Practice Start Test

Motion in a Plane

NEET Physics · 49 questions · Page 4 of 5 · Click an option or "Show Solution" to reveal answer

Q31
A particle has initial velocity (3i^+4j^)\left( {3\widehat i + 4\widehat j} \right) and has acceleration (0.4j^+0.3j^).\left( {0.4\widehat j + 0.3\widehat j} \right). Its speed after 10 s is
A 7 units
B 727\sqrt 2 units
C 8.5 units
D 10 units
Correct Answer
Option B
Solution
u→=3i^+4j^\overrightarrow u = 3\widehat i + 4\widehat j

,

a→=0.4i^+0.3j^\overrightarrow a = 0.4\widehat i + 0.3\widehat j
⇒ux=3 units, uy=4 units\Rightarrow {u_x} = 3\,units,\,{u_y} = 4\,units
ax=0.4 units, ay=0.3 units{a_x} = 0.4\,units,\,{a_y} = 0.3\,units

∴\therefore v x = u x + a x ×\times 10 = 3 + 4 = 7 units and v y = 4 + 0.3 ×\times 10 = 4 + 3 = 7 units

∴v=vx2+vy2=72\therefore v = \sqrt {v_x^2 + v_y^2} = 7\sqrt 2

units

Q32
Six vectors, a→\overrightarrow a through f→\overrightarrow f have the magnitudes and directions indicated in the figure. Which of the following statements is true ?
A b→+c→=f→\overrightarrow b + \overrightarrow c = \overrightarrow f
B d→+c→=f→\overrightarrow d + \overrightarrow c = \overrightarrow f
C d→+e→=f→\overrightarrow d + \overrightarrow e = \overrightarrow f
D b→+e→=f→\overrightarrow b + \overrightarrow e = \overrightarrow f
Correct Answer
Option C
Solution

Using the law of vector addition, (

d→+e→\overrightarrow d + \overrightarrow e

) is as shown in the fig.

∴d→+e→=f→\therefore \overrightarrow d + \overrightarrow e = \overrightarrow f
Q33
A particle shows distance - time curve as given in this figure. The maximum instaneous velocity of the particle is around the point
A D
B A
C B
D C
Correct Answer
Option D
Solution

The slope of the graph

dsdt{{ds} \over {dt}}

is maximum at C and hence the instantaneous velocity is maximum at C.

Q34
A particle starting from the origin (0, 0) moves in a straight line in the (x, y) planes. Its coordinates at a later time are (3,3)\left( {\sqrt 3 ,3} \right). The path of the particle makes with the x-axes an angle of
A 45 o
B 60 o
C 0 o
D 30 o .
Correct Answer
Option B
Solution

Let θ\theta be the angle which the particle makes with an x-axis. From figure,

tan⁡θ=33=3\tan \theta = {3 \over {\sqrt 3 }} = \sqrt 3
⇒θ=tan⁡−1(3)=60∘\Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = 60^\circ
Q35
A→\overrightarrow A and B→\overrightarrow B are two vectors and θ\theta is the angle between them, if ∣A→×B→∣=3(A→.B→),\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \left( {\overrightarrow A .\overrightarrow B } \right), the value of θ\theta is
A 45 o
B 30 o
C 90 o
D 60 o
Correct Answer
Option D
Solution
∣A→×B→∣=3(A→.B→)\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \left( {\overrightarrow A .\overrightarrow B } \right)

∴\therefore

ABsin⁡θ=3ABcos⁡θAB\sin \theta = \sqrt 3 AB\cos \theta

or

tan⁡θ=3\tan \theta = \sqrt 3

or

θ=tan⁡−13=60∘\theta = {\tan ^{ - 1}}\sqrt 3 = 60^\circ

;

Q36
For angles of projection of a projectile at angle (45 o −- θ\theta ) and (45 o + θ\theta ), the horizontal range described by the projectile are in the ratio of
A 2 : 1
B 1 : 1
C 2 : 3
D 1 : 2.
Correct Answer
Option B
Solution

(45º – θ\theta) & (45º + θ\theta) are complementary angles as 45º – θ\theta + 45º + θ\theta = 90º.

We know that if angle of projection of two projectiles make complementary angles, their ranges are equal.

In this case also, the range will be same.

So the ratio is 1 : 1.

