MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 16

Newton’s second law of motion says that the mass of object times its acceleration is equal to the net force on the object. Which of the following gives the correct units for force?

0%
$kg.m/s^{2}$
0%
$kg.m^{2}/s^{2}$
0%
$kg/m.s^{2}$
0%
$kg.m^{2}/s$
0%
None of those answer

Explanation

The SI unit of mass is

$kg$ .

The SI unit of acceleration is

$m/s^2$ .

We know,

$\text{Force}=\text{mass}\times \text{acceleration}$

The unit of force is

$kg.m/s^2$ .

A body of mass

$10\ kg$ undergoes a change in velocity of

$10\ m/s$ in

$5\ s$ . What is the force acting on it?

0%
$100\ N$
0%
$25\ N$
0%
$50\ N$
0%
$20\ N$

Explanation

Acceleration

$a = \dfrac{\text{velocity}}{\text{time}}$

Given ,

Velocity

$= 10\ m/s$

Time

$= 5s$

Mass

$= 10\ kg$

$a= \dfrac{10}{5}\ m/s^{2}$

$= 2\ m/s^{2}$

Force = mass

$\times$ acceleration

$= 10 \times 2 = 20 N$

When a bus starts suddenly, the passengers are pushed back. This is an example of which of the following?

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Newton's first law
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Newton's second law
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Newton's third law
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None of Newton's laws

A force produces an acceleration of

$5.0 \:cm \:s^{-2}$ when it acts on a body of mass 20 g. Find the force acting on the body.

0%
$2\times 10^{-3}N$
0%
$4\times 10^{-3}N$
0%
$1.0\times 10^{-3}N$
0%
$5\times 10^{-3}N$

In the above figure, a locomotive has broken through the wall of a train station. During the collision, what can be said about the force exerted by the locomotive on the wall?

0%
The force exerted by the locomotive on the wall was larger than the force the wall could exert on the locomotive.
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The force exerted by the locomotive on the wall was the same in magnitude as the force exerted by the wall on the locomotive.
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The force exerted by the locomotive on the wall was less than the force exerted by the wall on the locomotive.
0%
The wall cannot be said to “exert” a force; after all, it broke.

An athlete jumps straight up in the air to catch a ball. If he took

$0.4$ seconds to jump up and

$0.4$ seconds to come back down, how high did he jump? Take

$g = 9.8\ m/s^{2}$ .

0%
$0.784\ m$
0%
$1.568\ m$
0%
$2.352\ m$
0%
$3.136\ m$

Explanation

Given,

Time taken by the athlete to jump up

$=$ Time taken by the athlete to come down

$t=0.4\ s$

Distance travelled by the athlete while moving up

$=$ distance travelled by the athlete while coming down

Let the distance travelled while moving up or down be

$h$

While coming down,

Initial velocity,

$u=0$

Acceleration due to gravity,

$g=9.8\ m/s^2$

Using Newton's second equation of motion :

$s = ut + \dfrac{1}{2} at^{2}$

$h = ut + \dfrac{1}{2} gt^{2}$

$\therefore h = (0\times 0.4) +\dfrac{1}{2} \times 9.8 (0.4)^{2}$

$h =\dfrac{1}{2} \times 9.8\times 0.16$

$h = 0.784\ m$

A gun of mass 1000kg fires a shell of mass 2 kg with a velocity of 200 m/s Calculate velocity of recoll of the gun

0%
$4 m/s$
0%
$0.2 m/s$
0%
$0.4 m/s$
0%
None of these

Explanation

Hint: Use Law of conservation of momentum.

${m_1}{u_1} + {m_2}{u_2}+ ……… {m_n}{u_n} = {m_1}{v_1} + {m_2}{v_2}+ ……… {m_n}{v_n}$

Step 1: Note the given values

Let,

$M$ is the mass of the gun

$m$ = mass of the shell

$v$ = velocity of shell

$V$ = velocity of recoil of gun

Given,

$M$ =

$1000 Kg$

$m$ =

$2 Kg$

$v$ =

$200 m/s$

$V$ =

$?$

Step 2 : Applying law of conservation of momentum.

