Chapter 4 Motion of Plane Questions and Answers: NCERT Solutions for Class 11th Physics (Partnership Physics )

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Class 11 Physics (Partnership Physics ) Chapter 4: Motion of Plane - Questions and Answers of NCERT Book Solutions.

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1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Ans:
Scalar: Mass, volume, density, angular frequency, number of moles, speed.
Vector: Acceleration, angular velocity, velocity, displacement.
A scalar quantity is specified by its magnitude. Mass, volume, density, angular frequency, number of moles, speed are some of the scalar physical quantities.
A vector quantity is specified by its magnitude and the direction associated with it.
Acceleration, angular velocity, velocity, displacement belong to this category.

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2. Pick out the two scalar quantities in the following list:
Force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Ans:
Work and current are examples of scalar quantities.
Work done is said to be the dot product of force and displacement. As the dot product of two quantities is always a scalar, work is considered as a scalar physical quantity.
Current is described by its magnitude. Its direction is not considered.
Thus, it is a scalar quantity.

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3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Ans: Impulse
It is given by the product of force and time. As force is a vector quantity, its product with time gives a vector quantity.

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4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: (a) adding any two scalars.
(b) adding a scalar to a vector of the same dimension s,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.

Ans:
(a) Not Meaningful.
The addition of two scalar quantities will be meaningful only if they both represent the same physical quantity.
(b) Not Meaningful.
The addition of a vector quantity with a scalar quantity is considered not meaningful.
(c) Meaningful.
A scalar can be multiplied with a vector. Force is multiplied with time to give impulse.
(d) Meaningful.
A scalar, respective of the physical quantity, can be multiplied with another scalar having the same or different dimensions.
(e) Not Meaningful.
The addition of two vector quantities is considered meaningful only if they both represent the same physical quantity.
(f) Meaningful
A component of a vector can be added to the same vector as both of them have the same dimensions.

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5. Read each statement below carefully and state with reasons, if it is true or false:
(a) The magnitude of a vector is always a scalar,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the displacement vector of a particle.
(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
(e) Three vectors not lying mm a plane can never add up to give a null vector.

Ans:
(a) True.
The magnitude of a vector is a number. So, it is a scalar.
(b) False.
Each component of a vector is a vector.
(c) False.
The total path length is scalar, whereas displacement is a vector quantity.
So, the total path length is greater than the magnitude of displacement. It is equal to the magnitude of displacement only when a particle is moving in a straight line.
(d) True.
It is because the total path length is always greater than or equal to the magnitude of displacement of a particle.
(e) True.
Three vectors, which do not lie in a plane, can’t be represented by the sides of a triangle taken in the same order.

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6. Establish the following vector inequalities geometrically or otherwise:
(a) |a+b|⩽|a|+|b||a+b|⩽|a|+|b|
(b) |a+b|⩾|a|−|b||a+b|⩾|a|−|b|
(c) |a−b|⩽|a|−|b||a−b|⩽|a|−|b|
(d) |a−b|⩾|a|−|b||a−b|⩾|a|−|b|
When does the equality sign above apply?

Ans:
a) Let a→a→ and b→b→ be represented by the adjacent sides of a parallelogram OMNPOMNP as shown below:
|OM−→−|=|a→| ...(i)|OM→|=|a→| ...(i)
|MN−→−|=|OP−→−|=|b→| ...(ii)|MN→|=|OP→|=|b→| ...(ii)
|ON−→−|=|a→+b→| ...(iii)|ON→|=|a→+b→| ...(iii)
As each side is smaller than the sum of the other two sides in a triangle,
In ΔOMNΔOMN,
ON<(OM+MN)ON<(OM+MN)
|a→+b→|<|a→|+|b→| ...(iv)
|a→+b→|<|a→|+|b→| ...(iv)
If a⃗ a→ and b⃗ b→ act along a straight line in the same direction, then:
|a→+b→|=|a→|+|b→| ...(v)|a→+b→|=|a→|+|b→| ...(v)
Combine equations (iv)(iv) and (v)(v)
|a→+b→|⩽|a→|+|b→||a→+b→|⩽|a→|+|b→|
b) Let a⃗ a→ and b⃗ b→ be represented by the adjacent sides of a parallelogramOMNPOMNP, as shown below:
(Image will be uploaded soon)
|OM−→−|=|a→| ...(i)|OM→|=|a→| ...(i)
|MN−→−|=|OP−→−|=|b→| ...(ii)|MN→|=|OP→|=|b→| ...(ii)
|ON−→−|=|a→+b→| ...(iii)|ON→|=|a→+b→| ...(iii)
As each side is smaller than the sum of the other two sides in a triangle,
In ΔOMNΔOMN,
ON+MN>OMON+MN>OM
ON+OM>MNON+OM>MN
|ON|>|OM−→−−OM−→−|(∵OP=MN)|ON|>|OM→−OM→|(∵OP=MN)
If a⃗ a→ and b⃗ b→ act along a straight line in the same direction, then:
|a→+b→|=||a→|−|b→|| ...(v)
|a→+b→|=||a→|−|b→|| ...(v)
Combine equations (iv)(iv) and (v)(v)
|a→+b→|⩾||a→|−|b→|||a→+b→|⩾||a→|−|b→||
c) Let a→a→ and b→b→ be represented by the adjacent sides of a parallelogram PQRSPQRS:
(Image will be uploaded soon)
|OR−→−|=|PS−→|=|b→| ...(i)|OR→|=|PS→|=|b→| ...(i)
|OP¯¯¯¯¯¯¯|=|a→| ...(ii)|OP¯|=|a→| ...(ii)
As each side is smaller than the sum of the other two sides in a triangle,
In ΔOPSΔOPS,

