Sunday, August 17, 2014
Physics (Motion in one dimension)
Equation of motion
Distance and displacement
Distance: For a moving body, length of actual path followed. It is scalar quantity.
Displacement: shortest distance from initial position to final position with direction. It is a vector quantity.
Distance travelled ≥ displacement covered.
Note: Total displacement covered during certain time may be zero but distance travelled is never zero.
Distance and displacement
Distance: For a moving body, length of actual path followed. It is scalar quantity.
Displacement: shortest distance from initial position to final position with direction. It is a vector quantity.
Distance travelled ≥ displacement covered.
Note: Total displacement covered during certain time may be zero but distance travelled is never zero.
Note: A body is moving in circular path of radius R and it displaces certain angle θ in time t as in fig
OA + AB = OB
AB = OB - OA
ǀABǀ = ǀOB - OAǀ
= √(OB2 - 2OA OB cos θ + OA2)
= √(2R2 - 2R2 cos θ)
= √2R √(1 - cos θ)
= √2R √(1 - 1 + 2 sin2 θ/2)
Displacement, AB = 2R sin θ/2 = 2R sin wt/2
& Distance = R θ = wtR , where θ is in radian.
Q. A vehicle is moving along a circular track of radius 200 m at constant angular speed π / 30 rad/sec. The displacement covered by vehicle in 10 s will be
(a) 100 π / 3 (b) 100 (c) 10 π / 3 m (d) 100 π
Ans: (b)
Hint: Displacement = 2R sin θ/2
Speed
Rate of change of distance. It is a scalar quantity.
Avg. speed = (total distance) / (total time) = Δs / Δt
velocity
Rate of change of displacement. It is a vector quantity.
Avg. velocity = (total displacement) / (total time) = Δs / Δt
Acceleration
Rate of change of velocity. It is a vector quantity.
Avg. acceleration (a) = Δv / Δt
OA + AB = OB
AB = OB - OA
ǀABǀ = ǀOB - OAǀ
= √(OB2 - 2OA OB cos θ + OA2)
= √(2R2 - 2R2 cos θ)
= √2R √(1 - cos θ)
= √2R √(1 - 1 + 2 sin2 θ/2)
Displacement, AB = 2R sin θ/2 = 2R sin wt/2
& Distance = R θ = wtR , where θ is in radian.
Q. A vehicle is moving along a circular track of radius 200 m at constant angular speed π / 30 rad/sec. The displacement covered by vehicle in 10 s will be
(a) 100 π / 3 (b) 100 (c) 10 π / 3 m (d) 100 π
Ans: (b)
Hint: Displacement = 2R sin θ/2
Speed
Rate of change of distance. It is a scalar quantity.
Avg. speed = (total distance) / (total time) = Δs / Δt
velocity
Rate of change of displacement. It is a vector quantity.
Avg. velocity = (total displacement) / (total time) = Δs / Δt
Acceleration
Rate of change of velocity. It is a vector quantity.
Avg. acceleration (a) = Δv / Δt
Q. An object travels displacement s with 20 m/s and returns with velocity 30 m/s. Find avg. velocity and speed.
Soln:
Avg. velocity = (total displacement) / (total time) = 0/t = 0
total distance = 2s
total time (T) = s/20 + s/30
So Avg. speed = {total distance (2s)} / {total time(T)} = 2s / {s/20 + s/30} = 24 m/s
Soln:
Avg. velocity = (total displacement) / (total time) = 0/t = 0
total distance = 2s
total time (T) = s/20 + s/30
So Avg. speed = {total distance (2s)} / {total time(T)} = 2s / {s/20 + s/30} = 24 m/s
