MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 10

A bullet of mass $0.01$ kg is fired from a gun of mass $5$ kg with velocity of $250$ m/s, calculate the speed with which the gun recoils.

0%
0.50 m/s
0%
0.25 m/s
0%
0.05 m/s
0%
0.025 m/s

Explanation

As the net external force on the gun bullet system is zero, the sum of the momentum of the gun and bullet system will be constant.

Now as the gun+bullet system was initially at rest, the initial momentum of the system =-0.

Hence, according to the law of conservation of momentum,

(the final momentum of the bullet fired) + (final momentum of the gun that recoils (opposite direction))

$=0$

Momentum of the bullet =

$\left(0.01 \right) \times 250 = 2.5 \ kgm/s$

Let

$v$ be the velocity of the recoil of the gun.

Momentum of the recoiling gun =

$\left(5 \right) \times v= 5v \ kgm/s$

Equating we have

$2.5+5 v=0$

$\Rightarrow v=-0.5 \ m/s$

The speed with which the gun recoils is

$0.50\ m/s$ in the opposite direction of speed of bullet.

Swimming is based on Newton's:

0%
First law of motion
0%
second law of motion
0%
third law of motion
0%
law of gravitation

In an elevator moving vertically up with an acceleration 'g' the force exerted on the floor by a passenger of mass M is

0%
$Mg$
0%
$0.5Mg$
0%
Zero
0%
$2Mg$

Explanation

$\textbf{Step 1 - Draw F.B.D of passenger}$

$\textbf{Step2:Apply Newton's second law of motion}$

Using newton's second law.

Apply

$\sum F = \sum ma$ is a vertical direction

$N - Mg = Ma$

$\Rightarrow N - Mg = Mg \Rightarrow N = 2Mg$

Hence option (D) correct.

The mass of a loaded truck

$10$ tonnes is going at a speed of

$20 \ kmph$ . A box of mass

$2$ tonnes slips from it and falls on the road. Find the final velocity of the truck (assume that the power of the engine is constant)?

0%
$2.5\ kmph$
0%
$25 \ kmph$
0%
$5.5 \ kmph$
0%
$3.5\ kmph$

Explanation

When the external force is zero then momentum remains constant.

Using momentum conservation,

Initial momentum

$(P_i)$ = Final momentum

$(P_f)$

Momentum = mass

$\times$ velocity

$P_i = 10 \times 20$ ; here units are taken same as given in the question

$P_f = 8 \times v$

Now equate momentum-

$10 \times 20 = 8 \times v$

$\Rightarrow v = 25\ kmph$

Two bodies have masses in the ratio

$3 : 4$ . When a force is applied on the first body, it moves with an acceleration of

$6 \ m/s^2$ . How much acceleration will the same force produce in the second body?

0%
$2.25 \ m/s^2$
0%
$4.5 \ m/s^2$
0%
$6 \ m/s^2$
0%
$5 \ m/s^2$

Explanation

$Force = mass \times acceleration.$
So, for a constant force,

$m_1a_1 = m_2a_2$

$\displaystyle \implies a_2 = \frac{m_1}{m_2} \times a_1 = \frac{3}{4} \times 6 = 4.5 m/s^2$

Suppose a body that is acted on by only two forces, is accelerated. For this situation, mark the correct statement :
( Initial Velocity is non zero and in same direction as acceleration )

0%
The body cannot move with constant speed
0%
The velocity can never be zero
0%
The sum of two forces cannot be zero
0%
The two forces must act in the same line

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 ?

0%
Light rifle
0%
Heavy rifle
0%
Both of them will hurt equally
0%
None of these

Explanation

The two identical bullets recoil with the same force, but the lighter one will accelerate faster than the heavier one.
We can use Newton's third law to explain better: If

$F_l\ and\ F_H$ are the forces exerted on the rifles, we have

$F_l = F_h$
or

$M_1<M_h$ and

$a_1 > a_h$ .
The lighter rifle hurts more than a heavier rifle as the acceleration is higher.

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

0%
Accelerated
0%
Uniform
0%
Retarded
0%
Along circular tracks

The inertia of an object tends to cause the object:

0%
$\text to\ increase\ its\ speed$
0%
$\text to\ decrease\ its\ speed$
0%
$\text to\ resist\ any\ change\ in\ its\ state\ of\ motion$
0%
$\text to\ decelerate\ due\ to\ friction$

STATEMENT - 1 : Gas ejected from rocket will never exert thrust on the rocket if the ejected gas and the rocket move in the same direction.
STATEMENT - 2 : To exert thrust on rocket in its direction of motion, the ejected gas (w.r.t. rocket) must move opposite to the velocity of rocket (w.r.t. ground).

0%
Statement-1 is True, Statement-2 is True ;Statement-2 is a correct explanation for Statement-1.
0%
Statement-1 is True, Statement-2 is True ;Statement-2 is NOT a correct explanation for Statement-1.
0%
Statement-1 is True, Statement-2 is False.
0%
Statement-1 is False, Statement-2 is True.

