MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 9

A cyclist is moving on a circular path with constant speed

$v$ . What is the change in its velocity after it has described an angle of

$30^{o}$

0%
$v\sqrt{2}$
0%
$\dfrac{v}{2}$
0%
$v(0.3\sqrt{3})$
0%
$None\ of\ these$

A body is standing on a merry-go-round which is rotating uniformly in a horizontal plane. His frame of reference is

0%
an inertial frame
0%
a non- inertial frame
0%
either inertial or non- inertial
0%
neither inertial nor non- inertial

A ball of mass

$150g$ moving with an acceleration

$20m/s^2$ is hit by force , which acts on it for

$0.1sec.$ The impulsive force is

0%
$0.5$ N-s
0%
$0.1$ N-s
0%
$0.3$ N-s
0%
$30$ N-s

A ball mass 1 kg moving with speed

$10{ ms }^{ -1 }$ strikers a wall and comes to rest. The impulse of the force on the ball is of magnitude

0%
$5kg{ ms }^{ -1 }$
0%
$10kg{ ms }^{ -1 }$
0%
$15kg{ ms }^{ -1 }$
0%
$20kg{ ms }^{ -1 }$

Explanation

Initial velocity

$u=10m/s$ and final velocity

$v=0(rest)$

so the change in momentum will be

$\Delta P=m\Delta V=m(v-u)=1kg(0-10)=-10Kgm/s$

So magnitude of the impulse will be

$10Kgm/s$

The limiting friction between two surfaces does not depend

0%
On the nature of two surfaces
0%
On normal reaction
0%
On the weight of the body
0%
on volume of the body

In the state of equilibrium
I) Net force is zero
II) Body at rest will continue to be in its state of rest
III) Body moving with uniform velocity will continue to be in its state of uniform velocity
IV) Body with linear momentum

$\vec{p}$ will continue to have the same momentum

$\vec{p}$ .

0%
All are correct
0%
I, II, III are correct
0%
II, III, IV are correct
0%
None of these

Explanation

$\begin{array}{l} \text { To the state of equilibrium, Body has no acceleration } \\ \text { means zero force. } \\ \text { and momentum will not change as } \\ \text { velocity does not changes as acceleration is } \\ \text { zero. } \\ \text { Hence Answer is A } \end{array}$

A uniform rod of mass M has an impulse applied at right angles to one end. If the other end begins to move with speed V, the magnitude of the impulse is

0%
$MV$
0%
$\dfrac {MV} {2}$
0%
$2MV$
0%
$\cfrac {2MV} {3}$

A bod of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make real the point B.The angle

$\theta$ at which the speed of the bob is half of that it A satisfies:

0%
$\theta=\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{4} < \theta<\frac{\pi}{2}$
0%
$\dfrac{\pi}{2} < \theta<\dfrac{3\pi}{4}$
0%
$\dfrac{3\pi}{4} < \theta<\pi$

Explanation

Use conservation of energy,

$\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgl \left( 1-\cos { \theta } \right) -------(i)$

where

$v_{o}=$ horizontal velocity

$mg(2l)=\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }-\cfrac { 1 }{ 2 } m{ v }_{ top }^{ 2 }-------(ii)$

Since

$v_{o}$ is just sufficient

$\cfrac { m{ v }_{ top }^{ 2 } }{ l } =T+mg\\ T=0,{ v }_{ top }=\sqrt { gl }$

Then equation

$(ii)$ ,

$\Longrightarrow { v }_{ o }=\sqrt { 5gl }$

According to the question,

$v=\cfrac { { v }_{ o } }{ 2 }$

From equation

$(i),$

$\cfrac { 1 }{ 2 } m5gl=\cfrac { 1 }{ 2 } m\left( \cfrac { 5gl }{ 4 } \right) +mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cfrac { 20mgl-5mgl }{ 8 } =mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cos { \theta } =\cfrac { 7 }{ 8 }$

Hence,

$\cfrac{3 \pi}{4} < \theta < \pi$ .