Q. Choose the current option.
(a) speed =ǀ velocity ǀ (b) speed > ǀ velocity ǀ (c) speed < ǀ velocity ǀ (d) speed ≥ ǀ velocity ǀ
Ans: (d)
Q. A person travels at constant velocity V1 due east and then he travels the same distance at constant velocity V2 due north. The avg. velocity will be
(a) √{(v12 + v22)/2} (b) ) 1/2 √(v12 + v22) (c) ) {√2(v1 v2) /(v1 + v2)} (d ) (2v1 v2) /(v1 + v2)
Ans: (c)
T = s/v1 + s/v2 = s(v1 + v2) / (v1 v2)
Vavg = (total displacement) / (total time)
= √2 s / {s (v1 + v2)/ (v1 v2)}
Vavg = {√2(v1 v2) /(v1 + v2)}
Q. Find out the speed of a particle if position of a particle varies like x = at2 and y = 3bt3
soln:
Vx = dx / dx = 2at i
Vy= dy / dt = 9 bt2 j
ǀVǀ = √ {(2at)2 + (9 bt2)2} & tan θ = Vy / Vx
= √ {(4a2t2 + 18 b2t4}
Motion under gravity
For upward motion For downward motion
v = u - gt v = u + gt
v2 = u 2 - 2gh v2 = u 2 + 2gh
h = ut - 1/2 gt2 h = ut + 1/2 gt2
ht = u - g/2 (2t - 1) ht = u + g/2 (2t - 1)
The value of g = 9.81 m/s2 = 981 cm/s2 = 32 ft/s2
Distance travelled by a body in a st. line in tth second is
St = u + a/2 (2t - 1)
Q. A body is thrown vertically upward from ground. While moving upward, the acceleration of the body without neglecting air resistance is
(a) equal to g (b) more than g (c) less than g (d) zero
Ans: (b)
Hint: F = mg + 6πȠrv
=> a = g + (6πȠrv)/m
=> a > g , when moves upward
(a) speed =ǀ velocity ǀ (b) speed > ǀ velocity ǀ (c) speed < ǀ velocity ǀ (d) speed ≥ ǀ velocity ǀ
Ans: (d)
Q. A person travels at constant velocity V1 due east and then he travels the same distance at constant velocity V2 due north. The avg. velocity will be
(a) √{(v12 + v22)/2} (b) ) 1/2 √(v12 + v22) (c) ) {√2(v1 v2) /(v1 + v2)} (d ) (2v1 v2) /(v1 + v2)
Ans: (c)
T = s/v1 + s/v2 = s(v1 + v2) / (v1 v2)
Vavg = (total displacement) / (total time)
= √2 s / {s (v1 + v2)/ (v1 v2)}
Vavg = {√2(v1 v2) /(v1 + v2)}
Q. Find out the speed of a particle if position of a particle varies like x = at2 and y = 3bt3
soln:
Vx = dx / dx = 2at i
Vy= dy / dt = 9 bt2 j
ǀVǀ = √ {(2at)2 + (9 bt2)2} & tan θ = Vy / Vx
= √ {(4a2t2 + 18 b2t4}
Motion under gravity
For upward motion For downward motion
v = u - gt v = u + gt
v2 = u 2 - 2gh v2 = u 2 + 2gh
h = ut - 1/2 gt2 h = ut + 1/2 gt2
ht = u - g/2 (2t - 1) ht = u + g/2 (2t - 1)
The value of g = 9.81 m/s2 = 981 cm/s2 = 32 ft/s2
Distance travelled by a body in a st. line in tth second is
St = u + a/2 (2t - 1)
Q. A body is thrown vertically upward from ground. While moving upward, the acceleration of the body without neglecting air resistance is
(a) equal to g (b) more than g (c) less than g (d) zero
Ans: (b)
Hint: F = mg + 6πȠrv
=> a = g + (6πȠrv)/m
=> a > g , when moves upward
F = mg - 6πȠrv
=> a = g - (6πȠrv)/m
=> a <g , when moves downward
When a body is released or dropped from the certain height h then,
=>Time taken, t = √(2h/g)
=>Velocity at height h' above ground, v = √{2g(h - h')}
=> Final velocity, v =√(2gh)
=> Height of the particle at any time h = H- g/2 (T - t)2
=> a = g - (6πȠrv)/m
=> a <g , when moves downward
When a body is released or dropped from the certain height h then,
=>Time taken, t = √(2h/g)
=>Velocity at height h' above ground, v = √{2g(h - h')}
=> Final velocity, v =√(2gh)
=> Height of the particle at any time h = H- g/2 (T - t)2
A body is dropped from a top of tower and falls freely
(i) distance covered by it after n sec = 1/2 g n2 i.e. α n2
(ii) distance covered in nth sec = 0 + 1/2 g (2n - 1) i.e. α (2n - 1)
(iii) velocity of body after n sec = 0 + gn i.e. α n
(i) distance covered by it after n sec = 1/2 g n2 i.e. α n2
(ii) distance covered in nth sec = 0 + 1/2 g (2n - 1) i.e. α (2n - 1)
(iii) velocity of body after n sec = 0 + gn i.e. α n
Q. A body is released from a height 45m. The distance covered by it in last 0.6 sec is
(a) 8.6 m (b)7.2 m (c) 16.2 m (d) 18.2 m
Ans: (c)
Hint: t = √(2h/g) =√(2 * 45 /10) = 3 sec.
h = ut + 1/2 gt2 = 0 + 1/2 * 10 * 2.42 = 28.8 m
distance covered in last 0.6 sec = (45 - 28.8) = 16.2 m
Q. A body is thrown vertically upward from earth surface with velocity 'u', which can attain a maximum height 'H'. What is its value at 3H/4 above earth?