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7. Given a+b+c+d=0a+b+c+d=0, which of the following statements are correct:
(a) a,b,ca,b,c, and dd must each be a null vector,
(b) The magnitude of (a+c)(a+c) equals the magnitude of (b+d)(b+d)
(c) The magnitude of a can never be greater than the sum of the magnitudes of bb, cc, and dd,
(d) b+cb+c must lie in the plane of aa and dd if aa and dd are not collinear, and in the line of aa and dd.
if they are collinear?

Ans:
(a) Incorrect
To make a→+b→+c→+d→=0a→+b→+c→+d→=0, it is not necessary to have all four vectors as null vectors. There are many other combinations which will give the sum zero.
(b) Correct
a→+b→+c→+d→=0a→+b→+c→+d→=0
a→+c→=−(b→+d→)a→+c→=−(b→+d→)
Take modulus on both sides:
|a→+c→|=|−(b→+d→)|=|(b→+d→)||a→+c→|=|−(b→+d→)|=|(b→+d→)|
So, the magnitude of (a⃗ +c⃗ )(a→+c→) is the same as the magnitude of (b⃗ +d⃗ )(b→+d→)
(c) Correct
a→+b→+c→+d→=0a→+b→+c→+d→=0
a→=−(b→+c→+d→)a→=−(b→+c→+d→)
Take modulus on both sides:M
a⃗ =|(b⃗ +c⃗ +d⃗ )|a→=|(b→+c→+d→)|
a⃗ ⩽|b⃗ |+|c⃗ |+|d⃗ | ...(i)a→⩽|b→|+|c→|+|d→| ...(i)
(b⃗ +c⃗ +d⃗ )(b→+c→+d→) is the sum of vectors b⃗ b→, c⃗ c→ and d⃗ d→. The magnitude of (b⃗ +c⃗ +d⃗ )(b→+c→+d→) is less than, or equal to the sum of the magnitudes of b⃗ b→, c⃗ c→ and d⃗ .d→. So, the magnitude of a⃗ a→ cannot be greater than the sum of the magnitudes of b⃗ b→, c⃗ c→ and d→.d→. Equation (i)(i) shows that the magnitude of a→a→ is equal to or less than the sum of the magnitudes of b⃗ b→, c⃗ c→ and d⃗ d→
(d) Correct
For, a→+b→+c→+d→=0a→+b→+c→+d→=0
a→+(b→+c→)+d→=0a→+(b→+c→)+d→=0
The resultant sum of the vectors a⃗ , (b⃗ +c⃗ )a→, (b→+c→) and d⃗ d→ is zero only if (b⃗ +c⃗ )(b→+c→) lie in the same plane as a⃗ a→ and d⃗ d→
If a⃗ a→ and d→d→ are collinear, then (b→+c→)(b→+c→) is in the line of a→a→ and d→d→. This is true in this case and the vector sum of all the vectors will be zero.

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8. Three girls skating on a circular ice ground of radius 200m200m start from a point PP on the edge of the ground and reach a point QQ diametrically opposite to PP following different paths as shown in Fig. 4.20 . What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Ans:
The magnitudes of displacements are equal to the diameter of the ground.
Radius of the ground =200m=200m
Diameter of the ground =2×200=400m=2×200=400m
So, the magnitude of the displacement for each girl is 400m400mwhich is equal to the actual length of the path skated by girl BB.

