Laws of Motion Test

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Laws of Motion Test - 38

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Question 1
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In the given figure a block of weight $$10 N$$ is resting on a horizontal surface. The coefficient of static friction between the block and the surface is $$\mu_s = 0.4$$. A force of $$3.5 N$$ will keep the block in uniform motion, once it has been set in motion. A horizontal force of $$3 N$$ is applied to the block, then the block will

A
move over the surface with constant velocity.

B
move having accelerated motion over the surface.

C
will not move.

D
first it will move with a constant velocity for some time and then it will have an accelerated motion.

Solution

$$f_s^{max}=\mu N = 0.4\times 10N = 4N$$
The applied force is less than max $$f_s^{max}$$, so the block would not move.
Question 2
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A bob hangs from a rigid support by an inextensible string of length $$\ell$$. If it is displaced through a distance $$\ell$$ (from the lowest position) keeping the string straight, & then released. The speed of the bob at the lowest position is:

A
$$\sqrt{g\ell}$$

B
$$\sqrt{3g\ell}$$

C
$$\sqrt{2g\ell}$$

D
$$\sqrt{5g\ell}$$

Solution

Distance= $$l$$
change in height= $$lcos60 = l/2$$

From energy conservation,
PE at max height= KE at lowest point= $$mgl/2= mv^2/2$$
$$v=\sqrt{gl}$$
Question 3
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A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for At =0.01 sec, the impulse of the impact force is
(take 9 = 10 $$m/s^2$$)

A
1.3 N - s

B
1.06 N - s

C
1300 N - s

D
105 N - s

Solution

Given : $$m = 0.05$$ kg $$h = 10$$ m
Kinetic energy of ball just before collision $$E_i = mgh = 0.05\times 10\times 10 = 5$$ J
Thus initial momentum of the ball $$P_i = \sqrt{2mE_i} = \sqrt{2(0.05)(5)} =0.707$$ N-s (downwards)
Kinetic energy of ball just after collision $$E_f = \dfrac{E_i}{4} =\dfrac{5}{4} = 1.25$$ J
Thus final momentum of the ball $$P_f = \sqrt{2mE_f} = \sqrt{2(0.05)(1.25)} =0.353$$ N-s (upwards)
$$\therefore$$ Impulse $$I = \Delta P = P_f- (-P_i) = 0.353 + 0.707 = 1.06$$ N-s
Question 4
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A metal block is resting on a rough wooden surface. A horizontal force applied to the block is increased uniformly. Which of the following curves correctly represents velocity of the block ?

A

B

C

D

Solution

The force on the block starts increasing uniformly which states that the acceleration is increasing uniformly. The block starts moving after some time when the force is greater than the friction acting into play. The $$velocity$$ $$-$$ $$time$$ graph for a uniformly accelerated body is a curve and for this case it is the curve starting after time $$t$$
Question 5
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A block of mass $$2 kg$$ is placed on the floor. The coefficient of static friction is $$0.4$$. Force of $$2.8N$$ is applied on the block. The force of friction between the block and the floor is (Taken $$g = 10m/s^2$$)

A
2.8 N

B
8 N

C
2.0 N

D
zero

Solution

The maximum static frictional force is $$\mu mg=0.4\times2\times10=8N$$, which is more than the applied force.
So, when a force of $$2.8N$$ is applied, static frictional force is also $$2.8$$ N, as it is a self-adjusting force.
Question 6
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A particle slides from rest from the topmost point of a vertical circle of radius ($$r$$) along a smooth chord making an angle $$(\theta)$$ with the vertical the time of decent is

A
Least for $$\theta=0^{\circ}$$

B
Maximum for $$\theta=0^{\circ}$$

C
Least for $$\theta=45^{\circ}$$

D
Independent of $$\theta$$

Solution

Force on particle along the cord = $$mg\cos { \theta }$$
Distance travelled by the particle =$$\quad \dfrac { d }{ \cos { \theta } }$$, where $$d$$ is the diameter or the vertical circle.

$$s = \dfrac { a{ t }^{ 2 } }{ 2 }$$

$$\Rightarrow\ t =\sqrt { \dfrac { 2s }{ a } }$$.

Hence,$$\ t$$ is independent of $$\theta$$.
Question 7
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A body of mass M is kept on a rough horizontal surface (friction coefficient =$$\mu$$ ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on A is F where

A
$$F=Mg$$

B
$$F=\mu Mg$$

C
$$Mg\leq F\leq Mg\sqrt {1+\mu^2}$$

D
$$Mg\geq F\geq Mg\sqrt {1-\mu^2}$$

Solution

Force applied by any surface is resultant of friction force and Normal force.
0 $$\le $$ Friction force (f) $$\le $$ $$\mu \times N$$
where, N is Normal force applied by surface.
Normal force $$=$$ Mg (perpendicular to surface)
$$\therefore$$ Resultant force (F) $$=$$ $$\sqrt { { \left( N \right) }^{ 2 }+{ f }^{ 2 } } $$
$${ F }_{ min }$$ $$=$$ N $$=$$ Mg when f $$=$$ 0
$${ F }_{ max }$$ $$=$$ = $$\sqrt { { N }^{ 2 }+{ f }^{ 2 } } $$ when f = $$\mu$$mg
$$=$$ $$\sqrt { { \left( Mg \right) }^{ 2 }+\left( \mu Mg \right) ^{ 2 } } $$
$$=$$ $$Mg\sqrt { 1+{ \mu }^{ 2 } } $$
$$\therefore$$ $$Mg\quad \leq F\quad \leq Mg\sqrt { 1+{ \mu }^{ 2 } } $$
Question 8
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A hammer weighing 3 kg, moving with a velocity of 10 m/s, strikes against the head of a spike and drives it into a block of wood. If the hammer comes to rest in 0.025 s, the impulse associated with the spike will be :

A
30 Ns

B
-30 Ns

C
15 Ns

D
-15 Ns

Solution

Initial momentum of the hammer $$P_i = mu = 3\times 10 = 30 \ kg \ m/s$$
Final momentum of the hammer $$P_f = 0$$
Change in momentum of hammer $$\Delta P = P_f - P_i = 0-30 = -30 \ kg \ m/s$$
Impulse acting on the hammer $$I_{hammer} = \Delta P = -30 \ Ns$$
Impulse acting on the spike $$I_{spike} = -I_{hammer} = 30 \ N \ s$$
Question 9
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A ball comes back after an elastic collision with a wall, If initial momentum of the ball is P, the impulse on the wall after the collision is :

A
$$P$$

B
$$2P$$

C
$$zero$$

D
$$-2P$$

Solution

Initial momentum of the ball $$=P$$
Final momentum of the ball $$=-P$$
Change in momentum $$=-2P$$
Impulse on the wall is equal to the change in momentum imparted on the wall $$=2P$$
Question 10
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A bead is free to slide down a smooth wire tightly stretched between points $$A$$ and $$B$$ on a verticle circle of radius $$R$$. If the bead starts from rest at '$$A$$', the highest point on the circle, its velocity when it arrives at $$B$$ is

A
$$2\sqrt{gR}$$

B
$$2\sqrt{gR} \cos{\theta}$$

C
$$\displaystyle \dfrac{2\sqrt{R}}{g}$$

D
None of the above

Solution

$$a= g\cos{\theta}\\AB = 2R\cos{\theta}\Rightarrow v^2 =u^2 + 2as \\v^2 = 0 + 2 (g\cos{\theta}) 2R\cos{\theta}\Rightarrow v^2 = 4gR \cos ^{2 }{\theta}\\ \Rightarrow v = 2\sqrt{ gR} \cos{\theta}$$