(a) u/4 (b) 3/4 u (c) u/2 (d) √3/2 u
Ans: (c)
Hint: u' = √{2g(h - h') or put h' = 3H/4
v' = √{2g(H - 3H/4)} = = √{2gH/4} = √(2gH)/2 = u/2
Relative Velocity
It is a velocity of one body w.r.t. another body when both the bodies are in motion. Relative velocity of a w.r.t. B is
VAB = VA - VB
VAB = √{VA2 + VB2 + 2VAVB cos(180 - θ)}
Cases:
(1) when A & B are moving in same direction.
θ = 0o or cos 0 = 1
VAB = VA - VB
(2) when A & B are moving in opposite direction.
It θ = 180o
VAB = VA + VB
Q. A man walks in rain with a velocity of 5 km/h. The rain drops strike at him at an angle of 45o with the horizontal. The downward velocity of the raindrops will be
(a) 5 km/h (b) 4 km/h (c) 3 km/h (d) 1 km/h
Ans: (a)
(a) 8.6 m (b)7.2 m (c) 16.2 m (d) 18.2 m
Ans: (c)
Hint: t = √(2h/g) =√(2 * 45 /10) = 3 sec.
h = ut + 1/2 gt2 = 0 + 1/2 * 10 * 2.42 = 28.8 m
distance covered in last 0.6 sec = (45 - 28.8) = 16.2 m
Q. A body is thrown vertically upward from earth surface with velocity 'u', which can attain a maximum height 'H'. What is its value at 3H/4 above earth?
(a) u/4 (b) 3/4 u (c) u/2 (d) √3/2 u
Ans: (c)
Hint: u' = √{2g(h - h') or put h' = 3H/4
v' = √{2g(H - 3H/4)} = = √{2gH/4} = √(2gH)/2 = u/2
Relative Velocity
It is a velocity of one body w.r.t. another body when both the bodies are in motion. Relative velocity of a w.r.t. B is
VAB = VA - VB
VAB = √{VA2 + VB2 + 2VAVB cos(180 - θ)}
Cases:
(1) when A & B are moving in same direction.
θ = 0o or cos 0 = 1
VAB = VA - VB
(2) when A & B are moving in opposite direction.
It θ = 180o
VAB = VA + VB
Q. A man walks in rain with a velocity of 5 km/h. The rain drops strike at him at an angle of 45o with the horizontal. The downward velocity of the raindrops will be
(a) 5 km/h (b) 4 km/h (c) 3 km/h (d) 1 km/h
Ans: (a)
Q. A river is flowing from west to east at a speed of 5 m/min. In what direction should a man on south bank of the river capable of swimming at 10 m/ min in still water should swim to cross the river
(i) in shortest time (ii) along shortest path
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h in least possible time. The time taken to cross the river is
(a) 20 min (b) 12 min (c) 15 min (d) 20 min
Ans: (b)
t = d / (v cos θ)
tmin = d / v = 1/5 hr = 12 min
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h along shortest route. The time taken to cross the river will be
a) 10 min (b) 12 min (c) 15 min (d) 20 min
Ans: (c)
Q. A person has to start swimming at right angle to the stream of river flowing at velocity u in order to cross that river in least time.
Soln: vel. across water = √(v2 + u2)
least time taken = AB/v = BC / u = AC /√( v2 + u2)
Q. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high when the stone was dropped. Find the time when the stone hits the ground. (g = 10 m/s2).
(a) 2.7 s (b) 3.7 s (c) -2.7 s (d) 4.7 s
Ans: (b)
Soln: Taking vertically upwards as +ve X-axis. (↑+ve ↓-ve)
x = ut + 1/2 at2
-50 = 5 × t + 1/2 × (-10) × t2
or 5 t2 - 5t - 50 = 0
or t2 - t - 10 = 0
t = [+ 1 ± √{1 - 4×1× (-10)} ] / 2 = (1 ± √41) / 2
=> t = 3.75, & t = - 2.7 s
Q. A heavy truck and a car are in motion and both have same k.E. Brakes are applied to produce equal retarding force. Which will travel a smaller distance before coming to rest.
