Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 46

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Question 1
1 / -0

According to Newton's Third Law of Motion, action and reaction are equal and opposite. The resultant force is not zero because:

A
They act at different times

B
They act on different bodies

C
Their lines of action are different

D
They act on the same body

Solution

According to Newton's third law of motion,
If two bodies are in equilibrium, the force exerted by a body on the other body will be the same as the force exerted by the latter on the first one and that will be equal in magnitude and opposite in direction.
So, the force will not cancel out as it acts on different bodies.
Question 2
1 / -0

A driver first accelerates his car at a rate of $$\displaystyle 1.8{ m }/{ { s }^{ 2 } }$$ and then at a rate of $$\displaystyle 1.2{ m }/{ { s }^{ 2 } }$$. The ratio of the forces exerted by the engines respectively will be equal to

A
2 : 3

B
1 : 2

C
2 : 1

D
3 : 2

Solution

$$\underline{\text{First situation:}}$$
Acceleration $$a_1 = 1.8\ m/s^2$$
$$F_1 = ma_1=m\times 1.8$$

$$\underline{\text{Second situation:}}$$
Acceleration $$a_2 = 1.2\ m/s^2$$
$$ F_2 = ma_2=m\times 1.2$$

So, $$ \dfrac{F_1}{F_2} = \dfrac{a_1}{a_2}= \dfrac{1.8}{1.2}= 3:2$$
Question 3
1 / -0

The rocket engine lifts a rocket from the earth, because hot gases:

A
push it against the air with very high velocity

B
push it against the earth with very high velocity

C
heat up the air which lifts the rocket with very high velocity

D
react against rocket and push it up with very high velocity

Solution

When the rocket gas pushes it against the earth with high velocity, there is production of reaction force which creates lift for the rocket.
Question 4
1 / -0

A bullet of mass 0.01 kg is fired from a gun weighing 5.0 kg. If the initial speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.

A
-0.50 m/s

B
- 0.25 m/s

C
+ 0.05 m/s

D
+0.25 m/s

Solution

Let the recoil velocity of gun be $$V$$.
Mass of gun $$M_{gun} = 5.0 \ kg$$
Mass of bullet $$M_{bullet} = 0.01 \ kg$$
Velocity of bullet $$V_{bullet} = 250 \ m/s$$
Initial momentum of the system   $$P_i = 0$$
Applying the conservation of linear momentum :   $$P_f = P_i$$
Or    $$M_{gun}V + M_{bullet}V_{bullet} = 0$$
Or    $$5\times V+0.01\times 250 = 0$$
$$\implies \ V = -0.5 \ m/s$$
Question 5
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If the momentum of a body is doubled, the kinetic energy is

A
Halved

B
Unchanged

C
Doubled

D
Becomes 4 times

Solution

The correct answer is option (D).

Hint: Relate the formula of momentum and kinetic energy of any body.

Step 1: Find the relation between momentum and kinetic energy.
By formula,
$$p=mv$$ (where, $$p$$ is the momentum of the body)
$$v=\dfrac{p}{m}$$
Also,
$$K.E.=\dfrac{1}{2}mv^2$$
Substituting the value of $$v$$ in the above equation, we get,
$$K.E.=\dfrac{1}{2}m{(\dfrac{p}{m}})^2$$

$$K.E.=\dfrac{1}{2}m\dfrac{p^2}{m^2}$$

$$K.E.=\dfrac{p^2}{2m}$$
Hence, $$K.E.\propto p^2$$

Step 2: Calculate the final kinetic energy.
From the above relation, we can write,
$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{p_2^2}$$
According to the question, $$p_2=2p_1$$
$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{(2p_2)^2}$$

$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{4p_1^2}$$

$$K.E._2=4K.E._1$$

Hence, the kinetic energy becomes four times.
The correct answer is option (D).
Question 6
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The mass of balloon and its contents is M. It is descending with an acceleration a. By how much the mass should be decreased, keeping the volume constant, so that the balloon starts ascending with the same acceleration?