$MV = mv$

Putting the values in above equation,

$1000 \times V = 2 \times 200$

$V = \dfrac{400}{1000}$

$V = 0.4 m/s$

Hence, Velocity of recoil of gun is $0.4 m/s$ and correct option is C.

Which of these examples can be explained by Newtons first law of motion ?

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It is easier to push an empty shopping cart than a full shopping cart
0%
It takes a smaller force to roll a marble than a bowling ball
0%
When a person steps forward, he exerts a backward force on ground
0%
A runner does not come to a stop immediately upon crossing the finish line

A bullet of mass A and velocity B is fired into a block of wood of mass C. If loss of any mass and friction be neglected, then velocity of the system must be

0%
$\dfrac{AB}{A+C}$
0%
$\dfrac{A + C}{B+C}$
0%
$\dfrac{AC}{B+C}$
0%
$\dfrac{A+B}{AC}$

Explanation

Given:
The bullet strikes the block and after collision they become a system and

$(Mass)_{bullet} = A$

$(Velocity)_{bullet} = B$

$(Mass)_{block} = C$

$(Velocity)_{block} = 0$
Momentum before impact = Momentum after impact

$AB + 0 =(A+C)(v)_{system }$

$(v)_{system}=\dfrac{AB}{A+C}$
The velocity of the system

$(v)_{system}=\dfrac{AB}{A+C}$

A shell of mass 100 g moving with the speed of

$50 m s^{-1}$ along a straight line bursts into two pieces. The piece of mass 60 g starts moving horizontally with a speed of

$50 m s^{-1}$ . The other piece moves with a speed of

0%
$50 m s^{-1}$ opposite to the first piece.
0%
$100 m s^{-1}$ in vertically downward direction.
0%
$50 m s^{-1}$ in the same direction as the first piece
0%
zero (remains at rest)

$120 N$ of force is acting on a body with

$6 ms^{-2}$ acceleration, Its mass is _____________.

0%
$0.05 Kg$
0%
$20 Kg$
0%
$200Kg$
0%
$720 Kg$

Explanation

According to newton second law

$F(force)=m(mass) \times a(acceleration)$

$F=120N$ $a=6ms^{-2}$

Mass of the body is,

$m = \frac{F}{a}$

$= \frac{{120}}{6}$

$= 20\;{\rm{kg}}$

An engine of mass

$5\times {10}^{4}\ kg$ pulls a coach of mass

$4\times {10}^{4}\ kg$ . A resistance of

$1\ N$ per

$100\ kg$ is acting on both the coach and the engine and if the driving force of the engine is

$4500\ N$ then the acceleration of the engine and the tension in the coupling will respectively be_________

0%
The acceleration of the system (coach $+$ engine) is $0.04{m/s}^{2}$
0%
The acceleration of the system (coach $+$ engine) is $0.05{m/s}^{2}$
0%
The tension in the coupling is $2000\ N$ .
0%
The tension in the coupling is $2400\ N$ .

Explanation

$Driving\, force=4500\ N$

Opposing force Resistance

$f$

$f =\dfrac { { \left( { 5+4 } \right) { \times { 10 }^{ 4 } } } }{ { 100 } }$

$f=900N$

Resultant force:

$F_R = 4500-900$

$\Rightarrow F_R=3600\ N$

Mass of engine and coach

$=9 \times { 10^{ 4 } }\ kg$

According to Newton's law

$F = ma$

$\therefore 3600=9\times { 10^{ 4 } }\times a$

$\Rightarrow a=\dfrac { { 3600 } }{ { 9\times { { 10 }^{ 4 } } } }$

$\Rightarrow a =0.04\, m/{ s^{ 2 } } \\$

Now considering the eqilibrium of the coach only

We have,

$\left( { T-R } \right) =4\times { 10^{ 4 } }\times 0.04\, \, \, \, \left( { F=ma } \right)$

