Laws of Motion Test

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Laws of Motion Test - 30

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Question 1
1 / -0

A block of mass m is in contact with the cart as shown in figure

The coefficient of static friction between the block and the cart is . The acceleration of the cart that will prevent the block from falling satisfies

A
$$\alpha > \frac{mg}{\mu }$$

B
$$\alpha > \frac{g}{\mu m}$$

C
$$\alpha \geq \frac{g}{\mu }$$

D
$$\alpha < \frac{g}{\mu }$$

Solution

$$\textbf{Step}$$:
Given a block of mass $$m$$ is in contact with the cart.
Here the coefficient of static friction between the block and the cart is $$\mu$$.
Here the figure shows all the forces acting on the block of mass $$m$$
The pseudo force $$=m \alpha$$
Frictional force $$f_s \geq mg$$
Her $$N=m \alpha$$
So, frictional force $$\mu_N \geq mg$$
$$\implies \mu m \alpha \geq mg$$
$$\implies \alpha \geq \dfrac{g}{\mu}$$
Question 2
1 / -0

Instantaneous power of constant force acting on a particle moving in a straight line under the action of this force

A
is constant

B
increases linearly with time

C
decreases linearly with time

D
either increases or decreases linearly with time

Solution

Power is defined as rate of work done
$$\implies$$Power, $$P = F.v$$
Even though the force is constant, due to the force the velocity increases because of which its power increases.
Also from equations of motion we know that $$v = u + at$$
$$\implies$$ instantaneous power increases linearly with time
Question 3
1 / -0

In the following figure all surfaces are assumed to be frictionless and pulley is assumed to be ideal. The block 'A' is projected towards the pulley 'P' with an initial velocity $$u_0$$ then select incorrect option

A
the string would become tight at $$t=\dfrac{2u_0}{g}$$

B
the distance travelled by 'A' before the string is taut is $$\dfrac{u_0^2}{g}$$

C
the distance travelled by 'B' before string is taut is $$\dfrac{2u_0^2}{g}$$

D
the common speed of the blocks just after the string is taut is $$\left [ \dfrac{n+2}{n+1} \right ]u_0$$

Solution

When the string is taut again both the block would travel the same distance
hence $$u_0t=\dfrac{1}{2}gt^2$$ [as 'B' is falling freely]

hence $$t=\dfrac{2u_0}{g}$$ in this time 'A' and 'B' would travel a distance of $$\dfrac{2u_0^2}{g}$$

also when same impulse acts along the wire the change in momentum would be same for both blocks
.
$$J = nm\upsilon - nmu_0$$ (for block A)

$$-J = m\upsilon - m(2u_0)$$ (for block B)

$$\therefore nm\upsilon - nmu_0= 2mu_0-m\upsilon$$

$$(n+1)\upsilon=(n+2)u_0$$

$$\nu=\left [ \dfrac{n+2}{n+1} \right ]u_0$$
Question 4
1 / -0

A body is moving in a vertical circle of radius '$$r$$' by a string. If the ratio of maximum to minimum speeds is $$\sqrt{3}:1$$ , the ratio of maximum to minimum tensions in the string is:

A
3 : 1

B
5 : 1

C
7 : 1

D
9 : 1

Solution

Let $$v_1$$ and $$v_2$$ be the velocities at the lowest and the highest points.
$$\therefore \dfrac {v_1}{v_2}=\dfrac {\sqrt 3}{1}$$
Conserving energy between the lowest and highest points gives
$${v_1}^2={v_2}^2+2gh$$
$$\Rightarrow 3{v_2}^2={v_2}^2+2gh$$
$$\Rightarrow {v_2}^2=gh$$
$$\Rightarrow {v_1}^2=3gh$$
$$h$$ is distance between the lowest and the highest points which is $$2r$$.
$$h=2r$$

Tensions at the lowest and highest point
$$T_1=\dfrac {m{v_1}^2}{r}+mg$$

$$T_2=\dfrac {m{v_2}^2}{r}-mg$$

$$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {{v_1}^2/r+g}{{v_2}^2/r-g}$$

$$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {6gr/r+g}{2gr/r-g}$$

$$\Rightarrow \dfrac {T_1}{T_2}=7/1$$
Question 5
1 / -0

A body of mass $$5\ kg$$ undergoes a change in speed from $$30$$ to $$40\ m/s$$. Its momentum would increase by:

A
$$50\ kg m/s$$

B
$$75\ kg m/s$$

C
$$150\ kg m/s$$

D
$$350\ kg m/s$$

Solution

Given that the mass of the body is $$5\ kg$$ and moves with initial speed $$u = 30\ m/s$$,final speed $$v=40\ m/s$$.
Thus, initial momentum $$=mu= 5\times 30 =150\ kgm/s$$ and
final momentum $$ =mv= 5\times 40 =200\ kgm/s$$
Hence, the change in momentum $$=200-150=50\ kgm/s$$.
Question 6
1 / -0

