Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

What is the recoil velocity of the gun of mass $$8$$ kg when a bullet of mass $$10$$ g is fired from it with a velocity of $$400$$ m/s ?

A
$$5$$ m/s

B
$$2$$ m/s

C
$$50$$ m/s

D
$$0.5$$ m/s

Solution

Given :
mass of the gun $$m_g = 8$$ kg
mass of the bullet $$m_b = 0.01$$ kg
velocity of the bullet $$v_b = 400$$ m/s
recoil velocity of the gun $$V_g = ?$$
As no external force is acting on the system, the total linear momentum of the system is conserved.
Conservation of linear momentum : $$P_i = P_f$$
Initial momentum $$P_i = 0$$ as the gun and the bullet do not move.
Final momentum $$P_f = m_g V_g + m_b v_b$$
$$\therefore$$ $$0 = m_g V_g + m_b v_b$$
$$\Rightarrow 0 =8V_g + (0.01)(400)$$
$$\Rightarrow |V_g| =0.5$$ m/s
Question 2
1 / -0

What is the momentum of a 100000 kg truck whose velocity is 2 m/s?

A
$$2 \times 10^5 kg m/s$$

B
$$1 \times 10^5 kg m/s$$

C
4 kg m/s

D
none of these

Solution

Given:
Mass of the truck $$m = 100000\ kg = 10^{5}\ kg$$
Velocity of the truck $$v=2\ m/s$$
We know that, $$Momentum(P)= Mass(m) \times Velocity(v)$$
$$\therefore P= 2 \times 10^5\ kg m/s$$
Question 3
1 / -0

A boy sitting on the top most berth in the compartment of a train which is just going to stop on the railway station, drops an apple aiming at the open hand of his brother situated vertically below his hands at a distance of about $$2 \ m$$. The apple will fall

A
In the hand of his brother

B
Slightly away from the hand of his brother in the direction of the motion of the train.

C
Slightly away from the hands of his brother in the direction opposite to the direction of the motion of the train.

D
None of these

Solution

Given, that the train is going to stop at a railway station, which means that the train is under retardation or is accelerating opposite to the direction of motion.

The boy sitting on the topmost berth drops an apple aiming at the open hand of his brother. Let at this instant, the velocity of the train be $$v$$. Since, the boy, his brother and the apple are in the same frame, they are also moving with velocity $$v$$. Hence, when the apple is released, its velocity will be $$v$$ in the direction of motion of train.

Now, the apple falls downward only under the influence of gravitational force. Since, there are no forces acting in the horizontal direction, it will move with constant velocity $$v$$ in the horizontal direction. Let it takes time $$t$$ to come down a height of $$2 \ m$$. The horizontal distance it covered is given by , $$d=vt$$

Since, the train is under retardation, which means that the velocity of the train is decreasing. Hence, horizontal distance, the boy's hands will cover in time $$t$$ will be less than $$vt$$

Hence, the apple will fall slightly away from the hand of his brother in the direction of motion of the train.
Question 4
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Equal force F acts on isolated bodies X and Y as shown in figure. The mass of Y is three times that of X. The magnitude of the acceleration of X is ..........

A
Three times that of Y

B
1/3 that of Y

C
Nine times that of Y

D
1/9 that of Y

Solution

$$F=ma$$
$$a=\dfrac {F}{m}$$
Let mass of X is $$m$$, so mass of Y will be $$3m$$

The acceleration of X, $$A_X=\dfrac{F}{m} $$ and acceleration of Y, $$A_Y=\dfrac{F}{3m}$$.

hence, The magnitude of the acceleration of X is Three times that of Y

Option A
Question 5
1 / -0

Physical independence of force is a consequence of _____.

A
first law of motion

B
second law of motion

C
third law of motion

D
all ofthese laws

Solution
Question 6
1 / -0

A man fires a bullet of mass $$200\ g$$ at a speed of $$5\ m/s$$. The gun is of mass $$1\ kg$$. The velocity with which the gun will recoil backward is:

A
$$3\ ms^{-1}$$

B
$$2\ ms^{-1}$$

C
$$1\ ms^{-1}$$

D
$$0.01\ ms^{-1}$$

Solution

Given,
Mass of the bullet, $$m_b=200\ g=0.2\ kg$$
Initial velocity of bullet, $$v_{b1}=0$$
Muzzle velocity of bullet, $$v_{b2}=5\ m/s$$
Mass of the gun, $$m_g=1\ kg$$
Initial velocity of gun, $$v_{g1}=0$$
Final velocity of gun, $$v_{g2}$$

