Laws of Motion Test

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Laws of Motion Test - 56

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Question 1
1 / -0

A cyclist is moving on a circular path with constant speed $$v$$. What is the change in its velocity after it has described an angle of $$30^{o}$$

A
$$v\sqrt{2}$$

B
$$\dfrac{v}{2}$$

C
$$v(0.3\sqrt{3})$$

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

Choose the incorrect statement

A
The product of force and time for which it acts is called impulse

B
Change in linear momentum per unit time is called impulse

C
Change in momentum in small interval of time is called impulse

D
Change in linear momentum in a given time is called impulse

Solution
Question 3
1 / -0

The magnitude of the impulse developed by a mass of $$0.2$$kg which changes its velocity from $$5\hat{i}+3\hat{j}+7\hat{k}$$ m/s to $$2\hat{i}+3\hat{j}+\hat{k}$$

A
$$2.7$$N-s

B
$$1.3$$N-s

C
$$0.9$$N-s

D
$$3.6$$N-s

Solution

Given,
$$m=0.2kg$$
$$\vec v_i=5\hat i+3\hat j+7\hat k$$
$$\vec v_f=2\hat i+3\hat j+\hat k$$
Impulse, $$I=m(\vec v_i-\vec v_f)$$
$$\vec I=0.2[(5\hat i+3\hat j+7\hat k)-(2\hat i+3\hat j+\hat k)]$$
$$\vec I=0.2(3\hat i+0\hat j+6\hat k)$$
Magnitude, $$|\vec I|=0.2\sqrt{3^2+0^2+6^2}$$
$$I=1.3Ns$$
The correct option is B.
Question 4
1 / -0

In the state of equilibrium
I) Net force is zero
II) Body at rest will continue to be in its state of rest
III) Body moving with uniform velocity will continue to be in its state of uniform velocity
IV) Body with linear momentum $$\vec{p}$$ will continue to have the same momentum $$\vec{p}$$.

A
All are correct

B
I, II, III are correct

C
II, III, IV are correct

D
None of these

Solution

$$ \begin{array}{l} \text { To the state of equilibrium, Body has no acceleration } \\ \text { means zero force. } \\ \text { and momentum will not change as } \\ \text { velocity does not changes as acceleration is } \\ \text { zero. } \\ \text { Hence Answer is A } \end{array} $$
Question 5
1 / -0

A ball mass 1 kg moving with speed $$10{ ms }^{ -1 }$$ strikers a wall and comes to rest. The impulse of the force on the ball is of magnitude

A
$$5kg{ ms }^{ -1 }$$

B
$$10kg{ ms }^{ -1 }$$

C
$$15kg{ ms }^{ -1 }$$

D
$$20kg{ ms }^{ -1 }$$

Solution

Initial velocity $$u=10m/s$$ and final velocity $$v=0(rest)$$
so the change in momentum will be $$\Delta P=m\Delta V=m(v-u)=1kg(0-10)=-10Kgm/s$$
So magnitude of the impulse will be $$10Kgm/s$$
Question 6
1 / -0

If force acting on a system is zero. the quantity which remains constant

A
acceleration

B
linear momentum

C
potential energy

D
Kinetic energy

Solution

According to Newton's second law: $$F=\dfrac{\Delta P}{\Delta t}$$, where $$\Delta P$$ is the change in linear momentum and $$\Delta t$$ is the change in time.
When $$F=0$$, $$\Delta P=0$$.
So, when the force acting on the system is zero, the linear momentum is not changing, or it is a constant.
Question 7
1 / -0

A uniform rod of mass M has an impulse applied at right angles to one end. If the other end begins to move with speed V, the magnitude of the impulse is

A
$$MV$$

B
$$\dfrac {MV} {2}$$

C
$$2MV$$

D
$$\cfrac {2MV} {3}$$

Solution
Question 8
1 / -0

A bod of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make real the point B.The angle $$\theta$$ at which the speed of the bob is half of that it A satisfies:

A
$$\theta=\dfrac{\pi}{4}$$

B
$$\dfrac{\pi}{4} < \theta<\frac{\pi}{2}$$

C
$$\dfrac{\pi}{2} < \theta<\dfrac{3\pi}{4}$$

D
$$\dfrac{3\pi}{4} < \theta<\pi$$

Solution

Use conservation of energy,
$$\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgl \left( 1-\cos { \theta } \right) -------(i)$$
where $$v_{o}=$$horizontal velocity
$$mg(2l)=\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }-\cfrac { 1 }{ 2 } m{ v }_{ top }^{ 2 }-------(ii)$$
Since $$v_{o}$$ is just sufficient
$$\cfrac { m{ v }_{ top }^{ 2 } }{ l } =T+mg\\ T=0,{ v }_{ top }=\sqrt { gl } $$
Then equation $$(ii)$$,
$$\Longrightarrow { v }_{ o }=\sqrt { 5gl } $$
According to the question, $$v=\cfrac { { v }_{ o } }{ 2 } $$
From equation $$(i),$$
$$\cfrac { 1 }{ 2 } m5gl=\cfrac { 1 }{ 2 } m\left( \cfrac { 5gl }{ 4 } \right) +mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cfrac { 20mgl-5mgl }{ 8 } =mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cos { \theta } =\cfrac { 7 }{ 8 } $$
Hence, $$\cfrac{3 \pi}{4} < \theta < \pi$$.
Question 9
1 / -0

A bob is suspended from the ceiling of a car which is accelerating uniformly on a horizontal road
choose the wrong statement

A
the bob stays a constant angle $$\theta$$ with vertical with respect to car in equilibirum

B
the magnitude linear momentum of a bob as seen from the rod is increasing with time

C
linear momentum is conserved only in car reference of frame

D
linear momentum is conserved in car as well as ground reference of frame

Solution

The force acting on the bob as seen from the ground is the torsion, the gravitational attraction mg. the car is accelerating became its engine is pushing the ground backward and the friction is pushing the car forward. The force of friction is responsible for an increase in linear momentum of the system car and bub.
Hence, solve.
Question 10
1 / -0

Identical constant forces push two identical cars A and B continuously from a starting line to a finish line. The cars move on a frictionless horizontal surface. If car A is initially at rest and car-B is initially moving right with speed $$Vo.$$ Choose the correct statement

A
Car- A has the large change in momentum

B
Car- B has the large change in momentum

C
Both cars has the same change in momentum

D
No enough information is given to decide

Solution

Identical constant forces push two identical cars A and B continuously from a starting line to a finish line. The cars move on a friction less horizontal surface. If car A is initially at rest and car-B is initially moving right with speed of car A has the large change in momentum.
so the correct option is A.

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Laws of Motion Test - 56

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Laws of Motion Test - 56

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SHARING IS CARING

Question 1

$$v\sqrt{2}$$

$$\dfrac{v}{2}$$

$$v(0.3\sqrt{3})$$

$$None\ of\ these$$

Solution

Question 2

The product of force and time for which it acts is called impulse

Change in linear momentum per unit time is called impulse

Change in momentum in small interval of time is called impulse

Change in linear momentum in a given time is called impulse

Solution

Question 3

$$2.7$$N-s

$$1.3$$N-s

$$0.9$$N-s

$$3.6$$N-s

Solution

Question 4

All are correct

I, II, III are correct

II, III, IV are correct

None of these

Solution

Question 5

$$5kg{ ms }^{ -1 }$$

$$10kg{ ms }^{ -1 }$$

$$15kg{ ms }^{ -1 }$$

$$20kg{ ms }^{ -1 }$$

Solution

Question 6

acceleration

linear momentum

potential energy

Kinetic energy

Solution

Question 7

$$MV$$

$$\dfrac {MV} {2}$$

$$2MV$$

$$\cfrac {2MV} {3}$$

Solution

Question 8

$$\theta=\dfrac{\pi}{4}$$

$$\dfrac{\pi}{4} < \theta<\frac{\pi}{2}$$

$$\dfrac{\pi}{2} < \theta<\dfrac{3\pi}{4}$$

$$\dfrac{3\pi}{4} < \theta<\pi$$

Solution

Question 9

the bob stays a constant angle $$\theta$$ with vertical with respect to car in equilibirum

the magnitude linear momentum of a bob as seen from the rod is increasing with time

linear momentum is conserved only in car reference of frame

linear momentum is conserved in car as well as ground reference of frame

Solution

Question 10

Car- A has the large change in momentum

Car- B has the large change in momentum

Both cars has the same change in momentum

No enough information is given to decide

Solution

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