(a) car (b) truck (c) both travel through same distance (d) None
Ans: (c)
(i) in shortest time (ii) along shortest path
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h in least possible time. The time taken to cross the river is
(a) 20 min (b) 12 min (c) 15 min (d) 20 min
Ans: (b)
t = d / (v cos θ)
tmin = d / v = 1/5 hr = 12 min
Q. A person who can swim in still water at 5 km/h crosses a river, 1 km wide flowing at 3 km/h along shortest route. The time taken to cross the river will be
a) 10 min (b) 12 min (c) 15 min (d) 20 min
Ans: (c)
Q. A person has to start swimming at right angle to the stream of river flowing at velocity u in order to cross that river in least time.
Soln: vel. across water = √(v2 + u2)
least time taken = AB/v = BC / u = AC /√( v2 + u2)
Q. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high when the stone was dropped. Find the time when the stone hits the ground. (g = 10 m/s2).
(a) 2.7 s (b) 3.7 s (c) -2.7 s (d) 4.7 s
Ans: (b)
Soln: Taking vertically upwards as +ve X-axis. (↑+ve ↓-ve)
x = ut + 1/2 at2
-50 = 5 × t + 1/2 × (-10) × t2
or 5 t2 - 5t - 50 = 0
or t2 - t - 10 = 0
t = [+ 1 ± √{1 - 4×1× (-10)} ] / 2 = (1 ± √41) / 2
=> t = 3.75, & t = - 2.7 s
Q. A heavy truck and a car are in motion and both have same k.E. Brakes are applied to produce equal retarding force. Which will travel a smaller distance before coming to rest.
(a) car (b) truck (c) both travel through same distance (d) None
Ans: (c)
Note: The stopping distance of a car is x when its speed is v, what will be the stopping distance when its speed is increased to nv during same time.
V2 = u2 + 2ax => a = v2 / 2x & x = v2 / 2a
when V = nv
then x' = v2 / 2a (n2 v2 ) / {2( v2/2x)} = n2 x
Q. The shortest distance over which a car can be brought to rest by applying breaks is 1m. If its speed is made 3 times, the car can be brought to rest over a distance of
(a) 9 m (b) 6 m (c) 3 m (d) 1 m
Ans: (a)
Hint: x' α n2 x
Note: A body is dropped from the top of a tower of height 'h' m. if simultaneously another ball thrown upward with velocity 'v'. After what time do they cross each other?
=> h = h1 + h2
or h = (0 + 1/2 gt2) + (vt - 1/2 gt2)
t = h/v
h2 = h - g/2 h2 / v2
Q. A ball is dropped from 100 m height. If simultaneously another ball thrown upward with velocity 50 m/s. When and where would they meet each other.
(a) 1 sec, 20 m (b) 2 s, 80 m (c) 3 s, 80 m (d) 4 s, 80 m
Ans: (b )
Hint: h/v = 100/50 = 2
Q. A passenger is x m away from the bus moving with a constant acceleration a. What is the minimum velocity of the passenger so that he can catch the bus?
Soln: vt = x + 1/2 at2
=> t = {v ± √(v2 - 2ax)} / a
t is real so v2 - 2ax ≥ 0
So, Vmin = √(2ax)
V2 = u2 + 2ax => a = v2 / 2x & x = v2 / 2a
when V = nv
then x' = v2 / 2a (n2 v2 ) / {2( v2/2x)} = n2 x
Q. The shortest distance over which a car can be brought to rest by applying breaks is 1m. If its speed is made 3 times, the car can be brought to rest over a distance of
(a) 9 m (b) 6 m (c) 3 m (d) 1 m
Ans: (a)
Hint: x' α n2 x
Note: A body is dropped from the top of a tower of height 'h' m. if simultaneously another ball thrown upward with velocity 'v'. After what time do they cross each other?
=> h = h1 + h2
or h = (0 + 1/2 gt2) + (vt - 1/2 gt2)
t = h/v
h2 = h - g/2 h2 / v2
Q. A ball is dropped from 100 m height. If simultaneously another ball thrown upward with velocity 50 m/s. When and where would they meet each other.
(a) 1 sec, 20 m (b) 2 s, 80 m (c) 3 s, 80 m (d) 4 s, 80 m
Ans: (b )
Hint: h/v = 100/50 = 2
Q. A passenger is x m away from the bus moving with a constant acceleration a. What is the minimum velocity of the passenger so that he can catch the bus?
Soln: vt = x + 1/2 at2
=> t = {v ± √(v2 - 2ax)} / a
t is real so v2 - 2ax ≥ 0
So, Vmin = √(2ax)
Q. A ball is thrown upwards with speed u from a top of a tower and it reaches the ground after time t1. If same ball is thrown downwards from the same point with same speed u, it reaches to ground after time t2. What time does it take to reach to ground dropped from same point.
