Laws of Motion Test

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Laws of Motion Test - 37

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Question 1
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A hockey ball of mass 200 g travelling at $$10\ ms^{-1}$$ is struck by a hockey stick so as to return it along its original path with a velocity of $$5\ ms^{-1}$$. Calculate the change of momentum which occurred in the motion of the hockey ball by the force applied by the hockey stick:

A
$$-3\ Ns$$

B
$$3\ Ns$$

C
$$6\ Ns$$

D
$$-6\ Ns$$

Solution

Mass of ball(m)$$=$$200g$$=$$0.2kg
Initial velocity of ball(u)$$=$$10m$${s}^{-1}$$
Final velocity of ball(v)$$=$$5m$${s}^{-1}$$
Therefore, change in momentum ($$\triangle$$p) $$=$$ Final momentum - initial momentum $$=m(v-u)=0.2\times(-5-10)=-3\ kgm{s}^{-1}=-3\ Ns$$
Question 2
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A boy holds a pendulum in his hand while standing at the edge of a circular platform of radius $$r$$ rotating at an angular speed $$\omega$$. The pendulum will hang at an angle $$\theta$$ with the verticle such that

A
$$\;tan\,\theta=0$$

B
$$\;tan\,\theta=\displaystyle\frac{\omega^2r^2}{g}$$

C
$$\;tan\,\theta=\displaystyle\frac{r\omega^2}{g}$$

D
$$\;tan\,\theta=\displaystyle\frac{g}{\omega^2r}$$

Solution

A pseudo force $$(ma_r)$$ acts on the bob due to the rotation of circular platform.
In bob's frame, the bob is at rest (or equilibrium).
$$\therefore$$ $$T sin \theta = ma_r$$ where $$a_r = rw^2$$
$$\implies$$ $$T sin\theta = mrw^2$$ ........(1)

Also $$T cos\theta =mg$$ ..........(2)
Divide (2) by (1) we get, $$tan\theta =\dfrac{rw^2}{g}$$
Question 3
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Figures $$I, II, III\ and\ IV$$ depicts variation of force with time. In which situation impulse will be maximum?

A
$$I \&II$$

B
$$III \& I$$

C
$$III \& IV$$

D
$$Only\ IV$$

Solution

$$I$$. Area is $$0.25$$
$$II$$. Area is $$0.15$$
$$III$$. Area is $$0.5$$
$$IV$$. Area is $$0.5$$
Since area under the curve ( $$\Delta P$$ is impulse ), $$III$$ and $$IV$$ have maximum impulse.
Question 4
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A particle of mass $$'m'$$ describes a circle of radius $$(r)$$. The centripetal acceleration of the particle is $$\displaystyle\frac{4}{r^2}$$. The momentum of the particle:

A
$$\;\displaystyle\frac{2m}{r}$$

B
$$\;\displaystyle\frac{2m}{\sqrt{r}}$$

C
$$\;\displaystyle\frac{4m}{r}$$

D
$$\;\displaystyle\frac{4m}{\sqrt{r}}$$

Solution

Centripetal acceleration of the particle, $$a = \dfrac{v^2}{r}$$
$$\therefore$$ $$\dfrac{4}{r^2}= \dfrac{v^2}{r} $$ $$\implies v = \dfrac{2}{\sqrt{r}}$$
Thus the momentum of the particle, $$P = mv = \dfrac{2m}{\sqrt{r}}$$
Question 5
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A mass tied to a string moves in a vertical circle with a uniform speed of $$5\;m/s$$ as shown. At the point $$P$$ the string breaks. The mass will reach a height above $$P$$ of nearly: $$(g=10\ m/s^{2})$$

A
$$\;1\;m$$

B
$$\;0.5\;m$$

C
$$\;1.75\;m$$

D
$$\;1.25\;m$$

Solution

Given : $$u =5 m/s$$ $$a = - g = -10 m/s^2$$
When the string breaks at point P, from there onwards the mass has the free falling motion.
At the maximum height attained, the velocity of the mass is zero i.e $$v = 0 m/s$$
Using $$v^2 - u^2 = 2aS$$
$$\therefore$$ $$ 0 - 5^2 = 2 (-10) h$$ $$\implies h = 1.25$$ m
Question 6
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In a vertical circle of radius $$(r)$$, at what point in its path a particle may have tension equal to zero:

A
highest point

B
lowest point

C
at any point

D
at a point horizontal from the centre of radius

Solution

Let the tension in the string at the highest point be $$T$$
Minimum speed required by the particle at the highest point to complete the vertical circular motion is $$\sqrt{gr}$$
$$\therefore$$ $$\dfrac{mv^2}{r} =T + mg$$

OR $$\dfrac{m (gr)}{r} = T+mg$$ $$\implies T =0$$
Thus tension can be zero at the highest point.
Question 7
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A stone attached to one end of a string is whirled in a vertical circle. The tension in the string is maximum when

A
the string is horizontal

B
the string is vertical with the stone at highest position

C
the string is vertical with the stone at the lowest position

D
the string makes an angle of $$45^{\circ}$$ with the vertical

Solution

$$Answer:-$$ C
When the stone is at the bottom position forces acting on stone are:
Upward force tension (T), Downward forces are weight (mg) and also $$\dfrac{mv^2}{R}$$ centrifugal force
So balancing forces we get :
$$T=mg+\dfrac{mv^2}{R}$$
And minimum tension is obtained at highest point which will be:
$$T=mg-$$ $$\dfrac{mv^2}{R}$$
Question 8
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A particle is moving in a vertical circle, the tension in the string when passing through two position at angle $$30^{\circ}$$ and $$60^{\circ}$$ from vertical from lowest position are $$T_1$$ and $$T_2$$ respectively, then

A
$$T_1 = T_2$$

B
$$T_1 > T_2$$

C
$$T_1 < T_2$$

D
$$T_1\ge T_2$$

Solution

In equilibrium, $$T=mv^2/r+mg\cos\theta$$
Since, $$\theta$$ increases $$\cos\theta$$ and $$v$$ both will decrease.
Hence, for $$\theta=60^o$$, the tension will be less.
Thus, $$T_1>T_2$$.
Question 9
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A pendulum bob has a speed $$3\;m/s$$ while passing through its lowest position, length of the pendulum is $$0.5\;m$$ then its speed when it makes an angle of $$60^{\circ}$$ with the vertical is

A
$$\;2\;m/s$$

B
$$\;1\;m/s$$

C
$$\;4\;m/s$$

D
$$\;3\;m/s$$

Solution

$$\textbf{Step 1: Apply energy conservation}$$
In $$\Delta ABD$$,

$$AB = AD \cos \theta$$

$$\Rightarrow AB = l \cos \dfrac{\pi}{3} = \dfrac{l}{2}$$

$$\Rightarrow h = l - \dfrac{l}{2} = \dfrac{l}{2}$$

As there are no dissipative force energy will be conserved.
$$\Rightarrow (E)_i = E_f$$

$$\Rightarrow K_1 + U_1 = K_2 + U_2$$
Taking point $$C$$ is zero potential level

$$\Rightarrow \dfrac{1}{2} mu^2 = \dfrac{1}{2} mv^2 + (mgh)$$

$$\textbf{Step 2: Calculations}$$

$$\Rightarrow v = \sqrt{u^2 - 2gh}$$

Putting given values
$$v = \sqrt{(3)^2(m/s)^2 - 2 \times 9.8(m/s^2) \times \dfrac{0.5}{2}(m)} = 2 m/s$$

Hence $$A$$ is the correct option.
Question 10
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A rod of weight $$w$$ is supported by two parallel knife edges $$A$$ & $$B$$ and is in equilibrium in a horizontal position. The knives are at a distance $$d$$ from each other. The center of mass of the rod is at a distance $$x$$ from $$A.$$ The normal reactions at $$A$$ and $$B$$ will be

A
$$N_A \, =\, 2w (1\, -\, x/d),\, N_B\, =\, wx/d$$

B
$$N_A\, =\, w\, (1\, -\, x/d),\, N_B\, =\, wx/d$$

C
$$N_A\, =\, 2w(1\, -\, x/d), N_B\, =\, 2wx/d$$

D
$$N_A\, =\, w(2\, -\, x/d), N_B\, =\, wx/d$$

Solution

Rod is in the equilibrium,
$$\sum F_Y$$=0 $$\Rightarrow N_A + N_B = w$$ ......1

Taking moments about point A,
Moments clockwise = Moments Anticlockwise
$$w\times x = N_B \times d$$
$$\therefore N_B= \dfrac{wx}{d}$$ .......2
Solving equations 1 and 2,
$$N_A= w(1-\dfrac{x}{d})$$, $$ N_B= \dfrac{wx}{d}$$