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9. A cyclist starts from the centre OO of a circular park of radius 1 km1 km, reaches the edge PP of the park, then cycles along the circumference, and returns to the centre along QOQO as shown in the following figure. If the round trip takes 1010 min, what is the
(a) net displacement,
(b) average velocity, and
(c) average speed of the cyclist?

Ans 9:
(a) The cyclist comes to the starting point after cycling for 1010 minutes. So, his net displacement is zero.
(b) Average velocity = Net displacement Total time Total time = Net displacement Total time Total time
As the net displacement of the cyclist is zero, his average velocity is also zero.
(c) Average speed = Total path length Total time Average speed = Total path length Total time
Total path length =OP+PQ+QOTotal path length =OP+PQ+QO
=1+14(2π×1)+1=1+14(2π×1)+1
=2+12π3.570km=2+12π3.570km
Time taken =10min=10min
=1060=1060
=16h=16h
∴∴ Average speed =3.5701=21.42km/h

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10. On an open ground, a motorist follows a track that turns to his left by an angle of 600600 after every 500m500m. Starting from a given turn, specify the displacement of the total at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Ans:
The path is a regular hexagon with side 500m500m.
Let the motorist start from PP.
The motorist takes the third turn at SS.
∴∴ Magnitude of displacement =PS=PS
=PV+VS=PV+VS
=500+500=500+500
=1000m=1000m
Total path length =PQ+QR+RS=PQ+QR+RS
=500+500+500=500+500+500
=1500m=1500m
The motorist takes the sixth turn at PP, which is the starting point.
∴∴ Magnitude of displacement =0=0
Total path length =PQ+QR+RS+ST+TU+UP=PQ+QR+RS+ST+TU+UP
=500+500+500+500+500+500=500+500+500+500+500+500
=3000m=3000m
The motorist takes the eight turn at RR.
∴ Magnitude of displacement =PR∴ Magnitude of displacement =PR
=PQ2+QR2+(PQ)(QR)cos600−−−−−−−−−−−−−−−−−−−−−−−−−√=PQ2+QR2+(PQ)(QR)cos600
=5002+5002+(500)(500)cos600−−−−−−−−−−−−−−−−−−−−−−−−−√=5002+5002+(500)(500)cos600
=250000+250000+(500000×12)−−−−−−−−−−−−−−−−−−−−−−−−−−−√=250000+250000+(500000×12)
=866.03m=866.03m
β=tan−1(500×sin600500+500×cos600)β=tan−1(500×sin600500+500×cos600)
β=300β=300
Thus, the magnitude of displacement is 866.03m866.03m at an angle of 300300 with PRPR.
Total path length == Circumference of the hexagon +PQ++PQ+QRQR
=6×500+500+500=6×500+500+500
=4000m

Turn	Magnitude of Displacement	Total Path Length
Third	1000	1500
Sixth	0	3000
Eighth	866.03; 300	4000

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11. A passenger arriving in a new town wishes to go from the station to a hotel located 10km10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23km23km long and reaches the hotel in 28min28min. What is (a) the average speed of the taxi,
(b) the magnitude of average velocity? Are the two equal?

Ans:
(a) Total distance travelled =23km=23km
Total time taken =28min=2860h=28min=2860h
Average speed of the taxi = Total distance travelled Total time taken = Total distance travelled Total time taken
(b) Distance between the hotel and the station =10km==10km= Displacement of the car
∴ Average velocity =102860∴ Average velocity =102860
=21.43km/h

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12. Rain is falling vertically with a speed of 30ms−130ms−1. A woman rides a bicycle with a speed of 10ms−110ms−1 in the north to south direction. What is the direction in which she should hold her umbrella?

Ans:
The described situation is:
(Image will be uploaded soon
vc=vc= Velocity of the cyclist
vr=vr= Velocity of falling rain
To protect herself from the rain, the woman should hold the umbrella in the direction of the relative velocity of the rain with respect to the woman.
v=vc + vrv=vc + vr
=30+(−10)=30+(−10)
=20ms−1=20ms−1
tanθ=vcvrtan⁡θ=vcvr
tanθ=1030tan⁡θ=1030
θ=tan−1(13)θ=tan−1(13)
θ=tan−1(0.333)θ=tan−1(0.333)
θ≈180θ≈180
So, the woman should hold the umbrella toward the south, at an angle of 180180 with the vertical.

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13. A man can swim with a speed of 4.0km/h4.0km/h in still water. How long does he take to cross a river 1.0km1.0km wide if the river flows steadily at 3.0km/h3.0km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Ans:
Speed of the man vm=4kmh−1vm=4kmh−1
Width of the river =1km=1km
Time taken to cross the river = Width of the river Speed of the rivera = Width of the river Speed of the rivera
=14h=14h
=14×60=14×60
=15min=15min
Speed of the river, vr=3kmh−1vr=3kmh−1
Distance covered with flow of the river =vr×t=vr×t
=3×14=3×14
=34km=34km
=34×1000=34×1000
=750m

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14. In a harbour, wind is blowing at the speed of 72km/h72km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51km/h51km/h to the north, what is the direction of the flag on the mast of the boat?

