Laws of Motion Test

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Laws of Motion Test - 17

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Question 1
1 / -0

When a moving body is suddenly stopped, then:

A
Frictional force increases.

B
Friction force remains same as it was while the car was moving.

C
Friction force reduces but has a non zero value.

D
Frictional force reduces to zero as it is a self adjusting force.

Solution

Frictional force arises when there is a relative motion between two surfaces. When body is stopped means there is no relative motion and no resultant force is there hence frictional force acting becomes zero.
Question 2
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Impulse is equal to

A
momentum

B
change in momentum

C
rate of change of momentum

D
rate of change of force

Solution

Impulse $$(\vec{J})$$ is defined as the change in the momentum.
$$\therefore$$ $$\vec{J} = \Delta \vec{P} = m (\vec{v_2} - \vec{v_1})$$
Question 3
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A stone tied to a string is rotated in a vertical circle. The minimum speed with which the stone has to be rotated in order to complete the circle :

A
decreases with increasing the mass of the stone

B
is independent of the mass of the stone

C
decreases with increasing the length of the string

D
is independent of the length of the string

Solution

$$\textbf{Step 1: Energy Conservation [Ref. Fig.]}$$
$$\text{Dependence on m}$$
Let $$u$$ be the minimum speed required at $$B$$.
Apply energy conservation between point $$A$$ and $$B$$ we get
$$KE_A + PE_A = KE_B + PE_B$$
Taking point $$A$$ as zero potential level
$$\Rightarrow$$ $$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} mv^2 $$ $$....(1)$$

From equation $$1$$, we observe that mass $$m$$ is cancelled on both sides.
$$\Rightarrow u$$ is independent of $$m$$.

$$\textbf{Steo 2: Solving Equation}$$
$$\text{Dependence on R}$$
At highest point, Tension will be zero for minimum velocity
$$\Rightarrow$$ gravitational pull will provide necessary centripetal force

$$\Rightarrow mg = \dfrac{mv^2}{R}$$

$$\Rightarrow v^2 = Rg$$ .... Putting in equation $$(1)$$

We get, $$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} m (Rg)$$

$$\Rightarrow u = \sqrt{5Rg}$$

$$\Rightarrow u$$ is proportional to the square root of $$R$$.
Hence option $$B$$ is correct.
Question 4
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Momentum is a measure of:

A
weight

B
mass

C
quantity of motion

D
velocity

Solution

In order to stop a moving body we require a force and depends on the velocity as well as its mass. It was Newton in his second law who defined this new quantity as momentum.
Momentum is the product of mass and velocity of the body or in other words momentum is the measure of quantity of motion.
Question 5
1 / -0

In which of the following cases the net force is not equal to zero?

A
A kite skillfully held stationary in the sky.

B
A ball falling freely from a height.

C
A helicopter hovering above the ground.

D
A cork floating on the surface of water.

Solution

As ball is falling freely that means weight of ball is acting on it hence net force is not zero and in all other cases things are stationary so net force on them is zero.
Question 6
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To reduce the momentum of a given body to half its original value then the velocity must be __________.

A
reduced to half

B
reduced to one-third

C
doubled

D
remain same

Solution

P = mv
so, $$ P_1 / P_2 = v_1 / v_2 $$
So, if $$ P_2 = 1/ 2 \space P_1$$,
then $$v_2 = 1/2\space v_1$$
So option A is correct
Question 7
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Action and reaction

A
always act on two different bodies

B
are equal in magnitude

C
act in opposite directions

D
All the above

Solution

Newton's third law of motion : For every action there is always an equal and opposite reaction acting on the other body.
Question 8
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Momentum has the same units as that of:

A
impulse

B
torque

C
moment of momentum

D
couple

Solution

Momentum = mass $$\times$$ velocity and its unit is $$Kgms^{-1}$$
Impulse of the force(J) =force $$\times$$ time and
$$=ma \times t$$
$$=m\left(\dfrac{u}{t}\right) \times t$$
Thus, impulse (J) is the change of momentum and its unit is $$Kgms^{-1}$$
Question 9
1 / -0

When a body is in equilibrium, which of the following must be true ?

A
There is no force acting on it.

B
The sum of all the forces acting on it must be zero.

C
The sum of all the forces acting on it is not equal to zero.

D
There are exactly two forces acting on the body.

Solution

A body is in equilibrium only when there is no net external force acting on it. So, in other words, the sum of all forces acting on it must be zero for a body to be in equilibrium.
Question 10
1 / -0

The momentum of a body of mass 0.5 kg dropped from a certain height (h), when reaches the ground is 10 N s. The value h is _____m. (take $$g=10 m s^{-2}$$)

A
20

B
40

C
10

D
80

Solution

Let the velocity of the body on reaching the ground be 'v'
Momentum of the body on reaching the ground ,
P$$= mv = 10 Ns$$
$$v = 20 m/s $$

Using equation of motion
$$ v^2 = u^2 + 2gh$$ (u is 0)
$$ 400 = 2gh$$
$$ h = 20 m$$

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Laws of Motion Test - 17

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Laws of Motion Test - 17

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SHARING IS CARING

Question 1

Frictional force increases.

Friction force remains same as it was while the car was moving.

Friction force reduces but has a non zero value.

Frictional force reduces to zero as it is a self adjusting force.

Solution

Question 2

momentum

change in momentum

rate of change of momentum

rate of change of force

Solution

Question 3

decreases with increasing the mass of the stone

is independent of the mass of the stone

decreases with increasing the length of the string

is independent of the length of the string

Solution

Question 4

weight

mass

quantity of motion

velocity

Solution

Question 5

A kite skillfully held stationary in the sky.

A ball falling freely from a height.

A helicopter hovering above the ground.

A cork floating on the surface of water.

Solution

Question 6

reduced to half

reduced to one-third

doubled

remain same

Solution

Question 7

always act on two different bodies

are equal in magnitude

act in opposite directions

All the above

Solution

Question 8

impulse

torque

moment of momentum

couple

Solution

Question 9

There is no force acting on it.

The sum of all the forces acting on it must be zero.

The sum of all the forces acting on it is not equal to zero.

There are exactly two forces acting on the body.

Solution

Question 10

20

40

10

80

Solution

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