A simple pendulum of length $$50 cm$$ is suspended from a fixed point $$O$$ and a nail is fixed at a point $$P$$ which is vertically below $$O$$ at some distance. The bob is released when string is horizontal. The bob reaches lowest position then it describes vertical circle whose centre coincides with point $$P$$. The minimum distance between $$O$$ and $$P$$ is:

A
20 cm

B
25 cm

C
30 cm

D
40 cm

Solution

When the ball just reaches the bottom, from conservation of energy, $$v^2=2gL$$
But when it reaches the bottom, it describes a vertical circle. hence $$v^2=5gr$$
where $$r$$ is the radius of the circle.
Hence $$\dfrac{r}{L}=\dfrac{2}{5}$$
$$\implies r=20cm$$
Hence minimum distance between O and P=$$50cm-20cm=30cm$$
Question 7
1 / -0

A mass $$0.1\ kg$$ is rotated in a vertical circle using the cord of length $$1\ m$$, when the cord makes an angle $$60^0$$ with the vertical, the speed of the mass is $$3\ m/s$$ the resultant radial acceleration of mass in that position is:

A
$$9m/s^{2}$$

B
$$4.9m/s^{2}$$

C
$$4.1m/s^{2}$$

D
zero

Solution

Radial acceleration at that position , $$a_r = \dfrac{v^2}{r} - g cos \theta $$
$$a_r = 9 - 4.9 = 4.1\ m/s^2 $$
Question 8
1 / -0

A passenger in a moving train tosses a coin which falls behind him. It means that the motion of the train is

A
Accelerated

B
Uniform

C
Retarded

D
Along circular tracks
Question 9
1 / -0

A rubber ball of mass 250 g hits a wall normally with a velocity $$10 m s^{-1}$$ and bounces back with a velocity of $$8 m s^{-1}$$. The impluse is _______N s.

A
$$-0.5$$

B
$$+0.5$$

C
$$-4.5$$

D
$$+4.5$$

Solution

The ball initially has 10 m/s in the positive x - direction and 8 m/s in the negative x-direction
v $$ = -8 m/s$$
$$ u = 10 m/s$$
Impulse is the change in momentum$$ = mv - mu $$
$$I = 0.25 \times (-8 - 10) N s $$
$$I = -4.5 Ns $$
Question 10
1 / -0

A stone of mass $$1 kg$$ tied to a light inextensible string of length $$L=\dfrac{10}{3}m$$ is whirling in a circular path of radius $$L$$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum is $$4$$ and if $$g$$ is taken to be $$10 \ m/s^2$$, then speed of stone at the highest point of the circle is

A
$$20 m/sec$$

B
$$10\sqrt{3} \ m/sec$$

C
$$5\sqrt{2} \ m/sec$$

D
$$10 m/sec$$

Solution

Let '$$v$$' be the velocity at the top point
Velocity at the bottom point is $$= v^2 + 2gd $$ $$ = v^2 + 4gr$$
Maximum tension in string appears when the stone is at the bottom of the circle.

$$ T_{max} = \dfrac{m(v^2 + 4gr)}{r} + mg $$

Minimum tension appears in a string when the stone is at the top of the circle.

$$ T_{min} = \dfrac{mv^2}{r} - mg$$

$$\dfrac{T_{max}}{T_{min}} = \dfrac{\dfrac{v^2 + 4gr}{r} + g}{\dfrac{v^2}{r} - g}$$ $$ = 4$$

On solving the equation we get, $$v $$$$= 10\ m/s $$

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Laws of Motion Test - 30

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Laws of Motion Test - 30

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Question 1

$$\alpha > \frac{mg}{\mu }$$

$$\alpha > \frac{g}{\mu m}$$

$$\alpha \geq \frac{g}{\mu }$$

$$\alpha < \frac{g}{\mu }$$

Solution

Question 2

is constant

increases linearly with time

decreases linearly with time

either increases or decreases linearly with time

Solution

Question 3

the string would become tight at $$t=\dfrac{2u_0}{g}$$

the distance travelled by 'A' before the string is taut is $$\dfrac{u_0^2}{g}$$

the distance travelled by 'B' before string is taut is $$\dfrac{2u_0^2}{g}$$

the common speed of the blocks just after the string is taut is $$\left [ \dfrac{n+2}{n+1} \right ]u_0$$

Solution

Question 4

3 : 1

5 : 1

7 : 1

9 : 1

Solution

Question 5

$$50\ kg m/s$$

$$75\ kg m/s$$

$$150\ kg m/s$$

$$350\ kg m/s$$

Solution

Question 6

20 cm

25 cm

30 cm

40 cm

Solution

Question 7

$$9m/s^{2}$$

$$4.9m/s^{2}$$

$$4.1m/s^{2}$$

zero

Solution

Question 8

Accelerated

Uniform

Retarded

Along circular tracks

Question 9

$$-0.5$$

$$+0.5$$

$$-4.5$$

$$+4.5$$

Solution

Question 10

$$20 m/sec$$

$$10\sqrt{3} \ m/sec$$

$$5\sqrt{2} \ m/sec$$

$$10 m/sec$$

Solution

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