Laws of Motion Test

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Laws of Motion Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

A body $$X$$ of mass $$5 kg$$ is moving with velocity $$20 m s^{-1}$$ while another body Y of mass $$20 kg$$ is moving with velocity $$5 m s^{-1}$$. Compare the momentum of the two bodies.

A
$$1 : 4$$

B
$$4 : 1$$

C
$$1 : 2$$

D
$$1 : 1$$

Solution

For body X,
mass: $$m_X=5\ kg$$
velocity: $$v_X= 20\ m/s$$
$$\therefore$$ momentum: $$p_X=m_X \times v_X=5 \times 20 = 100\ kg\ m/s$$

For body Y,
mass: $$m_Y=20\ kg$$
velocity: $$v_Y= 5\ m/s$$
$$\therefore$$ momentum: $$p_Y=m_Y \times v_Y = 20 \times 5 = 100\ kg\ m/s$$

So, $$p_X:p_Y=1:1$$
Question 2
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A toy car is tied to the end of an unstretched string of a length, $$a$$. When revolved, the toy car moves in a horizontal circle of radius $$2a$$ with time period, $$T$$. If it is now revolved in a horizontal circle of radius $$3a$$ with a period $$T'$$ with the same force, then

A
$$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$$

B
$$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$$

C
$$T' = T$$

D
$$\displaystyle T' = \dfrac{3}{2} T$$

Solution

$$\displaystyle m(2a)\left(\dfrac{2\pi}{T}\right)^2 = m(3a) \left(\dfrac{2\pi}{T'}\right)^2$$.

So, $$\dfrac{T'}{T}=\sqrt{\dfrac{3}{2}}$$
Question 3
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A truck has a width of $$2.8\ m$$. It is moving on a circular road of radius $$48.6\ m$$. The height of centre of mass is $$1.2\ m$$. The maximum speed so that it does not over turn will be nearly

A
$$24.3\ m/s$$

B
$$27.2\ m/s$$

C
$$13.4\ m/s$$

D
$$15.7\ m/s$$

Solution

$$\displaystyle \frac{mgl}{2} =\dfrac{mv^2}{R} h$$

(overturning torque $$=$$ restoring torque)

or $$\displaystyle v=\sqrt{\dfrac{gRl}{2h}}=\sqrt{\dfrac{10\times 50\times 1.4}{1.2}}$$ $$=24.3\ m/s$$
Question 4
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A car of mass $$600 kg$$ is moving with a speed of $$10 m s^{-1}$$ while a scooter of mass $$80 kg$$ is moving with a speed of $$50 m s^{-1}$$. Compare their momentum.

A
$$2 : 3$$

B
$$1 : 2$$

C
$$3 : 1$$

D
$$3 : 2$$

Solution

As we know that momentum is nothing but product of mass and velocity. Let us substitute the given values.

Given that mass of the car as 600 kg and velocity as 10 m/sec.

Hence momentum of the car is $$P=600\times 10=6000 kg m/sec$$.

We have mass of the scooter as 80 kg where as velocity is 50 m/sec.

Hence momentum is $$P=80\times 50=4000 kg m/sec$$.

Therefore, momentum are in the ratio $$3:2$$.
Question 5
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A body of mass m moving with a velocity v is acted upon by a force. Write expression for change in momentum when $$v\rightarrow c$$.

A
$$m \Delta v$$

B
$$\Delta \left( mv \right)$$

C
$$v \Delta\left(m \right)$$

D
$$mv$$

Solution

As we know that momentum (p) =MAASS$$\times$$VELOCITY(vector quantity)
hence chance in momentum($$\Delta$$p)=$${{m}_{2}{v}_{2}}-{{m}_{1}{v}_{1}}$$=$$\Delta$$(mv)
where $${m}_{1}$$=$${m}_{2}$$=m
$$\Delta$$v=$${v}_{2}-{v}_{1}$$
Question 6
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At a curved path of the road, the road bed is raised a little on the side away from the centre of the curved path. The slope of the road bed is given by