Soln:
h = ut1 + 1/2 gt12 = ut2 + 1/2 gt22
or 1/2 g (t12 - t22 ) = u (t2 - t1 )
=> u = g/2 (t1 - t2 )
So, h = g/2 (t1 - t2 )t2 + 1/2 gt2
= g/2 t1 t2 - g/2 t22 + 1/2 gt2
. : h = g/2 t1 t2
Soln:
h = ut1 + 1/2 gt12 = ut2 + 1/2 gt22
or 1/2 g (t12 - t22 ) = u (t2 - t1 )
=> u = g/2 (t1 - t2 )
So, h = g/2 (t1 - t2 )t2 + 1/2 gt2
= g/2 t1 t2 - g/2 t22 + 1/2 gt2
. : h = g/2 t1 t2
again , h = 0 + d/2 gt2
so, g/2 t1 t2 = 1/2 gt2
=> t = √ t1t2
Q. A body when thrown upward with a certain velocity reaches to ground after 8s. If the same body is thrown downward from the same point with same velocity, it reaches to ground in 2 s. How much time does it take to reach to ground when it is dropped from the same point?
(a) 2 s (b) 4 s (c) 12 s (d) 10 s
Ans: (b )
Hint: t = √ t1t2 = √(8 × 2) = 4
so, g/2 t1 t2 = 1/2 gt2
=> t = √ t1t2
Q. A body when thrown upward with a certain velocity reaches to ground after 8s. If the same body is thrown downward from the same point with same velocity, it reaches to ground in 2 s. How much time does it take to reach to ground when it is dropped from the same point?
(a) 2 s (b) 4 s (c) 12 s (d) 10 s
Ans: (b )
Hint: t = √ t1t2 = √(8 × 2) = 4
Physics (Vectors and Scalars)
Vector
Physical quantities having magnitude and direction e.g. displacement, velocity gradent.
Polar Vector: vectors having a staring point e.g. displament, force etc.
Axial Vector: vectors that represent rotational effect and act along the axis of rotation. It's diration is given by right hand screw rule. e.g. torque, angular velocity, angular momentum.
Scalar
Physical quantities having magnitude only. e.g. volume, speed, potential etc.
. Necessary condition for being vector: The physical quanties must have both magnitude and direction.
. Sufficient condition for being vector: The physical quantities should obey the rules of vector addition.
Q. Which of the following is a vector quantity.
(a) electric current
(b) electric potential
( c) electric flux
(d) current density
Ans : (d)
Q. Physical quantity having magnitude and direction
(a) may be a vector
(b) must be a vector
( c) is not a vector
Ans : (a)
Q. A scalar quantity
(a) may have magnitude and may have direction.
(b) must have magnitude only but not direction.
( c) may not have magnitude and direction.
Ans : (a)
--> All vectors must have magnitude and direction but all physical quantities having magnitude and direction are not vectors. e.g. electric current, time, pressure.
Unit Vector
A vector divided by its own magnitude. It is unit less and dimensionless.
â = a / ǀ a ǀ
Vector can be represented by vector name with arrowhead or bold letter.
a = a1i + a2j + a3k = (a1,a2,a3)
ǀ a ǀ = √(a12 + a22 + a32)
Q. Two forces 3N and 4N are acting perpendicular onto a body a man 5kg. The value of acceleration generated in a body is
(a) 4 m/s2 (b) )3 m/s2 (c) 2 m/s2 (d) 1 m/s2
ans: (d)
we know, F = √(F12 + F22) = √(32 + 42) = 5 --->> a = F/m = 5/5 = 1 m/s2
Physical quantities having magnitude and direction e.g. displacement, velocity gradent.
Polar Vector: vectors having a staring point e.g. displament, force etc.
Axial Vector: vectors that represent rotational effect and act along the axis of rotation. It's diration is given by right hand screw rule. e.g. torque, angular velocity, angular momentum.
Scalar
Physical quantities having magnitude only. e.g. volume, speed, potential etc.
. Necessary condition for being vector: The physical quanties must have both magnitude and direction.
. Sufficient condition for being vector: The physical quantities should obey the rules of vector addition.
Q. Which of the following is a vector quantity.
(a) electric current
(b) electric potential
( c) electric flux
(d) current density
Ans : (d)
Q. Physical quantity having magnitude and direction
(a) may be a vector
(b) must be a vector
( c) is not a vector
Ans : (a)
Q. A scalar quantity
(a) may have magnitude and may have direction.
