Motion in a Plane Test

Result

Motion in a Plane Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following quantities is dependent of the choice of orientation of the coordinate axes?

A

B
A_x+ B_y

C

D
Angle between

Solution

A vector, its magnitude and the angle between two vectors do not depend on the choice of the orientation of the coordinate axes. So angle between are independent of the orientation of the coordinate axes. But the quantity A_x + B_ydepends upon the magnitude of the components along x and y axes, so it will change with change in coordinate axes.
Question 2
4 / -1

Two vectors inclined at an angle θ have a resultantwhich makes an angle α withand angle β withLet the magnitudes of the vectors be represented by A, B and R respectively. Which of the following relations is not correct?

A

B
Rsinα = Bsin (α+β)

C
Asinα = Bsin β

D
Rsinβ = Asin (α + β)

Solution

Let OP and OQ represent two vectorsmaking an angle (α + β). Using the parallelogram method of vector addition,
Resultant vector,
SN is normal to OP and PM is normal to OS.
From the geometry of the figure,
OS² = ON²+ SN² = (OP + PN)² + SN²= (A + Bcos(α + β))²+ (Bsin(α + β))²
R² = A²+ B²+ 2ABcos(α + β)
In ΔOSN, SN = OSsinα = Rsinα and in ΔPSN,SN = PSsin(α + β) = Bsin(α + β)
Rsinα = Bsin(α + β) or R/sin(α + β)= B/sinα
Similarly,
PM = Asinα = Bsinβ
A/sinβ = B/sinα
Combining (i) and (ii), we get
R/sin(α + β) = A/sinβ = B/sinα
From eqn, (iii), Rsinβ = Asin(α + β)
Question 3
4 / -1

Vectorsinclude an angle θ between them. Ifrespectively subtend angles α and β with then (tanα + tanβ) is :

A

B

C

D

Solution
Question 4
4 / -1

A unit vector in the direction of resultant vector of

A

B

C

D

Solution
Question 5
4 / -1

In the question number 20, a unit vector perpendicular to the direction ofis

A

B

C

D

Solution

Unit vector perpendicular to the direction of
Question 6
4 / -1

Resultant of two vectorsis of magnitude P. Ifis reversed, then resultant is of magnitude Q. What is the value of P² + Q²?

A
2(A² + B²)

B
2(A²-B²)

C
A²-B²

D
A² + B²

Solution
Question 7
4 / -1

If then the angel betweenwill be

A
30^o

B
45^o

C
60^o

D
90^o

Solution

Let θ be the angle between the vectors
Question 8
4 / -1

A motorboat is racing towards the north at 25 kmh⁻¹ and the water current in that region is 10 kmh⁻¹ in the direction of 60^∘ east of south. The resultant velocity of the boat is:

A
11 kmh⁻¹

B
22 kmh⁻¹

C
33 kmh⁻¹

D
44 kmh⁻¹

Solution

velocity of the water current v_c = 10 km/s
velocity of the motorboat v_b = 25 km/h
angle between north and south east is 120^o

The resultant velocity of the boat is 22km/h⁻¹
Question 9
4 / -1

Ifis a vector of magnitude 5 units due east. What is the magnitude and direction of a vector -5?

A
5 units due east

B
25 units due west

C
5 units due west

D
25 units due east

Solution

= 5 units due east.
∴ −5= −5(5 units due east) = −25 units due east = 25 units due west
Question 10
4 / -1

The driver of a car moving towards a rocket launching pad with a speed of 6 ms⁻¹ observed that the rocket is moving with a speed of 10 ms⁻¹ the upward speed of the rocket as seen by the stationary observer is

A
4 ms⁻¹

B
6 ms⁻¹

C
8 ms⁻¹

D
11 ms⁻¹

Solution

Observed speed, v =
= 11ms⁻¹