Science Ncert_exemplar 2019 Solutions for Class 9 Science Chapter 9 Force And Laws Of Motion are provided here with simple step-by-step explanations. These solutions for Force And Laws Of Motion are extremely popular among Class 9 students for Science Force And Laws Of Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science Ncert_exemplar 2019 Book of Class 9 Science Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Science Ncert_exemplar 2019 Solutions. All Science Ncert_exemplar 2019 Solutions for class Class 9 Science are prepared by experts and are 100% accurate.

#### Page No 61:

#### Question 1:

Which of the following statement is * not *correct for an object moving along a straight path in an accelerated motion?

(a) Its speed keeps changing

(b) Its velocity always changes

(c) It always goes away from the earth

(d) A force is always acting on it

#### Answer:

For an object moving along a straight path in accelerated motion, it is not necessary that it always goes away from the earth.

Hence, the correct answer is option C.

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#### Question 2:

According to the third law of motion, action and reaction

(a) always act on the same body

(b) always act on different bodies in opposite directions

(c) have same magnitude and directions

(d) act on either body at normal to each other

#### Answer:

According to the third law of motion, action and reaction always act on different bodies in opposite directions.

Hence, the correct answer is option B.

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#### Question 3:

A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to

(a) exert larger force on the ball

(b) reduce the force exerted by the ball on hands

(c) increase the rate of change of momentum

(d) decrease the rate of change of momentum

#### Answer:

The goalkeeper pulls his hands backward after holding the ball to decrease the rate of change of momentum by increasing the time. By doing this, less force is exerted on his hands (Force is directly proportional to the rate of change of momentum).

Hence, the correct answer is option B and D.

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#### Question 4:

The inertia of an object tends to cause the object

(a) to increase its speed

(b) to decrease its speed

(c) to resist any change in its state of motion

(d) to decelerate due to friction

#### Answer:

The inertia of an object tends to cause the object to resist any change in its state of rest or motion.

Hence, the correct answer is option C.

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#### Question 5:

A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is

(a) accelerated

(b) uniform

(c) retarded

(d) along circular tracks

#### Answer:

If the coin falls behind the passenger that means the train is accelerated. When the coin is tossed it has the same velocity as that of the train but during the time it is in the air its velocity becomes less than that of the train (because the train is accelerated), so it falls behind the passenger.

Hence, the correct answer is option A.

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#### Question 6:

An object of mass 2 kg is sliding with a constant velocity of 4 m s^{–1 }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

#### Answer:

Given, mass *m* = 2 kg, velocity *v *= 4 ms^{-1}

As the object is moving with a constant velocity i.e., 4 ms^{-1} so the acceleration of the object is zero (*a* = 0) and according to the property of inertia if there is no external force acting on the body, then body remains as it is i.e., if the body is at rest, it remains in rest and if it is in motion, it remains in motion.

Hence, the correct answer is option B.

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#### Question 7:

Rocket works on the principle of conservation of

(a) mass

(b) energy

(c) momentum

(d) velocity

#### Answer:

Rocket works on the conservation of momentum. In a rocket, the fuel burns and produces gas at high temperature. These gases are ejected out of the rocket from a nozzle at the backside of the rocket. The ejecting gas exerts a forward force on the rocket which helps in accelerating.

Though the mass of gases escaping per second is very small and their momentum is very large due to their tremendous velocity of escape. An equal and opposite momentum is imparted to the rocket which despite its large mass builds up a high velocity.

Hence, the correct answer is option C.

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#### Question 8:

A water tanker filled up to $\frac{2}{3}$ of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would

(a) move backward

(b) move forward

(c) be unaffected

(d) rise upwards

#### Answer:

On the sudden application of the brake, the tanker will come in the state of rest but the water remains in the state of motion, so the water will move forward.

Hence, the correct answer is option B.

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#### Question 9:

There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?

#### Answer:

As the mass is a measure of inertia, the ball of the same shape and size, having more mass than other balls will have the highest inertia. Since, the steel has the greatest density and greatest mass, therefore, it has highest inertia.

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#### Question 10:

Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.

#### Answer:

When the train is stopped suddenly, then it comes in the state of rest but the balls remain in the state of motion. Due to the inertia of motion, the balls move in the forward direction.

As the balls are of the same size but of different materials that means their masses will be different. So, both the balls will move with different speeds.

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#### Question 11:

Two identical bullets are fired one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?

#### Answer:

As both the bullets are identical and are fired with the same force. So, to reduce recoil velocity, we use heavy rifle so that its heavy mass compensates the large momentum. But for light rifle, recoil velocity will be large due to small mass thus light rifle will hurt more to the shoulder.

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#### Question 12:

A horse continues to apply a force in order to move a cart with a constant speed. Explain why?

#### Answer:

When the cart starts moving, frictional force starts working on the wheel of the cart opposite to the motion. So, the horse needs to apply continuous force in the forward direction to maintain a constant speed.

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#### Question 13:

Suppose a ball of mass *m *is thrown vertically upward with an initial speed *v, *its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed *v *again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?

#### Answer:

The momentum of a system remains conserved if no external force acts on the system. In the given example, there is gravitational force acting on the ball which is an external force, so it is not an example of conservation of momentum.