(b) must have magnitude only but not direction.
( c) may not have magnitude and direction.
Ans : (a)
--> All vectors must have magnitude and direction but all physical quantities having magnitude and direction are not vectors. e.g. electric current, time, pressure.
Unit Vector
A vector divided by its own magnitude. It is unit less and dimensionless.
â = a / ǀ a ǀ
Vector can be represented by vector name with arrowhead or bold letter.
a = a1i + a2j + a3k = (a1,a2,a3)
ǀ a ǀ = √(a12 + a22 + a32)
Q. Two forces 3N and 4N are acting perpendicular onto a body a man 5kg. The value of acceleration generated in a body is
(a) 4 m/s2 (b) )3 m/s2 (c) 2 m/s2 (d) 1 m/s2
ans: (d)
we know, F = √(F12 + F22) = √(32 + 42) = 5 --->> a = F/m = 5/5 = 1 m/s2
Triangle law of vectors
R = A + B
R = A + B
Parallelogram Law of Vector
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
tanα = B Sinθ / (A + B cosθ) , with A
tanβ = A Sinθ / (B+ A cosθ) , with B (@ exchage A and B)
If A > B then α < β
If A < B then α > β
If A = B then α = β = θ/2
Note: The larger component makes smaller angle with the resultant.
Special cases of resultant
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
--> If θ = 0o, then R = A + B = Rmax
--> If θ = 90o, then R = √(A2 + B2)
--> If θ = 180o, then R = A - B or B - A = Rmin
Note: so, the limit of resultant of magnitude of A and B lies A-B ≤ R ≤ A+B
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
tanα = B Sinθ / (A + B cosθ) , with A
tanβ = A Sinθ / (B+ A cosθ) , with B (@ exchage A and B)
If A > B then α < β
If A < B then α > β
If A = B then α = β = θ/2
Note: The larger component makes smaller angle with the resultant.
Special cases of resultant
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
--> If θ = 0o, then R = A + B = Rmax
--> If θ = 90o, then R = √(A2 + B2)
--> If θ = 180o, then R = A - B or B - A = Rmin
Note: so, the limit of resultant of magnitude of A and B lies A-B ≤ R ≤ A+B
Q. Which of the following can not be resultant of 20 and 30 vectors
(a) 50 (b) 35 (c) 15 (d) 8
Ans: (d)
Hint: Result remains between the range 30-20 ≤ R ≤ 30+20
Q. The magnitude of sum of two vectors is equal to the difference of their magnitude. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (d)
Hint: ǀ a + b ǀ = ǀ a ǀ - ǀ b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab + b2
or, 2ab (1 + cosθ) = 0
or, cosθ = -1
=> θ = 180o
(a) 50 (b) 35 (c) 15 (d) 8
Ans: (d)
Hint: Result remains between the range 30-20 ≤ R ≤ 30+20
Q. The magnitude of sum of two vectors is equal to the difference of their magnitude. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (d)
Hint: ǀ a + b ǀ = ǀ a ǀ - ǀ b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab + b2
or, 2ab (1 + cosθ) = 0
or, cosθ = -1
=> θ = 180o
Q. The magnitude of sum of two vectors is equal to the magnitude of their difference. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (c)
Hint: ǀ a + b ǀ = ǀ a - b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab cosθ + b2
or, 4ab cosθ = 0
or, 4cosθ = 0
=> θ = 90o
Q. The sets of vectors are acting on a body, whose resultant can not be zero.
(a)10, 10, 10 (b) 10, 10, 20 (c) 10, 20, 20 (d) 10, 20, 40
Ans: (d)
Hint: To form a triangle, sum of two smaller sides should be ≥ larger side. (the sides 1, 2,5 can not form a triangle)
Note: A particle is travelling with a velocity v along certain direction If it turns θ, then the change in its velocity
ΔV2 = v12 + 2v1v2 cos(180-θ) + v22
= v2 + 2v2 cos(180-θ) + v2
= 2v2 (1 + cosθ)
= 4 v2 sin2 θ/2
=>ΔV = 2 v sinθ/2 (remember)
Q. A particle is moving eastward with a velocity of 5m/s. In 10s the velocity changes to 5m/s northward. What is average acceleration in this case.