Ans:
Velocity of the boat, vb=51kmh−1vb=51kmh−1
Velocity of the wind, vw=72kmh−1vw=72kmh−1
As the flag is fluttering in the north-east direction, the wind is blowing towards the north-east direction. When the ship begins sailing toward the north, the flag will start moving along the direction of the relative velocity of the wind with respect to the boat.
The angle between vwvw and (−vb)=900+450(−vb)=900+450
tanβ=51sin(90+45)72+51cos(90+45)tan⁡β=51sin⁡(90+45)72+51cos⁡(90+45)
=51sin(90+45)72+51(−cos45)=51sin⁡(90+45)72+51(−cos⁡45)
=51×12–√72−51×12–√=51×1272−51×12
=51722–√−51=51722−51
=5172×1.414−51=5172×1.414−51
=5150.800=5150.800
∴β=tan−1(1.0038)∴β=tan−1(1.0038)
β=45110β=45110
Angle with respect to the east direction is 45.110−450=45.110−450= 0.1100.110
So, the flag will flutter almost due east.

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15. The ceiling of a long hall is 25m25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40ms−140ms−1 can go without hitting the ceiling of the hall?

Ans:
Speed of the ball, 40ms−140ms−1
Maximum height, h=25mh=25m
In projectile motion, the maximum height reached, by a body projected at an angle θθ is:
h=u2sin2θ2gh=u2sin2θ2g
25=402sin2θ2×9.825=402sin2θ2×9.8
sin2θ=0.30625sin2θ=0.30625M
sinθ=0.5534sinθ=0.5534
θ=sin−1(0.5534)θ=sin−1(0.5534)
θ=33.60∘θ=33.60∘
The horizontal range is
R=u2sin2θgR=u2sin2θg
R = (40)2×sin2×33.609.8R = (40)2×sin2×33.609.8
R = 1600×sin67.29.8R = 1600×sin67.29.8
R = 1600×0.9229.8R = 1600×0.9229.8
R = 150.53mR

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16. A cricketer can throw a ball to a maximum horizontal distance of 100m100m.How much high above the ground can the cricketer throw the same ball?

Ans:
Maximum horizontal distance, R = 100mR = 100m
The cricketer will throw the ball to the maximum horizontal distance when the angle of projection is 450450, i.e., θ=450θ=450
The horizontal range for a projection velocity vv, is:
R=u2sin2θgR=u2sin⁡2θg
100=u2sin900g100=u2sin⁡900g
u2g=100 ...(i)u2g=100 ...(i)
The ball will reach the maximum height when it is thrown vertically upward. For this type of motion, the final velocity is zero at the maximum height HH.
Acceleration, a=−ga=−g
Use the third equation of motion:>
v2−u2=−2gHv2−u2=−2gH
H=12×u2gH=12×u2g
H=12×100H=12×100
H=50m

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17. A stone tied to the end of a string 80cm80cm long is whirled in a horizontal circle with a constant speed. If the stone makes 1414 revolutions in 25s25s, what is the magnitude and direction of acceleration of the stone?

Ans:
Length of the string, =80cm=0.8m=80cm=0.8m
Number of revolutions =14=14
Time taken =25s=25s
Frequency, v= Number of revolutions Time taken =1425Hzv= Number of revolutions Time taken =1425Hz
Angular frequency, ω=2πvω=2πv
=2×227×1425=2×227×1425
=8825rads−1=8825rads−1
Centripetal acceleration, ae=ω2rae=ω2r
=(8825)2×0.8=(8825)2×0.8
=9.91ms−2=9.91ms−2
The direction of centripetal acceleration is always along the string, towards the center, at all points.

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18. An aircraft executes a horizontal loop of radius 1.00km1.00km with a steady speed of 900km/h.900km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Ans:
Radius of the loop, r=1km=1000mr=1km=1000m
Speed of the aircraft, v=900kmh−1v=900kmh−1
=900×518=900×518m
=250ms−1=250ms−1
Centripetal acceleration, ae=v2rae=v2r
=(250)21000=(250)21000
=62.5ms−2=62.5ms−2
Acceleration due to gravity, g=9.8ms−2g=9.8ms−2
acg=6.251000acg=6.251000
ac=6.38gac=6.38g
The Centripetal acceleration is 6.386.38 times the acceleration due to gravity.