#### Page No 63:

#### Question 14:

Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in Figure. Calculate the acceleration and frictional force of the floor on the ball.

#### Answer:

$\mathrm{Given},\mathrm{mass},m=50\mathrm{g}=0.05\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{velocity},u=80\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{velocity},v=0\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Time}\mathrm{taken},t=8\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{first}\mathrm{equation}\mathrm{of}\mathrm{motion},v=u+at\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration},a=\frac{v-u}{t}=\frac{0-80}{8}=-10\mathrm{m}/{\mathrm{s}}^{2}(\mathrm{Negative}\mathrm{sign}\mathrm{shows}\mathrm{retardation})\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{frictional}\mathrm{force}\mathrm{of}\mathrm{the}\mathrm{floor}\mathrm{on}\mathrm{the}\mathrm{ball}\phantom{\rule{0ex}{0ex}}F=ma=0.05\times (-10)=-0.5\mathrm{N}(\mathrm{Negative}\mathrm{sign}\mathrm{shows}\mathrm{against}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{motion})$

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#### Question 15:

A truck of mass *M *is moved under a force *F*. If the truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change?

#### Answer:

$\mathrm{Let},\mathrm{initial}\mathrm{mass},{m}_{1}=M$

$\mathrm{Initial}\mathrm{Force},{F}_{1}=F\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question},\mathrm{new}\mathrm{mass},{m}_{2}=M+M=2M\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{new}\mathrm{force},{F}_{2}=\frac{F}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Change}\mathrm{in}\mathrm{acceleration}:\phantom{\rule{0ex}{0ex}}\frac{{a}_{1}}{{a}_{2}}=\frac{{F}_{1}}{{F}_{2}}\times \frac{{m}_{2}}{{m}_{1}}=\frac{F}{F/2}\times \frac{2M}{M}=\frac{2F}{F}\times \frac{2m}{m}=4\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}_{1}}{{a}_{2}}=4\Rightarrow 4{a}_{2}={a}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{2}=\frac{{a}_{1}}{4}=\frac{1}{4}\times {a}_{1}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{new}\mathrm{acceleration}\mathrm{will}\mathrm{be}\mathrm{one}-\mathrm{fourth}\mathrm{of}\mathrm{previous}\mathrm{acceleration}.\phantom{\rule{0ex}{0ex}}$

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#### Question 16:

Two friends on roller-skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other, who catches it, How will this activity affect the position of the two? Explain your answer.

#### Answer:

The separation between the two friends will increase. Initially, the momentum of both of them is zero as they are at rest. In order to conserve the momentum the one who throws the ball would move backward.

The second will experience a net force after catching the ball and therefore will move backward that is in the direction of the force.

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#### Question 17:

Water sprinkler used for grass lawns begins to rotate as soon as the water is supplied. Explain the principle on which it works.

#### Answer:

Water sprinkler works on Newton’s third law of motion. When the water is supplied, then force applies to the sprinkler due to which it rotates.

#### Page No 63:

#### Question 18:

Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 10^{3 }m s^{-1 }and gets embedded after travelling 5 cm. Calculate

(i) the resistive force exerted by the sand on the bullet

(ii) the time taken by the bullet to come to rest.

#### Answer:

i) If a body of mass (*m*), moving at velocity (*u*) accelerates uniformly at (*a*) for time *T*, so that its velocity changes to *v*, then

$\mathrm{initial}\mathrm{momentum},{p}_{1}=mu\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{final}\mathrm{momentum},{p}_{2}=mv\phantom{\rule{0ex}{0ex}}so,\phantom{\rule{0ex}{0ex}}\mathrm{Change}\mathrm{in}\mathrm{momentum}={p}_{2}-{p}_{1}=mv-mu=m(v-u)\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{the}\mathrm{second}\mathrm{law}\mathrm{of}\mathrm{motion},\mathrm{force}\left(F\right)\propto \frac{\mathrm{Change}\mathrm{in}\mathrm{momentum}}{\mathrm{Time}}\phantom{\rule{0ex}{0ex}}F\propto \frac{{p}_{2}-{p}_{1}}{t}\phantom{\rule{0ex}{0ex}}F\propto \frac{m\left(v-u\right)}{t}\phantom{\rule{0ex}{0ex}}F\propto ma\left(\therefore wherea=\frac{v-u}{t}\right)\phantom{\rule{0ex}{0ex}}F=kma\phantom{\rule{0ex}{0ex}}\mathrm{Here},k=1\phantom{\rule{0ex}{0ex}}\mathrm{so},F=ma\phantom{\rule{0ex}{0ex}}\mathrm{Given},m=10\mathrm{g}=0.01\mathrm{kg},u={10}^{3}\mathrm{m}/\mathrm{s},v=0\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}s=5\mathrm{cm}=0.05\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

ii) From the first equation of motion,

$v=u+at\phantom{\rule{0ex}{0ex}}\Rightarrow at=v-u\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{time}\mathrm{taken}\mathrm{by}\mathrm{bullet}\mathrm{to}\mathrm{come}\mathrm{to}\mathrm{rest},\phantom{\rule{0ex}{0ex}}t=\frac{v-u}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{0-{10}^{3}}{-{10}^{7}}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{-{10}^{3}}{-{10}^{7}}={10}^{3}\times {10}^{-7}\phantom{\rule{0ex}{0ex}}\Rightarrow t={10}^{-4}\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{force}\left(F\right)={10}^{5}\mathrm{N}\mathrm{and}\mathrm{time}\left(t\right)={10}^{-4}\mathrm{s}$

#### Page No 63:

#### Question 19:

Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 m s^{–2 }on a mass *m*_{1} and an acceleration of 24 m s^{–2 }on a mass *m*_{2}. What acceleration would the same force provide if both the masses are tied together?