A bob is suspended from the ceiling of a car which is accelerating uniformly on a horizontal road
choose the wrong statement

0%
the bob stays a constant angle $\theta$ with vertical with respect to car in equilibirum
0%
the magnitude linear momentum of a bob as seen from the rod is increasing with time
0%
linear momentum is conserved only in car reference of frame
0%
linear momentum is conserved in car as well as ground reference of frame

Choose the incorrect statement

0%
The product of force and time for which it acts is called impulse
0%
Change in linear momentum per unit time is called impulse
0%
Change in momentum in small interval of time is called impulse
0%
Change in linear momentum in a given time is called impulse

Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density=1.5 are connected by weightless wire and placed in a large tank of water.Under equilibrium the lighter cube will project above the water surface to a height of:

0%
50 cm
0%
25 cm
0%
10 cm
0%
Zero

A body of mass

$M$ hits normally a rigid wall with velocity

$V$ and bounces back with the same velocity. The impulse experienced by the body is

0%
$MV$
0%
$1.5MV$
0%
$2MV$
0%
zero

Explanation

$\textbf {Hint :}$ Impulse is defined as the change in momentum in very small time.

$\textbf{Step 1: Write initial and final momentum of body}$

Let initially the body is moving along positive x- axis with speed

$v$ .

So, momentum of the body initially

$\vec{p_i} = mv \ \hat{i}$

After bouncing back, the body moves along negative x-axis with same speed.

So, momentum of the body finally

$\vec{p_f} = -mv \ \hat{i}$

$\textbf{Step 2: Calculate impulse}$

Thus change in momentum of body

$\Delta \vec{p} = \vec{p_f} - \vec{p_i} = -2mv \ \hat{i}$

Impulse experienced

$\vec{I} = \Delta \vec{p} = -2mv \ \hat{i}$

$\implies \ |\vec{I}| = 2mv$

${\textbf{Correct option: C}}$

The magnitude of the impulse developed by a mass of

$0.2$ kg which changes its velocity from

$5\hat{i}+3\hat{j}+7\hat{k}$ m/s to

$2\hat{i}+3\hat{j}+\hat{k}$

0%
$2.7$ N-s
0%
$1.3$ N-s
0%
$0.9$ N-s
0%
$3.6$ N-s

Explanation

Given,

$m=0.2kg$

$\vec v_i=5\hat i+3\hat j+7\hat k$

$\vec v_f=2\hat i+3\hat j+\hat k$

Impulse,

$I=m(\vec v_i-\vec v_f)$

$\vec I=0.2[(5\hat i+3\hat j+7\hat k)-(2\hat i+3\hat j+\hat k)]$

$\vec I=0.2(3\hat i+0\hat j+6\hat k)$

Magnitude,

$|\vec I|=0.2\sqrt{3^2+0^2+6^2}$

$I=1.3Ns$

The correct option is B.

A road is

$10 m$ wide. Its radius of curvature is

$50 m$ . The outer edge is above the lower edge by a distance of

$1./5 m$ . This road is most suited for the velocity

0%
$2.5 m/sec$
0%
$4.5 m\sec$
0%
$6.5 m\sec$
0%
$8.5 m\sec$

Explanation

Hint:

· This question is directly formula based, in the question, radius of curvature is given, so here, in this case we need to use the formula of velocity directly.

· The equation of velocity is ${\rm{v = }}\sqrt {{\rm{gRtan}}\theta }$

Step 1:Resolve the forces.

Width of the road is given, ${\rm{L = 10cm}}$
Radius or curvature, ${\rm{R = 50cm}}$
The distance between the above and lower edge of the road, ${\rm{h = 1}}{\rm{.5cm}}$

By applying newtons second law of motion,

$\Sigma F_{net}= m{a_c}$

Here, ${a_c}$ is the centripetal acceleration.