A force of

$5\ N$ produces an acceleration of 8

$ms^{-2}$ on a mass

$m_1$ and an acceleration of 24

$m s^{-2}$ on a mass

$m_{2}$ . What acceleration (in

$ms^{-2}$ ) would the same force provide if both the masses are tied together?

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

From Newton's second law of motion, the applied force is directly proportional to the acceleration produced or

$F = m a$
The unit of force is newton in SI denoted by the symbol N.
Forces acting on a mass

$m_1$ with acceleration

$a_1$ and mass

$m_2$ with acceleration

$a_2$ are as given below.

$m_{1}=\dfrac{F}{a_{1}}=\dfrac{5}{8}kg.$

$m_{2}=\dfrac{F}{a_{2}}=\dfrac{5}{24}kg.$

$M=\left ( \dfrac{5}{8}+\dfrac{5}{24} \right )kg=\left ( \dfrac{5}{6} \right )kg$
Acceleration produced in M (when the two masses are tied together),

$a=\dfrac{F}{M}=\dfrac{5}{\dfrac{5}{6}}=6 \ m s^{-2}$

According to the Newton's third law of motion, action and reaction pair:

0%
always act on the same body
0%
always act on different bodies in opposite directions
0%
have same magnitude and directions
0%
act on either body at normal to each other

An apple falls from a tree because of gravitational attraction between the earth and apple. If

$F_{1}$ is the magnitude of force exerted by the earth on the apple and

$F_{2}$ is the magnitude of force exerted by apple on earth, then

0%
$F_{1}$ is very much greater than $F_{2}$
0%
$F_{2}$ is very much greater than $F_{1}$
0%
$F_{1}$ is only a little greater than $F_{2}$
0%
$F_{1}$ and $F_{2}$ are equal

Explanation

According to Newton's III law "Every action has an equal and an opposite reaction". An apple falling down has the force of gravity acting downward (

$F_{1}$ ) which is the action. The opposite reaction is the force the apple exerts on the earth (

$F_{2}$ ).

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. Is this an example of conservation of momentum?

0%
Yes
0%
No
0%
Sometimes
0%
None of these

During a planned manoeuvre in a space flight, a free-floating astronaut A pushes another free floating astronaut B, the mass of A being greater than that of B. Then, the magnitude of the force exerted by astronaut A on astronaut B is:

0%
equal to zero
0%
equal to the force exerted by B on A
0%
greater than the force exerted by B on A
0%
less than the force exerted by B on A

What are action-reaction forces?

0%
Those which act on the same body.
0%
Those which act on different bodies in different directions
0%
Those which act vertically.
0%
Those which act in the same direction

If action and reaction were to act on the same body then

0%
resultant would be zero
0%
body would not move at all
0%
both (a) and (b)
0%
none of these

A body of mass

$300\ g$ is at rest. What force in Newton will you have to apply to move it through

$200\ cm$ in

$10\ s$ ?

0%
Zero
0%
$12\ N$
0%
$1.2\ N$
0%
$0.012\ N$

Explanation

Given:

$S=200 cm=2 m, u=0\ m/s, t=10 s$

Using the formula :

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

$\Rightarrow 2=0+0.5 \times a \times 10^2$

$\Rightarrow a=0.04\ m/s^2$

$\underline{\text{Required force:}}$

Mass of the body

$m= 300\ g=0.3\ kg$

From Newton's second law:

$F=ma=0.3 \times 0.04=0.012\ N$

In a rocket, fuel burns at the rate of

$1\ kg/s$ . This fuel is ejected from the rocket with a velocity of

$60\ km/s$ . Force exerted on the rocket in

$N$ is:

0%
$60 N$
0%
$600 N$
0%
$6000 N$
0%
$60000 N$

Explanation

Overall momentum of the system is conserved.
Therefore, change in momentum of the rocket will be equal to that of the fuel.
Change in momentum of the fuel will be

$\Delta mv = 60000$
Hence, change in momentum of the rocket will also be:

$\bigtriangleup P = 60000$
Thrust force

$F =\dfrac{\bigtriangleup p}{\bigtriangleup t}$

$F = 60000\ N$

Which of the following requires a pushing force?

0%
Throwing a stone at a bird.
0%
Grabbing hold of a pencil.
0%
Leaves falling from a tree.
0%
A load lifted by a pulley.

Which of the following units is used to measure thrust?

0%
Dyne
0%
Pascal
0%
Newton per meter square
0%
Dyne per centimeter

A man is standing on a boat in still water. If he walks towards the shore the boat will :

0%
move away from the shore.
0%
remain stationary.
0%
move towards the shore.
0%
sink.

