Laws of Motion Test

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Laws of Motion Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

..... of Roads is done by raising the outer edges of the road slightly above the level of inner edge, to avoid accidents.

A
Finishing

B
Banking

C
Inclination

D
None of these

Solution

Banking of roads is done by raising the outer edges to avoid slipping at high speed or to avoid accidents.
Question 2
1 / -0

A horse continues to apply a force in order to move a cart with a constant
speed. This is because:

A
the force applied by the horse balances the gravitational force.

B
the force applied by the horse balances the force of friction.

C
Horse does not apply any force.

D
None of these

Solution

A horse is harnessed to a cart. If it exerts a force on ( or pulls ) the cart, then by the $$\text{Newton's Third Law of Motion} $$ the cart must then exert an equal and opposite force on the horse.

For the horse to move with constant velocity, hence zero net acceleration, the $$\text{frictional force}$$ acting on him from the surface and the mechanical force it applies on the art should and indeed balance each other.

So, B. The force applied by the horse balances the force of friction.
Question 3
1 / -0

A body of mass $$M$$ collides against a wall with velocity $$V$$ and rebounds with the same speed. Its magnitude of change in momentum is:

A
$$0$$

B
$$MV$$

C
$$2MV$$

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

Solution

Given that the body of mass M, velocity V rebounds with the same speed.
Change of momentum $$=M \times $$ (Change in velocity)
$$ = M(V - (-V)) = 2MV $$
Thus, the change in momentum of the rebounding ball is $$2MV$$.
Question 4
1 / -0

China wares are wrapped in straw or paper before packing. This is the application of concept of :

A
Impulse

B
Momentum

C
Acceleration

D
Force

Solution

By wrapping in straw or paper we can reduce the direct impulse to the main object, thus we can reduce damage.
so the answer is A.
Question 5
1 / -0

A particle of mass $$m$$ is projected with a velocity $$6\hat { i } +8\hat { j } $$. Then the magnitude of change in momentum when it just touches ground will be

A
$$0$$

B
$$12m$$

C
$$16m$$

D
$$20m$$

Solution

Initial velocity, $$\overrightarrow { v }_{ i }=6\hat { i } +8\hat { j } $$ and final velocity, $$\overrightarrow { v }_{ f }=6\hat { i } -8\hat { j } $$
change in momentum $$ =\overrightarrow\Delta p=m\left( \overrightarrow{ v }_{ f }-\overrightarrow{ v }_{ i } \right) =-16m\hat { j } $$
$$ \left| \Delta p \right| =16m$$
Question 6
1 / -0

A system of $$n$$ particles is free from any external force. Which of the following is true for the magnitude of the total momentum of the system?

A
It must be zero

B
It could be non-zero, but it must be constant

C
It could be non-zero, and it might not be constant

D
The answer depends on the nature of the internal forces in the system

Solution

The magnitude of the total momentum of the system is equal to total mass of all the n particles times the velocity of centre of mass.
Now, given their is no external force. Hence velocity of centre of mass won't change and hence momentum will be constant.
Question 7
1 / -0

S.I. unit of momentum is :

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

B
$$kg\ mg^{-2}$$

C
$$kg\ mg^2$$

D
$$kg\ m^{-1}s^{-1}$$

Solution

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 $$\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}$$
Question 8
1 / -0

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

A
200 N-s

B
250 N-s

C
500 N-s

D
1000 N-s

Solution

$$\displaystyle F = \frac{m(v-u)}{t}$$

$$ \displaystyle 100 = \frac{2 (v-u)}{10}$$

$$\Rightarrow (v-u) = 500 m s^{-1}$$

Change in momentum $$=m (v-u)$$

$$= 2 \times (500)$$

$$= 1000 \ N - s$$
Question 9
1 / -0

A pendulum hangs from the ceiling of a jeep moving with a speed, $$v$$, along a circle of radius, $$R$$. Find the angle with the vertical made by the pendulum.

A
$$0$$

B
$$\tan ^{-1}\displaystyle \frac{v^{2}}{Rg}$$

C
$$\tan ^{-1}\displaystyle \frac{Rg}{v^{2}}$$

D
None of the above

Solution

Considering accelerations,
$$a_{r}=\displaystyle \dfrac{v^{2}}{R}$$
$$\therefore tan\theta =\displaystyle \dfrac{v^2/R}{g}$$
$$=\displaystyle \dfrac{v^{2}}{Rg}$$
Question 10
1 / -0

A stone tied to a string of length $$L$$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed $$u$$.The magnitude of change in its velocity as it reaches a position, where the string is horizontal is

A
$$\sqrt{u^2 - 2gL}$$

B
$$\sqrt{2gL}$$

C
$$\sqrt{u^2 - gL}$$

D
$$\sqrt{2(u^2 - gL)}$$

Solution

Here $$v^2 - u^2 = -2gl$$ ....(i)

$$v^2 = u^2-2gl$$

Since the velocities are mutually perpendicular, change in velocity

$$\triangle{v} = \sqrt{u^2 + v^2}$$ ....(ii)

$$=\sqrt{u^2 + u^2 - 2gl}$$

(substituting the value of $$v^2$$ from (i))

or $$\triangle{v} = \sqrt{2(u^2 - gl)}$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 31

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Laws of Motion Test - 31

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SHARING IS CARING

Question 1

Finishing

Banking

Inclination

None of these

Solution

Question 2

the force applied by the horse balances the gravitational force.

the force applied by the horse balances the force of friction.

Horse does not apply any force.

None of these

Solution

Question 3

$$0$$

$$MV$$

$$2MV$$

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

Solution

Question 4

Impulse

Momentum

Acceleration

Force

Solution

Question 5

$$0$$

$$12m$$

$$16m$$

$$20m$$

Solution

Question 6

It must be zero

It could be non-zero, but it must be constant

It could be non-zero, and it might not be constant

The answer depends on the nature of the internal forces in the system

Solution

Question 7

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

$$kg\ mg^{-2}$$

$$kg\ mg^2$$

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

Solution

Question 8

200 N-s

250 N-s

500 N-s

1000 N-s

Solution

Question 9

$$0$$

$$\tan ^{-1}\displaystyle \frac{v^{2}}{Rg}$$

$$\tan ^{-1}\displaystyle \frac{Rg}{v^{2}}$$

None of the above

Solution

Question 10

$$\sqrt{u^2 - 2gL}$$

$$\sqrt{2gL}$$

$$\sqrt{u^2 - gL}$$

$$\sqrt{2(u^2 - gL)}$$

Solution

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