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19. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle m circular motion is always along the radius of the circle towards the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Ans:
(a) False
In circular motion, the net acceleration of a particle is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.
(b) True
At a point on a circular path, a particle appears to move tangentially to the circular path.
Thus, the velocity vector of the particle is always along the tangent at a point.
(c) True
In uniform circular motion, the acceleration vector points towards the centre of the circle. The average of these vectors over one cycle is a null vector.

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20. The position of a particle is given by r=3.0tiˆ−2.0t2jˆ+4.0kmr=3.0ti^−2.0t2j^+4.0km
Where tt is in seconds and the coefficients have the proper units for rr to be in metres.
(a) Find the vv and aa of the particle?
(b) What is the magnitude and direction of velocity of the particle at t=2.0s?

Ans:
v→(t)=(3.0iˆ−4.0jˆ):a→=−4.0jˆv→(t)=(3.0i^−4.0j^):a→=−4.0j^
The position of the particle is:
r→=3.0iˆ−2.0t2jˆ+4.0kˆr→=3.0i^−2.0t2j^+4.0k^
Velocity, v⃗ v→ of the particle is:
v→=dr→dtv→=dr→dt
v→=ddt(3.0tiˆ−2⋅Ot2jˆ+4.0k)v→=ddt(3.0ti^−2⋅Ot2j^+4.0k)
∴v→=3.0iˆ−4.0t∴v→=3.0i^−4.0t
Acceleration of the particle is:
a→=dv→dta→=dv→dt
a→=ddt(3.0iˆ−4.Ojˆ)a→=ddt(3.0i^−4.Oj^)
∴a→=−4.Ojˆ∴a→=−4.Oj^
The velocity vector, v⃗ =3.0i^−4.0tj^v→=3.0i^−4.0tj^
At t=2.0st=2.0s :
v→=3.0iˆ−8.0jˆv→=3.0i^−8.0j^
The magnitude of velocity is:
|v→|=32+(−8)2−−−−−−−−−√|v→|=32+(−8)2
|v→|=73−−√|v→|=73
|v→|=8.54m/s|v→|=8.54m/s
And θ=tan−1(vyvx)θ=tan−1(vyvx)
=tan−1(−83)=tan−1(−83)
=−tan−1(2.667)=−tan−1(2.667)
=−69.450=−69.450
The negative sign indicates that the direction of velocity is 8.54ms−18.54ms−1, 69.45069.450 below the x−x−axis.

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21. A particle starts from the origin at t=0st=0s with a velocity of 10.0710.07 and moves in the x−yx−y plane with a constant acceleration of (8.0iˆ+2.0jˆ)ms - 2(8.0i^+2.0j^)ms - 2
(a) At what time is the x−x− coordinate of the particle 16 m16 m? What is the y−y−coordinate of the particle at that time?
(b) What is the speed of the particle at the time?

Ans:
Velocity of the particle, v→=10.0jˆ ms−1v→=10.0j^ ms−1
Acceleration of the particle a→=(8.0iˆ+2.0jˆ)a→=(8.0i^+2.0j^)
But, a→=dv→dt=8.0iˆ+2.0jˆa→=dv→dt=8.0i^+2.0j^
dv→=(8.0iˆ+2.0jˆ)dtdv→=(8.0i^+2.0j^)dt
Integrate both sides:
v→(t)=8.0iˆ+2.0jˆ+u→v→(t)=8.0i^+2.0j^+u→
Where
u→=u→= Velocity vector of the particle at t=0t=0
v→=v→= Velocity vector of the particle at time ϕϕ
But v→=drdtv→=drdt
dr→=v→dt=(8.0tiˆ+2.0tjˆ+u)dtdr→=v→dt=(8.0ti^+2.0tj^+u)dt
Integrate the equations with: at t=0; r=0t=0; r=0 and at t=t; r=rt=t; r=r
r→=u→t+128.0t2iˆ+122.0t2jˆr→=u→t+128.0t2i^+122.0t2j^
=u→t + 4 \times 0t2iˆ + t2jˆ=u→t + 4 \times 0t2i^ + t2j^
=(10.0j)t+4.0t2iˆ+i2jˆ=(10.0j)t+4.0t2i^+i2j^
xiˆ+yˆ=4.0t2iˆ+(10t+t2)jˆxi^+y^=4.0t2i^+(10t+t2)j^
Equate the coefficients of iˆi^ and jˆj^:
x=4t2x=4t2
t=(x4)12t=(x4)12
And y=10t+t2y=10t+t2
When x=16mx=16m :,=
t=(164)12t=(164)12
t=2st=2s
∴y=10×2+(2)2=24m∴y=10×2+(2)2=24m
Velocity of the particle is:
v→(t)=8.0tiˆ+2.0tˆ+u→v→(t)=8.0ti^+2.0t^+u→
At t=2st=2s
v→(t)=8.0×2iˆ+2.0×2jˆ+10jˆv→(t)=8.0×2i^+2.0×2j^+10j^
=16iˆ+14jˆ=16i^+14j^
∴∴ Speed of the particle:
|v⃗ |=(16)2+(14)2−−−−−−−−−−−√|v→|=(16)2+(14)2
=256+196−−−−−−−−√=256+196
=452−−−√=452
=21.26ms−1