$\Rightarrow T-\dfrac { { 4\times { { 10 }^{ 4 } } } }{ { 100 } } =4\times { 10^{ 4 } }\times 0.04$

$\Rightarrow T=(4\times { 10^{ 4 } }\times 0.04)\ +\ (4\times { 10^{ 2 } })$

$\Rightarrow T =1600+400$

$\Rightarrow T=2000\ N$

Hence, Option A and C are answer.

A ball rests upon a flat piece of paper on a table top. The paper is pulled horizontally but quickly towards right as shown. Relative to its initial position with respect to the table, the ball
I. Remains stationary if there is no friction between the paper and the ball.
II. Moves to the left and starts rolling backwards, i.e. to the left if there is a friction between the paper and the ball.
III. Moves forward, i.e., in the direction in which the paper is pulled.
Here, the correct statements is/ are

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I and II only
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III only
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I only
0%
II only

Which of the following cannot be explained using Newton's third law of motion?

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Working of rocket in the fire crackers
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Revolution of satellite around a planet
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Functioning of satellite launch vehicle
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Walking of a person on ground

A bullet of mass

$50$ gram is fired from a

$5kg$ gun with a velocity of

$1km/s$ , the speed of recoil of the gun is

0%
$5m/s$
0%
$1m/s$
0%
$0.5m/s$
0%
$-10m/s$

In CID serial (on TV ), Abhijeet fires a bullet of mass 100g with a speed of 100

${ ms }^{ -1 }$ on a soft plywood of thickness 4 cm. The bullet emerges with 10% of its intial KE. Find the emergent speed of the bullet.

0%
31.62 m/s
0%
30.52 m/s
0%
32.75 m/s
0%
33.84 m/s

A bullet is of mass

$5g$ is fired at a velocity

$900m{s}^{-1}$ from a rifle of mass

$2.5kg$ . What is the recoil velocity of the rifle?

0%
$9m{s}^{-1}$
0%
$180m{s}^{-1}$
0%
$900m{s}^{-1}$
0%
$1.8m{s}^{-1}$

A bomb of mass

$16kg$ at rest explodes into two pieces of masses

$4kg$ and

$12kg$ . The velocity of the

$12kg$ mass is

$4m{s}^{-1}$ . The kinetic energy of the other mass is

0%
$192J$
0%
$96J$
0%
$144J$
0%
$288J$

A target of mass

$M$ is moving with a velocity

$u$ . Lead shots, each of mass

$m$ , are fired on it in the opposite direction with a velocity

$v$ . Find the number of bullets required to stop the target.

0%
$\dfrac{Mu}{mv}$
0%
$\dfrac{mv}{Mu}$
0%
$\dfrac{(M+m)u}{v}$
0%
$\dfrac{mu(M+m)}{v}$

Explanation

Let

$n$ be the number of lead shots required to stop the target.

Given:

Mass of the target =

$M$

Velocity of the target =

$u$

Mass of each shot =

$m$

Velocity of each shot =

$v$

Initial momentum of the target and the shots is

$P_1=Mu-nmv$ (the minus sign indicates that the shots are moving in the opposite direction)

Final momentum of the target and the shots is

$P_2=0$

From the conservation of momentum,

$P_1=P_2$

$\Rightarrow Mu-nmv=0$

$\Rightarrow n=\dfrac{Mu}{mv}$

If the action and reaction forces are always equal in magnitude, then these forces

0%
will produce accelerations of equal magnitudes.
0%
may not produce accelerations of equal magnitudes.
0%
produce velocities of equal magnitudes.
0%
will not produce accelerations of equal magnitudes

Explanation

According to newton's 3rd law of motion, Every action has equal and opposite reaction. Action and reaction always act in pairs, and on different bodies.

