Laws of Motion Test

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Laws of Motion Test - 49

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Question 1
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A rigid ball of mass m strikes a rigid wall at $$60^0$$ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be:

A
mV

B
2 mV

C
$$\dfrac{mV}{2}$$

D
$$\dfrac{mV}{3}$$

Solution

Given, $$P_i= P_f= mV$$
Change in momentum of the ball = $$\overline {P}_f-\overline{P}_i$$
=$$(-P_{fx}\hat{i}-P_{fy}\hat{j})-(P_{ix}\hat{i}-P_{iy}\hat{j})$$

=$$-\hat{i}(P_{fx}+P_{ix})-\hat{j}(P_{fy}-P_{iy})$$

=$$-2P_{ix}\hat{i}= -mV\hat{i} $$ $$ [ P_{fy}-P_{iy}=0]$$

Impulse imparted by the wall = change in the momentum of the ball = mV.
Question 2
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A house is on built on the top of a hill with $$45^0$$ slope. Due to the sliding of material and sand from top to the bottom of hill, the slope angle has been reduced.
If the coefficient of static friction between sand particles is 0.75, what is the final angle attained by hill? ($$ tan^{-1}0.75 = 37^0$$)

A
$$8^0$$

B
$$45^0$$

C
$$37^0$$

D
$$30^0$$

Solution

In equilibrium, when sliding stops

$$N=mg\cos\theta…(1)$$

And $$f=mg\sin\theta=\mu N…(1)$$

Dividing $$(2)$$ by$$(1)\\ \Rightarrow \mu=\tan\theta=0.75=\cfrac{3}{4}\\ \Rightarrow \theta=37°$$
Question 3
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A man revolves a stone of mass m tied to the end of a string in a vertical circle of radius R, The net force at the lowest and height points of the circle directed vertical downwards are
Here $$T_1, T_2$$ and $$v_1, v_2$$ denote the tension in the string and the speed of the stone at the lowest and highest points, respectively.

A
Lowest point: $$mg - T_1$$ Highest point : $$mg + T_2$$

B
Lowest point: $$mg + T_1$$ Highest point : $$mg - T_2$$

C
Lowest point: $$mg + T_1 -\frac{mv^2}{R}$$ Highest point: $$mg - T_2 +\frac{mv^2}{r}$$

D
Lowest point: $$mg - T_1 -\frac{mv^2}{R}$$ Highest point: $$mg + T +\frac{mv^2}{r}$$

Solution
Question 4
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For the conditions of the equilibrium of the body, i.e. the rigid body only the external forces defines the equilibrium. Because the internal forces cancels out so not to be considered.

A
The first part of the statement is false and other part is true

B
The first part of the statement is false and other part is false too

C
The first part of the statement is true and other part is false

D
The first part of the statement is true and other part is true too

Solution

Solution:
For the conditions of the equilibrium of the body, i.e. the rigid body both the external forces and internal forces defines the equilibrium. so the 1st statement is true and 2nd is false
Hence the correct opt: C
Question 5
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Two wires $$AC$$ and $$BC$$ are tied at $$C$$ of a small sphere of mass $$5\ kg$$, which revolves at a constant speed $$v$$ in the horizontal plane with the speed $$v$$ of radius $$1.6\ m$$. Find the minimum value of $$v$$.

A
$$4\ m{s}^{-1}$$

B
$$2\ m{s}^{-1}$$

C
$$2.5\ m{s}^{-1}$$

D
None of these

Solution

From $$FBD$$
$$T_1\cos 30^0+T_2\cos 45^0=mg$$
$$T_1\sin 30^0+T_2\sin 45^0=\dfrac{mv^2}{r}$$
$$T_1\sin 30^0-T_1\cos 30^0=\dfrac{mv^2}{R}-mg$$ $$\because\{\sin45^0=\cos 45^0\}$$
$$T_1=\dfrac{\dfrac{mv^2}{R}-mg}{\left[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\right]}$$

as it a string tension, So it can't be negative
$$T_1 > 0$$ i.e $$\dfrac{mv^2}{R}-mg \ge 0$$
$$\dfrac{v_2}{r}-g \ge 0$$
$$v\ge \sqrt{gR}$$
$$V_{min}\sqrt{gr}=\sqrt{10\times 1.6}=\sqrt{16}=4ms^{-1}$$
Hence (A) is the correct options.
Question 6
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Which of the following is correct?

A
The application of the conditions of the equilibrium of the body is valid only in the 2D

B
The application of the conditions of the equilibrium of the body is valid only in the 3D

C
The application of the conditions of the equilibrium of the body is valid only in the 1D

D
The application of the conditions of the equilibrium of the body is valid throughout

Solution

The application of the conditions of the equilibrium of the body is valid throughout. This means that the conditions are irrespective of the dimensions. The conditions are the basic rules that defines the equilibrium of the body and thus are applicable in any dimension of the real axis.
Question 7
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A particle of mass 1 kg is suspended by means of a string of length L=2 m. The string makes $$6/\pi$$ rps around a vertical axis through the fixed end. The tension in the string is

A
72 N

B
36 N

C
288 N

D
10 N

Solution

The tension is the centripetal force = $$mw^2R$$

The angular velocity is w=$$(6/\pi)(2 \pi) =12$$ rads/s

Substituting we get, Tension = $$1 \times 144 \times 2 = 288 N$$

Thus the correct options is (c)
Question 8
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A mass attached to a string that is itself attached to the ceiling swings back and forth. If the bob is observed to be moving upward at a given instance, as shown to the right, which arrow best depicts the direction of the net force acting on the bob at that instant

A
A

B
B

C
C

D
D

Solution

Two forces act on the bob. (i) is the tension in the string and the (ii) mg sin $$\theta$$, which will be tangential to the path. The resultant of both the forces will be along vector C

Thus the correct option is (c)
Question 9
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Two blocks of same mass ($$4$$kg) are placed according to diagram. Initial velocities of bodies are $$4$$ m/s and $$2$$ m/s and the string is taut. Find the impulse on $$4$$ kg when the string again becomes taut.

A
$$24$$ N-s

B
$$6$$ N-s

C
$$4$$ N-s

D
$$2$$ N-s

Solution

$$s=4t=2t+5t^2$$
$$5t^2=2t$$
$$t=0.4$$
$$v=2+gt=6$$m/s
$$4\times 4\times 4\times 6=8$$v
$$v=5$$
$$J=4(v-4)=4$$.
Question 10
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Momentum of an object changes from $$5\ kg\ m/s$$ to $$15\ kg\ m/s$$ in $$2$$ seconds. What is the force applied on the object?

A
$$10\ N$$

B
$$5\ N$$

C
$$20\ N$$

D
$$40\ N$$

Solution

$$F = \dfrac{\Delta p}{\Delta t} = \dfrac{(p2 - p1)}{t} = \dfrac{(15-5)}{2} = 5\ N$$