The components of the force will be:

${\rm{Nsin}}\theta {\rm{ = }}\dfrac{{m{v^2}}}{{\rm{R}}}$ ….....(1)

${\rm{Ncos}}\theta {\rm{ = mg}}$ ……(2)

Now, when we divide equation (1) by (2), we get,

${\rm{tan}}\theta {\rm{ = }}\dfrac{{{v^2}}}{{gR}}$

Step 2: Apply the formula of velocity

${\rm{v = }}\sqrt {{\rm{gRtan}}\theta }$ …..(3)

Also, we know that ${\rm{tan}}\theta {\rm{ = }}\dfrac{{\rm{h}}}{{\rm{L}}}$

Now, we put the values in equation (3),

$\Rightarrow {\rm{v = }}\sqrt {10 \times 50 \times \dfrac{{1.5}}{{10}}}$

$\Rightarrow {\rm{v = }}\sqrt {50 \times 1.5}$

$\Rightarrow {\rm{v = }}\sqrt {75}$

$\Rightarrow {\rm{v = 8}}{\rm{.5m/sec}}$

Hence, option (D), ${\rm{8}}{\rm{.5m/sec}}$ is the correct answer.

THIS QUESTION IS NOT CLEARED.

A particle of mass

$m$ is moving a uniform velocity

${v}_{1}$ . It is given an impulse such that its velocity becomes

${v}_{2}$ . The impulse is equal to

0%
$m\left[ \left| { v }_{ 2 } \right| -\left| { v }_{ 1 } \right| \right]$
0%
$1/2m[{v}_{1}^{2}-{v}_{1}^{2}]$
0%
$m[{v}_{1}+{v}_{2}]$
0%
$m[{v}_{2}-{v}_{1}]$

Explanation

Given that,

Mass = $m$

Initial velocity = $v_ {1}$

Final velocity= $v_ {2}$

We know that,

Impulse is the change in momentum.

So, the impulse is

$I=\Delta p$ .....(I)

We know that,

$\Delta p=m\Delta v$

Put the value of $\Delta p$ in equation (I)

$I =m\Delta v$

$\Delta v$ is the change in velocity

The impulse will be

$I=m (v_ {2}-v_ {1})$

The impulse will be $I= m (v_ {2}-v_ {1})$

Hence, this is the required solution.

If force acting on a system is zero. the quantity which remains constant

0%
acceleration
0%
linear momentum
0%
potential energy
0%
Kinetic energy

Explanation

According to Newton's second law:

$F=\dfrac{\Delta P}{\Delta t}$ , where

$\Delta P$ is the change in linear momentum and

$\Delta t$ is the change in time.

When

$F=0$ ,

$\Delta P=0$ .

So, when the force acting on the system is zero, the linear momentum is not changing, or it is a constant.

A particle of mass m is made to move with uniform speed

${ v }_{ 0 }$ along the perimeter of regular hexagon. The magnitude of impulse at each corner of the hexagon is :

0%
$2{ mv }_{ 0 }sin\dfrac { \pi }{ 6 }$
0%
${ mv }_{ 0 }sin\dfrac { \pi }{ 6 }$
0%
${ mv }_{ 0 }sin\dfrac { \pi }{ 3 }$
0%
$2{ mv }_{ 0 }sin\dfrac { \pi }{ 3 }$

A particle is moving in a vertical circle. The tension in the string when passing through two positions at angles

${30}^{o}$ and

${60}^{o}$ from vertical (lowest position) are

${T}_{1}$ and

${T}_{2}$ respectively, then

0%
${T}_{1}={T}_{2}$
0%
${T}_{2}> {T}_{1}$
0%
${T}_{1}> {T}_{2}$
0%
tension in the string always remains the same