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22. iˆi^ and jˆj^ are unit vectors along x−x− and yy-axis respectively. What is the magnitude and direction of the vectors iˆ+jˆi^+j^ and iˆ.jˆi^.j^? What are the components of a vector a=2iˆ+3jˆa=2i^+3j^ along the directions of iˆ+jˆi^+j^ and iˆ−jˆi^−j^?

Ans:
Consider a vector P→P→
P→=iˆ+jˆ
P→=i^+j^
Pxiˆ+Pyjˆ=iˆ+jˆ
Pxi^+Pyj^=i^+j^
Compare the components on both sides:
Px = Py = 1
Px = Py = 1
∣∣∣P→∣∣∣=Px2+Py2−−−−−−−−√
|P→|=Px2+Py2
∣∣∣P→∣∣∣=12+12−−−−−−√
|P→|=12+12
∣∣∣P→∣∣∣=2–√ ...(i)
|P→|=2 ...(i)
So, the magnitude of the vector iˆ+jˆi^+j^ is 2–√2
Let θθ be the angle made by P→P→, with the x - x - axis.
So, tanθ = (PxPy)
tanθ = (PxPy)
θ = tan−1(11)
θ = tan−1(11)
θ = 45∘ ...(ii)
θ = 45∘ ...(ii)
So, the vector iˆ+jˆi^+j^ makes an angle of 450450 with the x - x - axis
Let θθ be the angle made by Q→Q→, with the x - x - axis.
Q→=iˆ−jˆQ→=i^−j^
Qxiˆ−Qyjˆ=iˆ−jˆQxi^−Qyj^=i^−j^
Qx+Qy=1Qx+Qy=1
|Q→|=Q2x+Q2y−−−−−−−√|Q→|=Qx2+Qy2
|Q→|=2–√|Q→|=2
So, the magnitude of the vector iˆ−jˆi^−j^ is 2–√2.
Let θθ be the angle made by the vector Q→Q→, with the x - x - axis.
∴tanθ=(QyQx)∴tan⁡θ=(QyQx)
θ=−tan−1(−11)θ=−tan−1(−11)
θ=−450θ=−450
So, the vector iˆ−jˆi^−j^ makes and angle of −450−450 with the axis.
Compare the coefficients of iˆi^ and jˆj^
A→=2iˆ+3jˆA→=2i^+3j^
Axiˆ + Ayjˆ = 2iˆ + 3jˆAxi^ + Ayj^ = 2i^ + 3j^
Let A¯¯¯¯A¯ make an angle θθ with the x - x - axis
∴tanθ=(AxAy)∴tan⁡θ=(AxAy)
θ=tan−1(32)θ=tan−1(32)
=tan−1(1.5)=tan−1(1.5)
=56.310=56.310
Angle between (2i^+3j^)(2i^+3j^) and (i^+j^)(i^+j^)
θ=56.31−45=11.310θ=56.31−45=11.310
A⃗ A→, along the direction of P⃗ P→ making an angle θ′θ′
∴tanθ=(AxAy)∴tan⁡θ=(AxAy)
tanθ=(Acosθ)Ptan⁡θ=(Acos⁡θ)P
tanθ=(Acos11.31)(iˆ+jˆ)2–√tan⁡θ=(Acos⁡11.31)(i^+j^)2
tanθ=13−−√×0.98062–√(iˆ+jˆ)tan⁡θ=13×0.98062(i^+j^)
tanθ=2.5(iˆ+jˆ)tan⁡θ=2.5(i^+j^)
tanθ=2510×2–√tan⁡θ=2510×2
tanθ=52–√…(v)tan⁡θ=52…(v)
Let θθ be the angle between (2i^+3j^)(2i^+3j^) and (i^+j^)(i^+j^)
θ′′=45+56.31=101.310θ′′=45+56.31=101.310
Component of vector AA, along the direction of QQ, making an angle θθ
=(Acosθ)Q⃗ =(Acos⁡θ)Q→
=(Acosθ)(i^−j^)2–√=(Acos⁡θ)(i^−j^)2
=13−−√cos(901.310)(iˆ + \hat j)2–√=13cos⁡(901.310)(i^ + \hat j)2
=−132−−−√sin11.300(iˆ−jˆ)=−132sin⁡11.300(i^−j^)
=−2.550×0.1961(iˆ−jˆ)=−2.550×0.1961(i^−j^)
=−0.5(iˆ−j)=−0.5(i^−j)
=−510×2–√=−510×2
=−12–√