A
$$\tan{\theta} = rg/v^2$$

B
$$\tan{\theta} = r/gv^2$$

C
$$\tan{\theta} = v^2/rg$$

D
$$\tan{\theta} = v^2g/r$$

Solution

As shown in the given diagram, the road bed has been raised by an angle $$\theta$$.
Applying Newton's laws in horizontal and vertical directions. we get
$$N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ r } $$ ... $$(I)$$ and $$N\cos { \theta } =mg$$ ... $$(II)$$
by dividing $$(I)$$ by $$(II)$$, we get
$$\tan { \theta } =\dfrac { { v }^{ 2 } }{ rg } $$
Question 7
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A small sphere is attached to a cord and rotates in a vertical circle about a point $$O$$. If the average speed of the sphere is increased, the cord is most likely to break at the orientation when the mass is at

A
Bottom point $$B$$

B
Top point $$A$$

C
Point $$D$$

D
Point $$C$$

Solution

Tension in the cord is maximum (for a given average speed of rotation) when the mass, $$m$$, is at the bottom points B, as the gravitational force is in the downward direction and tension of the cord is directly opposing it.
Question 8
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With what minimum speed $$v$$ must a small ball should be pushed inside a smooth vertical tube from a height $$h$$ so that it may reach the top of the tube? Radius of the tube is $$R$$.

A
$$\displaystyle \sqrt{g(R-h)}$$

B
$$\displaystyle \sqrt{g(2R-h)}$$

C
$$\displaystyle \sqrt{2g(2R-h)}$$

D
$$\displaystyle \sqrt{2g}$$

Solution

Hint: If there is no external force acting on a body then Mechanical Energy of the system remains conserved.
That is,
$$ME_i = ME_f$$
or, $$KE_i + PE_i = KE_f + PE_f$$

Step 1: Calculate Initial Mechanical Energy of the ball when it is at height $$h$$.
Let mass of ball is $$m$$When the ball is at height $$h$$, considering Potential Energy at bottom most point of tube to be zero, Potential Energy of ball is given as,
$$PE_i = mgh$$

Also when the ball is just pushed let initial velocity of ball is $$v$$, then Initial Kinetic Energy of ball is given by,
$$KE_i = \dfrac{1}{2}mv^{2}$$

Step 2: Calculate Final Mechanical Energy of the ball when it is at height $$h$$.
When the ball reaches the uppermost point that is at height $$(2R + 2d)$$ from surface, the Potential Energy of ball is given by,
$$PE_f = mg(2R+2d)$$

We want the minimum value of velocity with which ball should be pushed such that it reaches the uppermost point. So let us consider that ball stops at the uppermost point. That is,
$$KE_f = 0$$

Step 3: Apply Conservation of Mechanical Energy.
Since the surfaces are smooth and no external forces are involved, we can apply conservation of energy. Thus, we get
$$KE_i + PE_i = KE_f + PE_f$$
$$\Rightarrow \dfrac{1}{2}mv^{2} + mgh = 0 + mg(2R+2d)$$
$$\Rightarrow \dfrac{1}{2}v^{2} + gh = g(2R+2d)$$
$$\Rightarrow \dfrac{1}{2}v^{2} = g(2R+2d - h)$$
It is given that $$d<<R$$, thus we can neglect $$d$$ from Right Hand Side.
Therefore,
$$ \dfrac{1}{2}v^{2} = g(2R- h)$$
$$\Rightarrow v^{2} = 2g(2R- h)$$
$$\Rightarrow v = \sqrt{2g(2R- h)}$$
This is the required expression.

Thus, minimum velocity with which ball should be pushed is, $$v = \sqrt{2g(2R- h)}$$.
Option C is correct.
Question 9
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Find the maximum speed at which a truck can safely travel without toppling over, on a curve of radius $$250m$$. The height of the centre of gravity of the truck above the ground is $$1.5 m$$ and the distance between the wheels is $$1.5m$$, the truck being horizontal.