Q37
The vectors A→\overrightarrow A and B→\overrightarrow B are such that ∣A→+B→∣=∣A→−B→∣.\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A - \overrightarrow B } \right|. The angle between the two vectors is
A 45 o
B 90 o
C 60 o
D 75 o
Correct Answer
Option B
Solution
∣A→+B→∣2=∣A→−B→∣2{\left| {\overrightarrow A + \overrightarrow B } \right|^2} = {\left| {\overrightarrow A - \overrightarrow B } \right|^2}
=∣A→∣2+∣B→∣2+2A→B→=A2+B2+2ABcos⁡θ= {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} + 2\overrightarrow A \overrightarrow B = {A^2} + {B^2} + 2AB\cos \theta
=∣A→−B→∣2=∣A→∣2+∣B→∣2−2A→.B→= {\left| {\overrightarrow A - \overrightarrow B } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} - 2\overrightarrow A .\overrightarrow B
=A2+B2−2ABcos⁡θ= {A^2} + {B^2} - 2AB\cos \theta

So, A 2 + B 2 + 2AB cos θ\theta = A 2 + B 2 – 2AB cos θ\theta 4AB cosθ\theta = 0 ⇒\Rightarrow cosθ\theta = 0 θ\theta = 90º So, angle between A & B is 90º.

Q38
If a vector 2i^+3j^+8k^2\widehat i + 3\widehat j + 8\widehat k is perpendicular to the vector 4j^−4i^+αk^,4\widehat j - 4\widehat i + \alpha \widehat k, then the value of α\alpha is
A 1/2
B −- 1/2
C 1
D −- 1.
Correct Answer
Option B
Solution

For two vectors to be perpendicular to each other

A→.B→=0{\overrightarrow A .\overrightarrow B = 0}
(2i^+3j^+8k^).(4j^−4i^+αk^)=0\left( {2\widehat i + 3\widehat j + 8\widehat k} \right).\left( {4\widehat j - 4\widehat i + \alpha \widehat k} \right) = 0

–8 + 12 + 8α\alpha = 0

α=−48=−12\alpha = - {4 \over 8} = - {1 \over 2}
Q39
If the angle between the vectors A→\overrightarrow A and B→\overrightarrow B is θ\theta , the value of the product (B→×A→).A→\left( {\overrightarrow B \times \overrightarrow A } \right).\overrightarrow A is equal to
A BA 2 sinθ\theta
B BA 2 cosθ\theta
C BA 2 sinθ\theta cosθ\theta
D zero.
Correct Answer
Option D
Solution

Let

A→×B→=C→\overrightarrow A \times \overrightarrow B = \overrightarrow C

The cross product of

B→\overrightarrow B

and

A→\overrightarrow A

is perpendicular to the plane containing

A→\overrightarrow A

and

B→\overrightarrow B

i.e. perpendicular to

B→\overrightarrow B

If a dot product of this cross product and

A→\overrightarrow A

is taken, as the cross product is perpendicular to

A→\overrightarrow A

,

C→\overrightarrow C

×\times

A→\overrightarrow A

= 0 Therefore product of

(A→×B→)×A→=0\left( {\overrightarrow A \times \overrightarrow B } \right) \times \overrightarrow A = 0
Q40
Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v 1 . The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is
A av2+v12{a \over {\sqrt {{v^2} + {v_1}^2} }}
B av+v1{a \over {v + {v_1}}}
C av−v1{a \over {v - {v_1}}}
D av2−v12\sqrt {{a \over {{v^2} - {v_1}^2}}}
Correct Answer
Option D
Solution

Velocity of A relative to B is given by

vA/B→=vA→−vB→=v→−v1→\overrightarrow {{v_{A/B}}} = \overrightarrow {{v_A}} - \overrightarrow {{v_B}} = \overrightarrow v - \overrightarrow {{v_1}}

...(i) By taking x-components of equation (i), we get

0=vsin⁡θ−v1⇒sin⁡θ=v1v0 = v\sin \theta - {v_1} \Rightarrow \sin \theta = {{{v_1}} \over v}

...(ii) By taking Y-components of equation (i), we get

vy=vcos⁡θ{v_y} = v\cos \theta

...(iii) Time taken by boy at A to catch the boy at B is given by t =

Relative displacement along Y−axisRelative velocity along Y−axis{{{\mathop{\rm Relative}\nolimits} \,displacement\,along\,Y - axis} \over {{\mathop{\rm Relative}\nolimits} \,velocity\,along\,Y - axis}}

=

avcos⁡θ=av.1−sin⁡2θ=av.1−(v1v)2{a \over {v\cos \theta }} = {a \over {v.\sqrt {1 - {{\sin }^2}\theta } }} = {a \over {v.\sqrt {1 - {{\left( {{{{v_1}} \over v}} \right)}^2}} }}

[From equation (i)] =

av.v2−v12v2=av2−v12{a \over {v.\sqrt {{{{v^2} - v_1^2} \over {{v^2}}}} }} = {a \over {\sqrt {{v^2} - v_1^2} }}

=

a2v2−v12\sqrt {{{{a^2}} \over {{v^2} - v_1^2}}}
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