The negative sign indicated that the bullet and target were moving in opposite directions.

**Q. No. 15) Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in the figure below. Calculate the acceleration and frictional force of the floor on the ball.**

**Ans.** Acceleration (a) = (v-u)/t = (0 – 80)/8 = - 10 ms^{-2}

Force exerted by ball on the floor = ma = (50/1000) x (-10) = -0.5 N

Therefore, the force exerted by floor on the ball = 0.5 N (Newton’s third law of motion)

**Q. No. 16) Why does a fireman struggle to hold a hose-pipe?**

**Ans.** A fireman has to make a great effort to hold a hose-pipe to throw a stream of water on the fire to extinguish it. This is because the stream of water rushing through the hose-pipe in the forward direction with a large speed exerts a large force on the hose-pipe in the backward direction.

**Q. No. 17) Why all cars are provided with seat belts?**

**Ans.** With the application of brakes, the car slows down but our body tends to continue in the same state of motion because of its inertia. A sudden application of brakes may thus cause injury to us by impact or collision with the panels in front. Safety belts are worn to prevent such accidents. Safety belts exert a force on our body to make the forward motion slower.

**Q. No. 18) Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.**

**Ans.** The leaves tend to continue in the state of rest on account of inertia and hence they resist any change in their state of motion. If we vigorously shake its branch, the leaves that are loosely attached to the branch fall off.

**Q. No. 19) Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?**

**Ans.** This is because the sudden application of brakes stops the motion of the bus as well as our feet in contact with the floor of the bus. But the rest of our body opposes this stopping because of its inertia of motion.

The opposite happens when the bus accelerates from rest. We tend to fall backwards because the sudden start of the bus brings motion to the bus as well as to our feet in contact with the floor of the bus. But the rest of our body opposes this motion because of its inertia of rest.

**Q. No. 20) i. Why do fielders pull their hands gradually with the moving ball while holding a catch?**

**ii. Why are athletes made to fall either on a cushioned bed or on a sand bed in a high jump athletic event?**

**Ans.** i. This is done so that the fielder increases the time during which the high velocity of the moving ball decreases to zero. If the ball is stopped suddenly then its high velocity decreases to zero in a very short interval of time. Thus the rate of change of momentum of the ball will be large. And the fielder will have to apply large force for holding the catch that may hurt the palm of the fielder.

ii. This is done so as to increase the time of the athlete’s fall to stop after making the jump. This decreases the rate of change of momentum and hence the force.

**Q. No. 21) i. Deduce Newton’s Second law of motion mathematically.**

**Or,**

**Using the second law of motion, derive the relation between force and acceleration.**

**ii. Define momentum. Is momentum a scalar or a vector quantity? State its direction. Write its SI unit.**

**Ans.** i. Let us assume an object of mass, m is moving along a straight line with initial velocity, u. It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F throughout the time, t.

Initial momentum (p_{1}) = mu

Final momentum (p_{2}) = mv

Change of momentum = mv – mu = m(v-u)

Rate of change of momentum = m(v-u)/t

According to the Second law of motion,

Applied force ∝ rate of change of momentum

F ∝ m(v-u)/t

F ∝ ma

F = kma, if k = 1

F = ma.

ii. The momentum, p of an object is defined as the product of its mass, m, and velocity, v.

Momentum has both direction and magnitude, so it is a vector quantity.

Its direction is the same as that of velocity.

The SI unit of momentum is kg ms^{-1}.

**Q. No. 22) Derive the unit of force using the second law of motion and define it. A force of 5 N produces an acceleration of 8 ms ^{-2} on a mass m_{1} and an acceleration of 24 ms^{-2} on a mass m_{2}. What acceleration would the same force provide if both the masses are tied together?**

**Ans.** According to Newton’s Second law of motion,

Force = Mass x Acceleration

1 unit of force = 1 kg x 1 ms^{-2}

Unit of force is kg ms^{-2} or newton (N).

1 N is defined as the amount of force that produces an acceleration of 1 ms^{-2} in an object of 1 kg mass.

m_{1} = F/a_{1} = 5/8 kg

m_{2} = F/a_{2} = 5/24 kg

New mass (M) = 5/8 + 5/24 = 5/6 kg

Acceleration produced in M,

a = F/M = 5/(5/6) = 6 ms^{-2 }

**Q. No. 23) A body of mass 5 kg is moving with an acceleration of 5 ms ^{-2}. What is the rate of change of momentum?**

**Ans.** According to Newton’s second law of motion,

Rate of change of momentum = applied force = ma = 5 x 5 = 25 N.

**Q. No. 24) What is the principle of working of a rocket?**

**Ans.** It is based on the principle of the law of conservation of momentum.

**Q. No. 25) A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100 ms ^{-1}. Calculate the velocity of the recoil of the gun.**

**Ans.**

Mass of bullet, m_{b} = 100 g = 0.1 kg

Initial velocity of bullet, u_{b}= 0 ms^{-1}

Final velocity of bullet, v_{b} = 100 ms^{-1}

Mass of gun, m_{g} = 20 kg

Initial velocity of gun, u_{g}= 0 ms^{-1}

Let Final velocity of gun = recoil velocity of gun = v_{g}

Initial momentum = m_{b}u_{b} + m_{g}u_{g} = 0.1x0 + 20x0 = 0 kg ms^{-1}

Final momentum = m_{b}v_{b} + m_{g}v_{g} = 0.1x100 + 20 x v_{g} = 10 + 20v_{g} kg ms^{-1}

According to the law of conservation of momentum,

Initial momentum = Final momentum

0 = 10 + 20v_{g}

v_{g}= -1/2 = -0.5 ms^{-1}

The negative sign shows that the direction of the recoil velocity of gun is opposite to the direction of the velocity of bullet.

**Q. No. 26) Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms ^{-2}. With what force does each person push the motorcar?**

**(Assume that all persons push the motorcar with the same muscular effort.)**

**Ans.**

Mass of motorcar (m) = 1200 kg

Let force applied by 1 person = F

A.T.Q.,

Force applied by 2 persons = Frictional force = 2F (since the car moves with uniform velocity, which means the force applied by 2 persons is balanced by the frictional force.)

Now, Force applied by 3 persons = 3F

Net force acting on the car = Force applied by 3 persons – Frictional Force = 3F – 2F = F

Therefore,

By the Second law of motion,

F = ma = 1200 x 0.2 = 240 N

*Watch Explanation of this question below:*

**Q. No. 27) A hammer of mass 500 g, moving at 50 ms ^{-1}, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?**

**Ans.** Mass of hammer (m) = 500 g = 500/1000 kg = 0.5 kg

Initial velocity of hammer (u) = 50 ms^{-1}^{ }

Final velocity of hammer (v) = 0 ms^{-1}

t = 0.01 s

By the Second law of motion,

Force applied by hammer on nail = ma = m (v- u)/t = 0.5 (0 - 50)/0.01 = - 2500N

Force of the nail on the hammer = 2500 N (By Newton's Third Law of motion).

*In addition to the above questions, also prepare the question given in the NCERT Textbook.*

**Watch the Detailed Explanation of this Chapter here:**

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