Explanation

From the Free Body Diagram we get,

$T= \dfrac{mv^2}{r}+ mg cos \theta$

$\theta= 30 ^0 \quad T = T_1= \dfrac{mv^2}{r}+ \dfrac{\sqrt{3}mg}{2}$

$\theta= 60 ^0 \quad T= T_2= \dfrac{mv^2}{r}+ \dfrac{mg}{2}$

From Above we get

$T_1 > T_2$

1263782_1287520_ans_6a9341e5be2d4687a8edae8f90075cc6.png

Two particles of masses

$m_1$ and

$m_2$ are moving velocities

$\vec{V_1}$ and

$\vec { V_2 }$ Momentum of the particle of mass

$m_1$ with respect to center of mass is

0%
Zero
0%
$\dfrac{m_1 m_2}{m_1 +m_2} (\vec{V_1} -\vec{V_2})$
0%
$\dfrac{m_2}{m_1 +m_2} (\vec{V_1} - \vec{V_2})$
0%
None of these

Explanation

Hence, option

$(D)$ is correct answer.
1380639_1276534_ans_90daebe5c8804509be0eaa759b1f0add.jpg

A rod of length

$l$ and mass

$m$ fixed at one end, it hanging vertically. The other end is now raised so that the rod makes an angle

${30}^{o}$ with horizontal line. The work done in this process will be:

0%
$3mgl/4$
0%
$mgl/2$
0%
$mgl/3$
0%
$mgl/4$

Explanation

Initial potential energy

$= \dfrac{-mgl}{2}\quad \text{(since we have take top point of rod as origin and axis is vertically upwards)}$

Final potential energy

$= \dfrac{ mg (l sin30^o)}{2}$

$=\dfrac{ mgl}{4}$

Work done is change in potential energy

Work done

$=\dfrac{3mgl}{4}$

A block of mass 200 kg is kept on a rough horizontal surface. It just begins to move when a force of 1000 N is applied. Calculate the coefficient of limiting static friction

0%
0.3
0%
0.4
0%
0.5
0%
Insufficient data

An object is subjected to a force in the north-east direction. To balance this force, a second force should be applied in the direction.

0%
North - East
0%
South
0%
South - West
0%
West

Explanation

An object is subjected to a force in the north-east direction. To balance this force, a second force should be applied in the south west direction.

The direction of the second force should be at

$180°$ .

1710906_1305167_ans_599744e991794e1bafc288401d552d1f.png

A glass marble whose mass is 20 gms falls from a height of 0.25 m and rebounds to a height of 1.6 m. If

$g = 9.8 m/s^2$ , find the impulse acting on the fall (in N - s)

0%
0.2
0%
0.4
0%
0.252
0%
0.534

The changes in a momentum of a particle,when the kinetic energy is increased by 0.1% will be:-

0%
0.05%
0%
0.1%
0%
1.0%
0%
10%

Explanation

$\begin{array}{l} The\, \, kineticenergyisE=\frac { 1 }{ 2 } m{ v^{ 2 } } \\ The\, \, \, momentum\, \, is\, \, p=mv \\ v=\sqrt { \frac { { 2E } }{ m } } \\ therefore, \\ \Rightarrow p=mv=m\sqrt { \frac { { 2E } }{ m } } =\sqrt { 2mE } \\ taking\, \log s, \\ \Rightarrow \log \, p=\frac { 1 }{ 2 } \left( { \log 2+\log m+\log \, E } \right) \\ differentiating: \\ \Rightarrow \frac { { \Delta \, p } }{ p } =\frac { 1 }{ 2 } \left( { \frac { { \Delta m } }{ m } } \right) +\frac { 1 }{ 2 } \left( { \frac { { \Delta E } }{ E } } \right) \\ but, \\ \Rightarrow \frac { { \Delta E } }{ E } =0.1 \\ so.\, therefore, \\ \Rightarrow \frac { { \Delta p } }{ p } =\frac { 1 }{ 2 } \times 0.1=0.05 \\ so\, the\, correct\, option\, is\, A. \end{array}$

Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10

$\mathrm { kg } \mathrm { mls }$ is given to the 5

$\mathrm { kg }$ mass. The velocity of the 10

$\mathrm { kg }$ mass just after the impulse, will be

0%
$\cfrac { 2 } { 3 } m / s$
0%
$\cfrac { 1 } { 3 } m / s$
0%
2 $\mathrm { m } / \mathrm { s }$
0%
zero

The acceleration of a body may be doubled by doubling its:

0%
mass
0%
momentum
0%
velocity
0%
force acting on it

Explanation

Acceleration

$a=\dfrac{Force}{mass}$

Also

$a=\dfrac{\Delta P}{\Delta t \times m}$ , Where,

$\Delta P$ is the change in momentum.