A
$$\displaystyle \frac { a }{ a+g } M$$

B
$$\displaystyle \frac { g }{ a+g } M$$

C
$$\displaystyle \frac { 2a }{ a+g } M$$

D
$$\displaystyle \frac { 2g }{ a+g } M$$

Solution

The forces acting on the balloon when the balloon is coming downwards with an acceleration a are
(i) weight, mg acting downwards and
(ii) the force of upthrust, F acting upwards. Since the accelerated movement is in the downward direction, the force Mg must be greater than F. So for the downwards accelerated movement of the block,

$$\displaystyle Mg-F=Ma$$..........(i)

Let m be the mass removed from the balloon, the weight of the balloon now becomes $$\displaystyle (M-m)g$$. Now for the upward accelerated movement with the same acceleration, we have

$$\displaystyle F-(M-m)g=(M-m)a$$.......(ii)

Solving equations, (i) and (ii)
$$ (Mg−Mg)−Mg+mg=Ma−ma$$
or $$m(g+a)=2Ma$$
$$m=\dfrac{2Ma}{g+a} .$$
Question 7
1 / -0

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

A
While the bus is moving, luggage tends to remain in inertia of motion state

B
When the bus stops, the luggage tends to resist the change and due to inertia of motion it move forward and may fall off

C
Both a and b are true

D
None of them is True

Solution

While the bus is moving, luggage tends to remain in inertia of motion state. When the bus stops, the luggage tends to resist the change and due to inertia of motion it moves forward and may fall off. That's why it is advised to tie any luggage kept on the roof of a bus with a rope.
Question 8
1 / -0

Two bodies of equal masses (M) moving with equal velocities (V) in opposite direction collide. Resultant velocity of the combination:

A
$$V$$

B
$$Va$$

C
$$-V$$

D
Zero

Solution

Given : $$m_1=m_2 = M$$ $$v_1 = V$$ $$v_2 = -V$$

Momentum of $$m_1$$ is $$P_1=MV$$
Momentum of $$m_2$$ is $$P_2=-MV$$
Their sum is $$P_R=P_1+P_2=MV-MV=0$$ . . . . (i)

Let us consider the resultant velocity of the system as $$V_R$$
Resultant mass of the system be $$M_R=m_1+m_2=2M$$
The Resultant momentum of the system $$P_R=M_RV_R=2MV_R$$ . . . . (ii)

Comparing (i) and (ii),
$$2MV_R=0$$
$$\Rightarrow V_R=0$$
Question 9
1 / -0

Mark the incorrect option:

A
Balanced forces cannot set any stationary body into motion

B
Balanced forces cannot change the speed of a moving body

C
Balanced forces change the shape and size of a soft body

D
None

Solution

Balanced forces result in zero net external force acting on the body and hence the body remains at rest or in equilibrium. Thus balanced forces can neither change the speed of the moving body nor set any stationary body into motion.
But balanced force can change the shape and size of a soft object for example balanced force acting on a piece of sponge.
Hence option D is correct.
Question 10
1 / -0

Two tugboats are moving a barge. Tugboat A exerts a force of $$3000$$ newtons on the barge. Tugboat B exerts a force of $$5000$$ newtons in the same direction. What is the combined force on the barge?

A
$$2000\ N$$

B
$$8000\ N$$

C
$$3000\ N$$

D
$$5000\ N$$

Solution

forces in the same direction combine by addition.
thus, $$3000N+5000N$$
$$=8000$$

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Re-Attempt Weekly Quiz Competition

Forces and Laws of Motion Test - 46

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Forces and Laws of Motion Test - 46

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Question 1

They act at different times

They act on different bodies

Their lines of action are different

They act on the same body

Solution

Question 2

2 : 3

1 : 2

2 : 1

3 : 2

Solution

Question 3

push it against the air with very high velocity

push it against the earth with very high velocity

heat up the air which lifts the rocket with very high velocity

react against rocket and push it up with very high velocity

Solution

Question 4

-0.50 m/s

- 0.25 m/s

+ 0.05 m/s

+0.25 m/s

Solution

Question 5

Halved

Unchanged

Doubled

Becomes 4 times

Solution

Question 6

$$\displaystyle \frac { a }{ a+g } M$$

$$\displaystyle \frac { g }{ a+g } M$$

$$\displaystyle \frac { 2a }{ a+g } M$$

$$\displaystyle \frac { 2g }{ a+g } M$$

Solution

Question 7

While the bus is moving, luggage tends to remain in inertia of motion state

When the bus stops, the luggage tends to resist the change and due to inertia of motion it move forward and may fall off

Both a and b are true

None of them is True

Solution

Question 8

$$V$$

$$Va$$

$$-V$$

Zero

Solution

Question 9

Balanced forces cannot set any stationary body into motion

Balanced forces cannot change the speed of a moving body

Balanced forces change the shape and size of a soft body

None

Solution

Question 10

$$2000\ N$$

$$8000\ N$$

$$3000\ N$$

$$5000\ N$$

Solution

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