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23. For any arbitrary motion in space, which of the following relations are true:
(a) vaverage = (12)(v(t1) + v(t2))vaverage = (12)(v(t1) + v(t2))
(b) vaverage = [r(t2) - r(t1)](t2 - t1)vaverage = [r(t2) - r(t1)](t2 - t1)
(c) v(t) = v(0) + atv(t) = v(0) + at
(d) r(t) = r(0) + v(0)t + (12)at2r(t) = r(0) + v(0)t + (12)at2
(e) aaverage = [v(t2) - v(t1)](t2 - t1)aaverage = [v(t2) - v(t1)](t2 - t1)
(The ‘average’ stands for average of the quantity over the time interval t1t1 to t2t2)

Answer: Given M = 2 x 1030 kg, r = 7 x 108 m
.-. Volume of Sun = 4/3πr3 x 3.14 x (7 x 108)3 = 1.437 x 1027 m3
As p = M/V, .’. p = 2 x 1030/1.437 x 1027= 1391.8 kg m-3 = 1.4 x 103 kg m-3
Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

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24. Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that:
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes

Ans:
(a) False. Energy is not conserved in inelastic collisions.
(b) False. Temperature can take negative values.
(c) False. Total path length is a scalar quantity and has the dimension of length.
(d) False. A scalar quantity like gravitational potential can vary from one point to another in space.
(e) True. The value of a scalar does not change for observers with different orientations of axes.

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25. An aircraft is flying at a height of 3400m3400m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0s10.0s apart is 300300, what is the speed of the aircraft?

Ans:
The positions of the observer and the aircraft are shown below:
(Image will be uploaded soon)
Height of the aircraft from ground, OR=3400mOR=3400m
Angle subtended between the positions, ∠POQ=300∠POQ=300
Time =10s=10s
In ΔPRO:ΔPRO:
tan150=PRORtan⁡150=PROR
PR=ORtan150PR=ORtan⁡150
PR=3400×tan150PR=3400×tan⁡150
ΔΔPRO is similar to ΔRQOΔRQO
∴PR=RQ∴PR=RQ
PQ=PR+RQPQ=PR+RQ
PQ=2PRPQ=2PR
PQ=2×3400tan150PQ=2×3400tan⁡150
PQ=6800×0.268PQ=6800×0.268
PQ=1822.4mPQ=1822.4m
∴∴ Speed of the aircraft =1822.410=182.24m/s

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26. A vector has magnitude and direction. Does it have a location in space? Can it vary with tune? Will two equal vectors aa and bb at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

Ans: No. A vector has no definite locations in space because it remains invariant when displaced in such a way that its magnitude and direction remains the same. A position vector has a definite location in space.
Yes. A vector can vary with time. The displacement vector of a particle moving with a certain velocity varies with time is an example.
No. Two equal vectors present at different locations in space need not produce the same physical effect. Consider the example of two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.

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27. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Ans:
No. A physical quantity having both magnitude and direction need not necessarily be a vector.
For example, current is a scalar quantity even though it has magnitude and direction. The requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.
No. The rotation of a body about an axis is not a vector quantity because it does not follow the law of vector addition. However, a rotation by a small angle follows the law of vector addition and is considered a vector.

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28. Can you associate vectors with
(a) the length of a wire bent into a loop,
(b) a plane area,
(c) a sphere? Explain.

Ans:
(a) No. A vector cannot be associated with the length of a wire bent into a loop as the length of a loop does not have a definite direction.
(b) Yes. An area vector can be associated with a plane area. The direction of this vector is represented by a normal drawn outward to the area.
(c) No. A vector cannot be associated with the volume of a sphere as it does not have a specific direction. A null vector can be associated with the area of a sphere.

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29. A bullet fired at an angle of 300300 with the horizontal hits the ground 3.0km3.0km a way. By adjusting its angle of projection, can one hope to hit a target 5.0km5.0km away? Assume the muzzle speed to the fixed, and neglect air resistance.