So to double the acceleration we need to

$double$ the

$force$ acting on it.

The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s. How many revolutions does it make during these second?

0%
400
0%
600
0%
500
0%
700

The tension in the string revolving in a vertical circle with a mass m at the end which is at the lowest position

0%
$\dfrac{mv^2}{r}$
0%
$\dfrac{mv^2}{r}-mg$
0%
$\dfrac{mv^2}{r}+mg$
0%
$mg$

Explanation

R.E.F image

$F_{c}$ Fore of centrifygal

$T=F_{c}+mg$

$=\frac{mv^{2}}{12}+mg$

so, option (c)

1167816_1339259_ans_7999f5f5e9ea4ca8bf04bb632276d1ac.png

When a simple pendulum is rotated in a vertical plane with constant angular velocity, centripetal force is:

0%
maximum at highest point
0%
maximum at lowest point
0%
same at all points
0%
zero

A car of mass

$1000 kg$ moves on a circular path with a constant speed of

$16 m/s$ .It is turned by

${ 90 }^{ \circ }$ after traveling

$628 m$ on the side. The centripetal force acting on the car is -

0%
$160 N$
0%
$320 N$
0%
$640 N$
0%
$1280 N$

Explanation

$m=1000kg\\ v=16m/s\\ \Longrightarrow 628=r\left( \cfrac { \pi }{ 2 } \right) \Longrightarrow r=400$

$\therefore$ Centripetal force

$=\cfrac{mv^{2}}{r}=\cfrac{(1000)(16)(16)}{400}=640N$

In the figure, at the free end of the light string, a force

$F$ is applied to keep the suspended mass of

$18$ \mathrm { kg }$$ at rest. Assuming

0%
$240 N$
0%
$120 M$
0%
$180 M$
0%
$240 M$

Explanation

$3T=180$

or,

$T=60 N$

Now,

$F=T=60 N$

Therefore ,total force on ceiling is

$4T=240 N$

Hence, Option

$A$ is correct.

A cyclist moves along a circular track at 144 km

$h^{-1}$ .What angle must he make with the vertical if the track is 1 km long?

0%
$40^o$
0%
$45^o$
0%
$35^o$
0%
none

A swing is at 1 m height above ground level in its lowermost position and 3 m in middle position.Its speed at lowest position of vertical circle is-

0%
$2 \sqrt{g}$
0%
$\sqrt{2g}$
0%
$\sqrt {g}$
0%
$\sqrt \frac {g}{2}$

An insect crawls up a hemispherical surface very slowly as shown in the figure. The coefficient of
friction between the surface and the insect is

$\dfrac { 1 }{ 3 }$ If the line joining the centre of the hemisphere surface to the insect makes an angle

$\propto$ with the vertical, the maximum possible value of

$\propto$ is given by

0%
$cot\propto =3$
0%
$tan\propto =3$
0%
$sec\propto =3$
0%
$cosec\propto =3$

A cone filled with water is whirled in a vertical circle of radius 4 m. What will be the time period of rotation so that water does not fall ?

0%
4 s
0%
6 s
0%
2 s
0%
1 s

A track for a certain motor sport event is in the form of a circle and banked at an angle

$\theta$ . For a car driven in a circle of radius r along the track at optimum speed the periodic time is --------

0%
$\sqrt{\dfrac{r}{g}}$
0%
$2 \pi \sqrt{r}{g}$
0%
$2 \pi \sqrt{rg tan \theta}$
0%
$2 \pi \sqrt{\dfrac{g tan \theta}{r}}$

Three blocks A,B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium mass of B is M, then :