While dusting a carpet, we give a sudden jerk or beat with a stick because :

0%
Inertia of rest keeps the dust in its position and the dust is removed by movement of carpet away.
0%
Inertia of motion removes dust.
0%
No inertia involved in process.
0%
None of the above

In a game of tug of wars, a condition of equilibrium exists. Both the teams pull the rope with a force of 10

$^4$ N. The tension in the rope is :

0%
10 $^4$ N
0%
10 $^5$ N
0%
0 N
0%
2 $\times$ 10 $^4$ N

Explanation

In a game of tug of wars, a condition of equilibrium exists. Both the teams pull the rope with a force of 10

$^4$ N. The tension in the rope is equal and opposite to the force with which both teams pull the rope i.e., 10

$^4$ N.

A force 100 N acts in a body mass 2 kg for 10 s. The change in the velocity of the body is :

0%
100 m s $^{-1}$
0%
250 m s $^{-1}$
0%
500 m s $^{-1}$
0%
1000 m s $^{-1}$

Explanation

Given :

$F = 100 \ N$

$m =2 \ kg$

$t = 10 \ s$

Change in momentum of the body

$m(\Delta v) = F.t$

where

$\Delta v$ is the change in velocity.

$\implies \ 2\times \Delta v = 100\times 10$

$\implies \ \Delta v = 500 \ m/s$

Two individual forces of magnitude

$F_1$ and

$F_2$ act on a body of mass 1 kg as shown in the figures (i) and (i). If product of net magnitudes of forces produced in fig (i) and (ii) is 27 N then find the magnitude of forces

$F_1$ and

$F_2$ . (assume

$F_1$ >

$F_2$ )

0%
3 N, 6 N
0%
6 N, 6 N
0%
6 N, 3 N
0%
3 N, 3 N

Explanation

From figure 1, net force F'

$= F_1 - F_2$
From figure 2 net force F"

$= F_1 + F_2$
Given

$(F_1-F_2) (F_1+ F_2) = 27 N$

$F_1^2 - F_2^2 = 27$
Factors of F

$=$ 27 are 1, 2, 7, 3, 9

$\Rightarrow$ If

$F_1 + F_2 = 9 N$

$F_1-F_2 = 3N$

$\therefore F_1 = 6N$ and

$F_2 = 3N$

S.I. unit of momentum is :

0%
$kg\ ms^{-1}$
0%
$kg\ mg^{-2}$
0%
$kg\ mg^2$
0%
$kg\ m^{-1}s^{-1}$

Explanation

Hint: Momentum $p=m\times v$

Correct opton is Option (A) $\bf{\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}}$

Explanation:

$\bullet$ Momentum depends upon the mass and velocity.

$\bullet$ In terms of equation, the momentum of an object is equal to the mass of the object times velocity of object .

$\bullet$ Momentum = $\bf\text { mass } \times \text { velocity }$

$\bullet$ The SI unit of momentum is

$\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}$

A loaded

$20,000 \ kg$ coal wagon is moving on a level track at

$6$ ms

$^{-1}$ . Suddenly

$5000 kg$ of coal is dropped out of the wagon. The final speed of the wagon is:

0%
$6$ ms $^{-1}$
0%
$8$ ms $^{-1}$
0%
$4.8$ ms $^{-1}$
0%
$4.5$ ms $^{-1}$

Explanation

Initial mass of the system

$m_i = 20000 \ kg$

Final mass of the system

$m_f = 20000-5000= 15000 \ kg$

Since no external force is acting on it, we can apply conservation of momentum,

$m_iv_i=m_fv_f$

$20000\times 6=15000\times v_f$

Final velocity,

$v_2=8\ m/s$

The time, in which a force of

$2\ N$ produces a change as the momentum of

$0.4\ kg ms^{-1}$ in the body whose mass is

$1\ kg$ is :

0%
$0.2\ s$
0%
$0.02\ s$
0%
$0.5\ s$
0%
$0.05\ s$

Explanation

According to newton's second equation of motion-

$F = \dfrac{ \Delta P }{ t }$

$F$ : External force

$\Delta P$ : Change in the momentum

$t$ : Time taken

$\Rightarrow 2 \displaystyle =\frac{0.4-0}{t}$

$\Rightarrow t = \dfrac{ 0.4 }{ 2 }$

$\Rightarrow t = 0.2\ s$

In case of rocket propulsion choose the correct options.

0%
Momentum of rocket always remains constant.
0%
Newton's third law is applied.
0%
If exhaust velocity and rate of burning of mass is kept constant, then acceleration of rocket will go on increasing.
0%
If exhaust velocity and rate of burning of mass is kept constant, then thrust force will be constant

Explanation

Newton's Second law

$\Rightarrow F=ma$
Newton's Third law

$\Rightarrow$ Every action has an equal and opposite reaction.
External force gravity acts on system. Therefore momentum of system is not conserved.
Mass keep on decreasing. Therefore acceleration will keep on increasing.

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 10 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 10 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.