#### Answer:

$\mathrm{We}\mathrm{know}\mathrm{that},\mathrm{SI}\mathrm{unit}\mathrm{of}\mathrm{mass},m=1kg\phantom{\rule{0ex}{0ex}}\mathrm{SI}\mathrm{unit}\mathrm{of}\mathrm{acceleration},a=1m{s}^{-2}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{from}\mathrm{Newton}\text{\'}\mathrm{s}\mathrm{second}\mathrm{law},F=ma\phantom{\rule{0ex}{0ex}}\mathrm{SI}\mathrm{unit}\mathrm{of}\mathrm{Force},F=1\mathrm{kg}\times 1{\mathrm{ms}}^{-2}\phantom{\rule{0ex}{0ex}}=1\mathrm{kg}-{\mathrm{ms}}^{-2}\phantom{\rule{0ex}{0ex}}1\mathrm{kg}-{\mathrm{ms}}^{-2}\mathrm{is}\mathrm{known}\mathrm{as}1\mathrm{newton}\left(\mathrm{N}\right).\phantom{\rule{0ex}{0ex}}\therefore 1\mathrm{newton}\left(\mathrm{N}\right)=1\mathrm{kg}-{\mathrm{ms}}^{-2}\phantom{\rule{0ex}{0ex}}1\mathrm{newton}\mathrm{can}\mathrm{be}\mathrm{defined}\mathrm{as},"\mathrm{the}\mathrm{force}\mathrm{is}\mathrm{said}\mathrm{to}\mathrm{be}1\mathrm{N}\mathrm{if}\mathrm{it}\mathrm{produces}1{\mathrm{ms}}^{-2}\phantom{\rule{0ex}{0ex}}\mathrm{acceleration}\mathrm{in}\mathrm{a}\mathrm{body}\mathrm{of}\mathrm{mass}1\mathrm{kg}."\phantom{\rule{0ex}{0ex}}\mathrm{Condition}\mathrm{I}:\mathrm{Given},\phantom{\rule{0ex}{0ex}}{F}_{1}=5\mathrm{N},{a}_{1}=8{\mathrm{ms}}^{-2}\mathrm{and}{m}_{1}=\mathrm{mass}\mathrm{of}\mathrm{first}\mathrm{body}\phantom{\rule{0ex}{0ex}}{F}_{2}=5\mathrm{N},{a}_{2}=24{\mathrm{ms}}^{-2}\mathrm{and}{m}_{2}=\mathrm{mass}\mathrm{of}\mathrm{second}\mathrm{body}\phantom{\rule{0ex}{0ex}}$

$\mathrm{From}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{second}\mathrm{law},\phantom{\rule{0ex}{0ex}}F=maor5={m}_{1}\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{1}=\frac{5}{8}kg\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Condition}\mathrm{II}\mathrm{Again},\mathrm{using}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{II}\mathrm{law},F=ma,5={m}_{2}\times 24\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=\frac{5}{24}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{masses}={m}_{1}+{m}_{2}=\frac{5}{8}+\frac{5}{24}=\frac{15+5}{24}=\frac{20}{24}=\frac{5}{6}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{is}\mathrm{produced}\mathrm{by}\mathrm{the}\mathrm{same}\mathrm{force}\mathrm{provided}\mathrm{if}\mathrm{both}\mathrm{the}\mathrm{masses}\mathrm{are}\mathrm{tied}\mathrm{together},\phantom{\rule{0ex}{0ex}}\Rightarrow F=same\phantom{\rule{0ex}{0ex}}\Rightarrow M={m}_{1}+{m}_{2},M=\mathrm{combined}\mathrm{mass}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{F}{({m}_{1}+{m}_{2})}\phantom{\rule{0ex}{0ex}}=\frac{5}{5/6}=6{\mathrm{ms}}^{-2}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{acceleration}\mathrm{is}6{\mathrm{ms}}^{-2}$

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#### Question 20:

What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically

(a) momentum versus velocity when mass is fixed.

(b) momentum versus mass when velocity is constant.

#### Answer:

Momentum is the product of the mass and velocity of the body.

$\mathrm{Momentum}=\mathrm{mass}\times \mathrm{velocity}\phantom{\rule{0ex}{0ex}}\Rightarrow p=mv$

SI unit of momentum is kg m/s.

(a) Momentum versus velocity when mass is fixed.

(b) Momentum versus mass when velocity is constant.

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