0%
M=2 m
0%
M < 2 m
0%
M > 2 m
0%
M=m

A ball of mass 1 kg strikes a heavy platform, elastically, moving upward with a velocity of 5 m/s. The speed of the ball just before the collision is 100 m/s downwards. Then the impulse imparted by the platform on the ball is :-

0%
15 N-s
0%
10 N-s
0%
20 N-s
0%
30 N-s

Explanation

According to the problem the mass of the ball is

$1\, kg$

The velocity of the ball initially is

$5\, m/s$ and the speed of the ball before collision is

$10\, m/s$

Therefore the total speed of the ball

$= 10+5=15 \,m/s$

Now we know that impulse is the change in momentum of ball.

Therefore impulse

$=15+15=30\,Ns$

The breaking strength of a string is 55 kg wt. The maximum permissible speed of a stone of mass 5 kg which revolved in a vertical circle of radius 4 m with the help of is the string is g=10

${s}^{2}$ .

0%
$10 \mathrm{m} / \mathrm{sec}$
0%
$15 \mathrm{m} / \mathrm{sec}$
0%
$20 \mathrm{m} / \mathrm{sec}$
0%
$25 \mathrm{m} / \mathrm{sec}$

Total impulse imposed by ball of mass 2 kg to the ball of mass 4 kg has the magnitude equal to

0%
10 N-s
0%
5.3 N-s
0%
40 N-s
0%
16 Ns

A simple pendulum of length 'l' carries a bob of mass 'm'. If the breaking strength of the string is 2 mg. The maximum angular amplitude from the vertical can be

0%
$0^{\circ}$
0%
$3 0^{\circ}$
0%
$6 0^{\circ}$
0%
$9 0^{\circ}$

A gun of mass M fires a bullet of mass m with a velocity v relative to the gun. The average force required to bring the gun to rest in 0.5 sec. is

0%
$\dfrac{2 \mathrm{Mmv}}{\mathrm{M}+\mathrm{m}}$
0%
$\dfrac{M m v}{2(M+m)}$
0%
$\dfrac{3 \mathrm{Mmv}}{2(\mathrm{M}+\mathrm{m})}$
0%
$\dfrac{\mathrm{Mmv}}{(\mathrm{M}+\mathrm{m})}$

Explanation

Velocity of gun

$=v_1$

Velocity of bullet

$=v-v_1$

Then,

$m(v-v_1)=Mv_1$

$v_1=\dfrac{mv}{m+M}$

Distance travelled by gun in

$0.5\,s$

$0=v_1-a\times 0.5$

$\implies a=2v_1$

Also,

$S=v_1\times 0.5-\dfrac 12 a0.5^2$

$\implies S=\dfrac{v_1}{4}$

From work energy theorem,

$F\times S=\dfrac 12 Mv_1^2-0$

$S=2Mv_1$

$F=\dfrac{2Mmv}{m+M}$

A ball of mass 250 g moving with 20 m/s strikes a vertical wall and rebounds along the same line with a velocity of 15 m/s. If the time of contact is 0.1 s, the force exerted by the wall on the ball is

0%
$87.5 N$
0%
$12.5 N$
0%
$-12.5 N$
0%
$-87.5 N$

Total impulse imposed by a ball of mass

$2 kg$ moving with the speed of

$10 m/s$ in rightward direction to the ball of mass

$4 kg$ moving with the speed of

$8 m/s$ in the same direction is (The collision is perfectly elastic)

0%
$10 Ns$
0%
$5.3 Ns$
0%
$40 Ns$
0%
$16 Ns$

Calculate the impulse necessary to stop the car of mass 500 kg travelling at 72 kmph.

0%
10000 Ns
0%
20000 Ns
0%
30000 Ns
0%
40000 Ns

For every action there exists an equal and opposite reaction

0%
True
0%
False

An impulse

$\vec { 1 }$ changes the velocity of a particle from

$\vec { v } _ { 1 }$ to

$\vec { v } _ { 2 }$ . Kinetic energy gained by the particle is

0%
$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$
0%
$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$
0%
$\vec { 1 } \cdot \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$
0%
$\vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$

Explanation

Kinetic energy of any moving particle is

$\frac { 1 }{ 2 } m{ v }^{ 2 }$

Impulse of any particle is change in momentum.