Ans:
No.
Range, R=3kmR=3km
Angle of projection, θ=300θ=300
Acceleration due to gravity, g=9.8m/s2g=9.8m/s2
Horizontal range for the projection velocity is:
R=u20sin2θgR=u02sin⁡2θg
3=u20gsin6003=u02gsin⁡600
u20g=23–√ ...(i)u02g=23 ...(i)
The maximum range (Rmax)(Rmax) is achieved by the bullet when it is fired 450450 with the horizontal.
Rmax=u20g ...(ii)
Rmax=u02g ...(ii)
Compare (i)(i) and (ii) (ii)
: Rmax=33–√Rmax=33
=2×1.732=2×1.732
=3.46km=3.46km
So, the bullet will not hit a target 5km5km away.

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30. A fighter plane flying horizontally at an altitude of 1.5km1.5km with speed 720kmh−1720kmh−1 passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600ms−1600ms−1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g=10ms−1g=10ms−1 ).

Ans:
Height of the fighter plane =1.5km=1500m=1.5km=1500m
Speed of the fighter plane, v=720kmh−1=200ms−1v=720kmh−1=200ms−1M
Let θθ be the angle with the vertical so that the shell hits the plane.
Muzzle velocity of the gun, u=600m/su=600m/s
Time taken by the shell to hit the plane =t=t
Horizontal distance travelled by the shell =uxt=uxt
Distance travelled by the plane =vt=vt
The shell hits the plane. So, these two distances should be equal.

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31. A cyclist 1 s1 s riding with a speed of 27kmh−127kmh−1. As he approaches a circular turn on the road of radius 80m80m, he applies brakes and reduces his speed at the constant rate of 0.50ms−10.50ms−1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Ans:
Speed of the cyclist, v=27kmh−1=7.5ms−1v=27kmh−1=7.5ms−1
Radius of the circular turn, r=80mr=80m
Centripetal acceleration is:
ac=v2rac=v2r
=(7.5)280=(7.5)280
=0.7ms−2=0.7ms−2
(Image will be uploaded soon)
Assume that the cyclist starts cycling from PP and moves toward QQ. At QQ, he applies the breaks and decelerates the speed of the bicycle by 0.5ms−2.0.5ms−2. This acceleration is along the tangent at QQ and opposite to the direction of motion of the cyclist.
As the angle between acac and arar is 900900, the resultant acceleration is:
a=a2c+a2r−−−−−−√a=ac2+ar2
=(0.7)2+(0.5)2−−−−−−−−−−−√=(0.7)2+(0.5)2
=0.74−−−−√=0.74
=0.85ms−1=0.85ms−1
tanθ=acartan⁡θ=acar
Where θθ is the angle of the resultant with the direction of velocity
=0.70.5=0.70.5
=1.4=1.4
θ=tan−1(1.4)θ=tan−1(1.4)
θ=

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32. (a) Show that for a projectile the angle between the velocity and the x−x−axis as a function of time is given by
θ(t)=tan−1(voy−gtvox)θ(t)=tan−1(voy−gtvox)
(b) Show that the projection angle θ0θ0 for a projectile launched from the origin is given by
θ0=tan−1(4hmR)θ0=tan−1(4hmR)
where the symbols have their usual meaning.

Ans:
Let voxvox and voyvoy be the initial components of the velocity of the projectile along horizontal (x)(x) and vertical (y)(y) directions.
Let vxvx and vyvy be the horizontal and vertical components of velocity at a point PP.
Time taken by the projectile to reach point P=tP=t
(a) Apply the first equation of motion along the vertical and horizontal directions.
vy=v0y - gtvy=v0y - gt
And vx=vaxvx=vax
∴tanθ=vyvx∴tan⁡θ=vyvx
tanθ=voy−gtvaxtan⁡θ=voy−gtvax<
θ=tan−1(vey−gtvox)θ=tan−1(vey−gtvox)
(b) Maximum vertical height, hm = u2osin22θ2g ...(i)hm = uo2sin22θ2g ...(i)
Horizontal range, R=u2osin22θg ...(ii)R=uo2sin22θg ...(ii)
Solve equation (i)(i) and equation (ii)(ii):
hmR=sin2θ2sin2θhmR=sin2θ2sin2θ
hmR=sinθ×sinθ2×2sinθcosθhmR=sin⁡θ×sin⁡θ2×2sin⁡θcos⁡θ
hmR=1sinθ4cosθhmR=1sin⁡θ4cos⁡θ
hmR=14tanθhmR=14tan⁡θ
tanθ=(4hmR)tan⁡θ=(4hmR)
θ=tan−1(4hmR)

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Last Updated on: Mar 20, 2024