(a) 1 m/s2 (b) √2 m/s2 (c) 1/√2 m/s2 (d) 2√2 m/s2
ans: (c)
Hint: Δv = 2v sin θ/2 = 2 * 5 * sin45o = 5√2
a = Δv/Δt = 5√2/10 = 1/√2
Q. A particle is revolving in a circle at constant speed of 10 m/s. Find the change in velocity in turning through 60o.
(a) 0 (b) 10√3 (c) 10 (d) 5√3
Ans: (c)
Hint: use formula Δv = 2v sin θ/2
Product of two vectors
Dot Product / scalar Product
a.b = a b cosθ
a.b = (a1i + a2j + a3k).( b1i + b2j + b3k) = a1b1 + a2b2 + a3b3
Cases:
when θ = 0, a.b = ab (max)
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (c)
Hint: ǀ a + b ǀ = ǀ a - b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab cosθ + b2
or, 4ab cosθ = 0
or, 4cosθ = 0
=> θ = 90o
Q. The sets of vectors are acting on a body, whose resultant can not be zero.
(a)10, 10, 10 (b) 10, 10, 20 (c) 10, 20, 20 (d) 10, 20, 40
Ans: (d)
Hint: To form a triangle, sum of two smaller sides should be ≥ larger side. (the sides 1, 2,5 can not form a triangle)
Note: A particle is travelling with a velocity v along certain direction If it turns θ, then the change in its velocity
ΔV2 = v12 + 2v1v2 cos(180-θ) + v22
= v2 + 2v2 cos(180-θ) + v2
= 2v2 (1 + cosθ)
= 4 v2 sin2 θ/2
=>ΔV = 2 v sinθ/2 (remember)
Q. A particle is moving eastward with a velocity of 5m/s. In 10s the velocity changes to 5m/s northward. What is average acceleration in this case.
(a) 1 m/s2 (b) √2 m/s2 (c) 1/√2 m/s2 (d) 2√2 m/s2
ans: (c)
Hint: Δv = 2v sin θ/2 = 2 * 5 * sin45o = 5√2
a = Δv/Δt = 5√2/10 = 1/√2
Q. A particle is revolving in a circle at constant speed of 10 m/s. Find the change in velocity in turning through 60o.
(a) 0 (b) 10√3 (c) 10 (d) 5√3
Ans: (c)
Hint: use formula Δv = 2v sin θ/2
Product of two vectors
Dot Product / scalar Product
a.b = a b cosθ
a.b = (a1i + a2j + a3k).( b1i + b2j + b3k) = a1b1 + a2b2 + a3b3
Cases:
when θ = 0, a.b = ab (max)
when θ = 90o, a.b = 0 (condition of perpendicularity)
when θ = 180o, a.b = -ab(min)
i.i =j.j = k.k = 1
Limit of product: -ab ≤ a.b ≤ ab
when θ = 180o, a.b = -ab(min)
i.i =j.j = k.k = 1
Limit of product: -ab ≤ a.b ≤ ab
Cross of product / vector product
a×b = ab sinθ n
n is unit vector perpendicular to the plane of a and b .
a×b = ab sinθ = (a1i + a2j + a3k) × ( b1i + b2j + b3k)
Cases:
when θ = 0o or 180o , a×b = 0 (null vector) (condition of parallelism)
when θ = 90o, a×b = ab ( max)
Limit of product: 0 ≤ a×b ≤ ab
Q. If the product of two vectors was found to be negative , then this product must be
(a) dot product
(b) cross product
(c) neither dot nor cross
(d) none
Ans: (a)
Note: Dot product may be negative , zero , +ve but cross product is always +ve or equal to zero in magnitude.
Q. If sum of vectors perpendicular to their vector difference, then these vectors
(a) must be equal (b) must be equal in magnitude (c)must be perpendicular (d) must be parallel
ans: (b)
Hint: (a+b).(a-b) = 0
a2 - a.b + a.b +b2 = 0
=> a = b
Angle between two vector
a.b = ab cosθ => cosθ = a.b / ab
ǀ a×b ǀ = ab sinθ => sinθ = ǀa×bǀ / ab
tanθ = ( ǀa×bǀ) / ( a.b)
Q. The dot product of two vectors is equal to the magnitude of their cross product . The angle between those vectors should be
(a) 30o (b) 45o (c) 60o (d) 90o
Ans: (b)
Hint: tanθ = ( ǀa×bǀ) / ( a.b)
Q. A person moves 30 m north, then 20 m east and finally 30√2 m south - west. What is his displacement from the original position?
s = s1 + s2 + s3
s1 = 30 j, s2 = 20 i
s3 = -30√2 sin 45o i - 30√2 sin 45o j
= -30 i - 30 j
so, s = 30 j + 20 i - 30 i - 30 j
= - 10 j
Resultant displacement = 10 m west.
a×b = ab sinθ n
n is unit vector perpendicular to the plane of a and b .
a×b = ab sinθ = (a1i + a2j + a3k) × ( b1i + b2j + b3k)
Cases:
when θ = 0o or 180o , a×b = 0 (null vector) (condition of parallelism)
when θ = 90o, a×b = ab ( max)
Limit of product: 0 ≤ a×b ≤ ab
Q. If the product of two vectors was found to be negative , then this product must be
(a) dot product
(b) cross product
(c) neither dot nor cross
(d) none
Ans: (a)
Note: Dot product may be negative , zero , +ve but cross product is always +ve or equal to zero in magnitude.