Kinetic energy gained by particle is,

$\frac { 1 }{ 2 } m{ { v }_{ f } }^{ 2 }-\frac { 1 }{ 2 } m{ { v }_{ i } }^{ 2 }\\ =\frac { 1 }{ 2 } m\left( { { v }_{ f } }^{ 2 }-{ { v }_{ i } }^{ 2 } \right) \\ =\frac { 1 }{ 2 } m\left( { v }_{ f }-{ v }_{ i } \right) \left( { v }_{ f }+{ v }_{ i } \right) \\ =\frac { 1 }{ 2 } \overrightarrow { I } \left( { v }_{ 2 }+{ v }_{ 1 } \right)$

Hence, correct option is

$(A)$

Who asserted that the natural state of body is the state of rest

0%
Newton
0%
Galileo
0%
Aristotle
0%
Einstein

A ball of mass

$1$

$\mathrm{kg}$ drops vertically on to the floor with a speed of

$25$

$\mathrm{m} / \mathrm{s}$ . It rebounds with an initial velocity of

$10$

$\mathrm{m} / \mathrm{s}$ . What impulse acts on the ball during contact?

0%
$35$ $\mathrm{kg} \mathrm{m} / \mathrm{s}$ downwards
0%
$35$ $\mathrm{kg} \mathrm{m} / \mathrm{s}$ upwards
0%
$30$ $\mathrm{kg} \mathrm{m} / \mathrm{s}$ downwards
0%
$30$ $\mathrm{kg} \mathrm{m} / \mathrm{s}$ upwards

Explanation

Given,

mass of ball is

$1kg$

Initial velocity is

$10m/s$

Final velocity is

$-25m/s$ (Rebounds)

We know that impulse is change in momentum therefore,

$\triangle P=m\left( { v }_{ f }-{ v }_{ i } \right)$

Substitute all value in above equation we will get

$\triangle P=1\times \left( -25-10 \right) \\ \triangle P=-35m/s$

Here, we have taken in upward direction as positive velocity and in downward direction is as negative velocity therefore, Impulse will act in downward direction with magnitude is

$35kgm/s$ hence, correct option is

$(A)$

A rubber ball of mass 50

$\mathrm { g }$ falls from a height of

$\mathrm { m }$ and rebounds to a height of .5

$\mathrm { m }$ . The impulse between the ball and the ground if the time during which they are in contact was

$0,1$ sec is

0%
$12.5 N - s$
0%
$11.8 \mathrm { N } - \mathrm { s }$
0%
$2.36 \mathrm { N } - \mathrm { s }$
0%
$3.76 \mathrm { N } - s$

Explanation

Given,

mass of ball is

$50g$

Height is

$1m$

Rebound at height

$0.5m$

When ball is falling from height

$1m$ then initial velocity will be zero. when it strike on ground then Final velocity is ,

$v^{ 2 }={ u }^{ 2 }+2gh\\ { v }^{ 2 }={ 0 }^{ 2 }+2\times 9.8\times 1\\ v=\sqrt { 19.6 } \\ v=4.4m/s$

Now when ball rebound then final velocity will be zero therefore initial velocity is

$v^{ 2 }={ u }^{ 2 }-2gh\\ { 0 }^{ 2 }={ u }^{ 2 }-2\times 9.8\times 0.5\\ u=\sqrt { 9.8 } \\ u=3.13m/s$

Time of contact

$(t)$ is

$0.1sec$

Impulse is change in momentum

$I=m(v-u)\\ I=\frac { 50 }{ 1000 } (4.4-(-3.13))\\ I=\frac { 50 }{ 1000 } (4.4+3.13)\\ I=0.376N-s$

Force exerted is

$F=\frac { 0.376 }{ 0.1 } =3.76N$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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