Let two masses

$m_1$ and

$m_2$ are pairs of masses applying force on each other ie

$F$ .

Acceleration of masses will be

$a_1=\dfrac F{m_1}$ and

$a_2=\dfrac F{m_2}$ Hence acceleration of masses also depends on the mass.

So, these forces may not produce accelerations of equal magnitudes.

Choose the correct statement from the following:
Weightlessness of an astronaut moving in a satellite is a situation of

0%
Zero $g$
0%
No gravity
0%
Zero mass
0%
Free fall

Rocket propulsion theory is analogues to-

0%
Newton`s third law
0%
Relative theory
0%
First law of thermodynamics
0%
none

A bullet weighing

$50$ gram leaves the gun with a velocity of

$30m/s$ .the ratio of their momentum is:

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:3$

Which of the following statement(s) can be explained by Newton's second law of motion?

0%
To stop a heavy body ( say truck), much greater force is needed than to stop a light body ( say motorcycle), in the same time, if they are moving with the same speed.
0%
For a given body, the greater speed, the greater the opposing force needed to stop the body in a certain time.
0%
To change the momentum ( given ) the force required is independent of time.
0%
The same forces acting on two different bodies for same time cause the same change in momentum in the bodies.

A truck is moving on a frictionless surface with uniform velocity of

$10\,m{s^{ - 1}}$ . A leak occurs in the water tank of the truck at the rate of

$2\,{kg^{ - 1}}$ . What is the speed of truck after

$50 \ s$ , if the mass of truck is

$100\,kg$ and mass of water in the truck initially was

$100\,kg$ ?

0%
$20\,m{s^{ - 1}}$
0%
$10\,m{s^{ - 1}}$
0%
$5\,m{s^{ - 1}}$
0%
None of these

A force of

$50\ N$ is required to push a car on a level road with constant speed of

$10\ ms^{-1}$ . The mass of the car is

$500\ kg$ . What forces should be applied to make the car accelerate at

$1\ ms^{-2}$ .

0%
$450\ N$
0%
$500\ N$
0%
$550\ N$
0%
$2500\ N$

Explanation

Hint: Newton's second law of motion suggests that the net force applied to an object produces a proportional acceleration.

Step 1: Obtain an expression for the net force acting on the car.

When the car travels with a constant speed, it is given that the force applied against friction is equal to

$50N$ . Now, the net force on the car to make it travel with an acceleration equal to

$1m{{s}^{-2}}$ can be represented as:

${{F}_{net}}=F'-50$

Let this be equation 1.

Here, ${{F}_{net}}$ denotes the net force on the car and $F'$ represents the force with which the car needs to be pushed for it to travel with an acceleration of $1m{{s}^{-2}}$ .

Step 2: Apply Newton's second law of motion to obtain the required force.

We are given that mass of the car is $500kg$ and the acceleration with which the car has to travel is $1m{{s}^{-2}}$ .

Newton's second law of motion suggests that the net force applied to an object produces a proportional acceleration. Mathematically,

${{F}_{net}}=ma$

Let this be equation 2.

Substituting the value of

$m$ ,

$a$ and equation 1 in equation 2,

$F'-50=ma$

$F'-50=500\times 1$

$F'=500+50$

$F'=550N$

Final step: Therefore, the force with which the car needs to be pushed for it to travel with an acceleration of

$1m{{s}^{-2}}$ is

$550N$ .

Principle of rocket is based on Newton's

$..........$ law of motion.

0%
first
0%
second
0%
third
0%
none

An object maintains its motion under the continuous application of ________force

0%
An unbalanced
0%
A balanced
0%
A strong
0%
A weak

Choose whether true or false:-
A bicycle stops if we stop pedalling.

0%
True
0%
False

Choose whether true or false:-
Unbalanced forces bring in motion.

0%
True
0%
False

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 16 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 16 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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