A
$$\displaystyle 15\:ms^{-1}$$

B
$$\displaystyle 25\:ms^{-1}$$

C
$$\displaystyle 30\:ms^{-1}$$

D
$$\displaystyle 35\:ms^{-1}$$

Solution

The Truck will topple about the outer wheels. let the height of the center of gravity be h, distance between the tyres be d, and radius of curvature be R, for toppling the maximum speed is given by,(equating the torque by gravity and torque by centrigal force about the outer wheel) $$\dfrac{m{ { v }_{ max } }^{ 2 } }{ R } h=mg\dfrac { d }{ 2 } $$
$$\Rightarrow { v }_{ max }=\sqrt { \dfrac { Rgd }{ 2h } } =35.35m/s$$
Question 10
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A bullet of mass $$50 g$$ moving with an initial velocity $$100 m {s}^{-1}$$, strikes a wooden block and comes to rest after penetrating a distance $$2 cm$$ in it. Calculate: (i) initial momentum of the bullet, (ii) final momentum of the bullet.

A
$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$5$$ $$kg$$ $$m$$ $${s}^{-1}$$

B
$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$0$$ $$kg$$ $$m$$ $${s}^{-1}$$

C
$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$2.5$$ $$kg$$ $$m$$ $${s}^{-1}$$

D
$$2.5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$0$$ $$kg$$ $$m$$ $${s}^{-1}$$

Solution

Mass, $$m=50\ g=0.05\ kg$$
Initial velocity, $$u=100\ m/s$$
Final velocity, $$v=0$$
(i) Initial momentum, $$p_i=0.05 \times 100=5\ kgm/s$$
(ii) Final momentum, $$p_f=0.05 \times 0 = 0$$

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Laws of Motion Test - 34

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Laws of Motion Test - 34

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SHARING IS CARING

Question 1

$$1 : 4$$

$$4 : 1$$

$$1 : 2$$

$$1 : 1$$

Solution

Question 2

$$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$$

$$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$$

$$T' = T$$

$$\displaystyle T' = \dfrac{3}{2} T$$

Solution

Question 3

$$24.3\ m/s$$

$$27.2\ m/s$$

$$13.4\ m/s$$

$$15.7\ m/s$$

Solution

Question 4

$$2 : 3$$

$$1 : 2$$

$$3 : 1$$

$$3 : 2$$

Solution

Question 5

$$m \Delta v$$

$$\Delta \left( mv \right)$$

$$v \Delta\left(m \right)$$

$$mv$$

Solution

Question 6

$$\tan{\theta} = rg/v^2$$

$$\tan{\theta} = r/gv^2$$

$$\tan{\theta} = v^2/rg$$

$$\tan{\theta} = v^2g/r$$

Solution

Question 7

Bottom point $$B$$

Top point $$A$$

Point $$D$$

Point $$C$$

Solution

Question 8

$$\displaystyle \sqrt{g(R-h)}$$

$$\displaystyle \sqrt{g(2R-h)}$$

$$\displaystyle \sqrt{2g(2R-h)}$$

$$\displaystyle \sqrt{2g}$$

Solution

Question 9

$$\displaystyle 15\:ms^{-1}$$

$$\displaystyle 25\:ms^{-1}$$

$$\displaystyle 30\:ms^{-1}$$

$$\displaystyle 35\:ms^{-1}$$

Solution

Question 10

$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$5$$ $$kg$$ $$m$$ $${s}^{-1}$$

$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$0$$ $$kg$$ $$m$$ $${s}^{-1}$$

$$5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$2.5$$ $$kg$$ $$m$$ $${s}^{-1}$$

$$2.5$$ $$kg$$ $$m$$ $${s}^{-1}$$, $$0$$ $$kg$$ $$m$$ $${s}^{-1}$$

Solution

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