Q. If sum of vectors perpendicular to their vector difference, then these vectors
(a) must be equal (b) must be equal in magnitude (c)must be perpendicular (d) must be parallel
ans: (b)
Hint: (a+b).(a-b) = 0
a2 - a.b + a.b +b2 = 0
=> a = b
Angle between two vector
a.b = ab cosθ => cosθ = a.b / ab
ǀ a×b ǀ = ab sinθ => sinθ = ǀa×bǀ / ab
tanθ = ( ǀa×bǀ) / ( a.b)
Q. The dot product of two vectors is equal to the magnitude of their cross product . The angle between those vectors should be
(a) 30o (b) 45o (c) 60o (d) 90o
Ans: (b)
Hint: tanθ = ( ǀa×bǀ) / ( a.b)
Q. A person moves 30 m north, then 20 m east and finally 30√2 m south - west. What is his displacement from the original position?
s = s1 + s2 + s3
s1 = 30 j, s2 = 20 i
s3 = -30√2 sin 45o i - 30√2 sin 45o j
= -30 i - 30 j
so, s = 30 j + 20 i - 30 i - 30 j
= - 10 j
Resultant displacement = 10 m west.
Q. If r = (2i + 2j) m and F = (3i + 2j) N. Find out whether the torque is clock wise or anti clock wise.
T = r × F
= i (0 - 0) - j (0 - 0) + k (4 - 6)
= - 2k
Here n = (- 2k) /(ǀ -2kǀ) = (- 2k) /2 = -k => clock wise.
If n = +ve =>anti close wise
If n = -ve => close wise
Lami's theorem
P/ sin β = Q /sin γ = R /sinα
Q. A mass of 1 kg is attached to the middle of a rope, which is being pulled from both ends in the opposite directions. Take g= 10 m/s2, what is the minimum pull required to completely straighten the rope.
(a) 5N (b) 10N (c) 20N (d) much larger than 20N
Ans: (d)
To strengthen the rope
γ = 180o , α = β = 90o
T1 / sinα = T2 / sin β = mg / sin γ
or T1 = T2 = mg sin 90o / sin 180o = ∞
Q. When forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular , then the particle remains stationary. If the force F1 is now removed, then acceleration of the particle is
(a) F1/m (b) F2 F3/m F1 (c) (F2 -F1)/m (d) F2 /m
Ans: (a)
F3 + F2 = F1 (to be stationary)
Resultant of ǀF3 + F2ǀ = F1
Acceleration = F1/ m
Note: All the gradients (e.g. temperature gradient, velocity gradient) are the vector quantities. Small angle 'dθ' is a vector quantity but the bigger angles are not vector quantities.
T = r × F
= i (0 - 0) - j (0 - 0) + k (4 - 6)
= - 2k
Here n = (- 2k) /(ǀ -2kǀ) = (- 2k) /2 = -k => clock wise.
If n = +ve =>anti close wise
If n = -ve => close wise
Lami's theorem
P/ sin β = Q /sin γ = R /sinα
Q. A mass of 1 kg is attached to the middle of a rope, which is being pulled from both ends in the opposite directions. Take g= 10 m/s2, what is the minimum pull required to completely straighten the rope.
(a) 5N (b) 10N (c) 20N (d) much larger than 20N
Ans: (d)
To strengthen the rope
γ = 180o , α = β = 90o
T1 / sinα = T2 / sin β = mg / sin γ
or T1 = T2 = mg sin 90o / sin 180o = ∞
Q. When forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular , then the particle remains stationary. If the force F1 is now removed, then acceleration of the particle is
(a) F1/m (b) F2 F3/m F1 (c) (F2 -F1)/m (d) F2 /m
Ans: (a)
F3 + F2 = F1 (to be stationary)
Resultant of ǀF3 + F2ǀ = F1
Acceleration = F1/ m
Note: All the gradients (e.g. temperature gradient, velocity gradient) are the vector quantities. Small angle 'dθ' is a vector quantity but the bigger angles are not vector quantities.
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