According to the law of conservation of momentum,
$$Initial\ momentum=Final\ momentum$$

$$m_bv_{b1}+m_gv_{g1}=m_bv_{b2}+m_gv_{g2}$$

$$0=m_bv_{b2}+m_gv_{g2}$$

$$\implies v_{g2}=\dfrac{m_b}{m_g}v_{b2}$$

$$ v_{g2}=\dfrac{0.2}{1}\times5$$

$$ v_{g2}=-1 \ m/s$$

The velocity with which the gun will recoil backward is $$1\ m/s$$
Question 7
1 / -0

A boat at rest has two persons of unequal weights seated at either end. If they walk across the boat and interchange their positions,

A
The boat will move in the direction of walking of the lighter person

B
The boat will move in the direction of walking of the heavier person

C
The boat will not move at all

D
The boat will oscillate about a mean position and stop once the persons have moved to the new positions

E
None of these

Solution

Answer is A.
The Centre of mass of the system should be at rest because no external force is acting on the system of the boat and 2 men.
When two men interchange their position, the center of mass of 2 men will move towards the heavier man but to keep it at its original position, the boat has to move towards the lighter man.
Question 8
1 / -0

A ball A of mass 2.5kg moving at 5 m/s collides with an identical ball B at rest. A stops and B starts moving. What is the magnitude of velocity of B after the collision?

A
5 m/s

B
12.5 m/s

C
2.5 m/s

D
10 m/s

Solution

Given :
$$m_A=m_B = 2.5$$ kg
$$u_A = 5$$ m/s
$$u_B = 0$$ m/s
$$v_A =0$$

Conservation of momentum:
Intial momentum $$=$$ Final momentum
$$m_Au_A + m_Bu_B = m_Av_A + m_Bv_B$$

$$\therefore$$ $$2.5(5) + 0 = 0 + 2.5v_B$$

$$\implies v_B = 5$$ m/s
Question 9
1 / -0

A stone is tied to the middle of a string and suspended from one end as shown in the given figure. Here S is the stone and O is the point of suspension. If we increase the pull at P gradually, the string will break

A
below the stone

B
at the point P itself

C
above the stone

D
nothing can be decided

Solution

When we increase force gradually, there is enough time that generates same tension in other side of stone.
So, we apply breaking stress on thread which is above the stone.
So, thread above the stone breaks.
Question 10
1 / -0

Consider two spring balances hooked as shown in the figure. We pull them in opposite directions. If the reading shown by A is 1.5 N, the reading shown by B will be :

A
$$1.5N$$

B
$$2.5N$$

C
$$3.0N$$

D
$$zero$$

Solution

When we apply force F on spring balance on one end, it gives reaction of same force F at another end and it reads force as F. Now, generally another end is fixed to a support.
So, in practice, there are always two forces present at both ends, which is of same magnitude.
So,
in this case, both forces are $$1.5N $$ on both ends reads $$1.5N$$

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Re-Attempt Weekly Quiz Competition

Forces and Laws of Motion Test - 45

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Forces and Laws of Motion Test - 45

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SHARING IS CARING

Question 1

$$5$$ m/s

$$2$$ m/s

$$50$$ m/s

$$0.5$$ m/s

Solution

Question 2

$$2 \times 10^5 kg m/s$$

$$1 \times 10^5 kg m/s$$

4 kg m/s

none of these

Solution

Question 3

In the hand of his brother

Slightly away from the hand of his brother in the direction of the motion of the train.

Slightly away from the hands of his brother in the direction opposite to the direction of the motion of the train.

None of these

Solution

Question 4

Three times that of Y

1/3 that of Y

Nine times that of Y

1/9 that of Y

Solution

Question 5

first law of motion

second law of motion

third law of motion

all ofthese laws

Solution

Question 6

$$3\ ms^{-1}$$

$$2\ ms^{-1}$$

$$1\ ms^{-1}$$

$$0.01\ ms^{-1}$$

Solution

Question 7

The boat will move in the direction of walking of the lighter person

The boat will move in the direction of walking of the heavier person

The boat will not move at all

The boat will oscillate about a mean position and stop once the persons have moved to the new positions

None of these

Solution

Question 8

5 m/s

12.5 m/s

2.5 m/s

10 m/s

Solution

Question 9

below the stone

at the point P itself

above the stone

nothing can be decided

Solution

Question 10

$$1.5N$$

$$2.5N$$

$$3.0N$$

$$zero$$

Solution

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