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PG Brainstormer - 1H (Solutions)

1.30 kph30 kph

5 kmv

Velocity of third car relative to first two is = v + 30

Time to cross the cars is 530v

= 4

60v + 30= 75

v = 45 kph.Ans. (D)

2. If B is taken at rest and motion of A is analyzed w.r.to B –vA = 50 kph

Distance AC = 10 cos 53°

A 30º

40º53º

B rest10 km

C

Time to reach point C is t = 50AC

= 600050 = 120 sec.

Ans. (B)

3. Quiet day = t1 = 2lv

Rough day = t2 = l l

v u v u = 2 22lv

v ut2 > t1Ans. (A)

4. vC 40kph vR 20 m/s

RCv = Rv – Cv

40 kph

20 m/s

tan =

54018

20

=

59

Ans. (C)

PG Brainstormer - 1 HRECTILINEAR MOTION, PROJECTILE & 2D MOTION & RELATIVE MOTION

An Ultimate Tool to understand advanced High School Physics by ASHISH ARORA Sir

Time Allowed 90 MinMaximum Marks 110SOLUTIONS

PHYS

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PG Brainstormer - 1H (Solutions)5. For particle to hit B

v = u cos

= cos–1

vu

Ans. (B)

6. In figure shown

3

31

2

2

v t

v tv t = 2

1

v tv t

( , )v tv1

v2

v2 t

Y

X

xy

=

v3

3t3

1 32

vv v =

2

1

vv

v1v3 = v2v1 2 – v2v3

v3 = 1 2

1 2

2v v

v v

Ans. (B)

7. (v + u) – (v – u)t = 6 = u (1 + t)v + u – vt + ut = u + utv = vt

2u = 6 u = 3 kphAns. (A)

8. drift x = cosd

v (u – v sin )

x is min when dxd = 0 =

dd (u sec – v tan )

u sec tan – v sec2 = 0u tan – v sec = 0u sin = v

= 1sin

vu

= 1 1sin

angle with stream direction = 1 1sin2

Ans. (C)

9. AB A BV V V

min dist BC = d sin ( – 45°)

– 45V V

A d B

d sin ( – 45º)

C

Ans. (B)

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PG Brainstormer - 1H (Solutions)

10.vtx =

2122

at

a x =

21

12

ut a t

a x

BA

2 – a x

aa

x ( , )u a1C21

22( )

vt at

a x

=

21

12

ut a t

a x

u = 2

v

a1 = +2a

Ans. (B)

11.dv = 10 … (1)

cosd

v = 15 … (2)

uv

A

from (1) and (2)

cos = 23

tan = 5

2

uv

vu =

25

Ans. (C)

12. N

E45º

V= 72

kph

VB = 51 kphV W

VWB

BV

51 kph

WBV

= WV

– BV

EastAns. (A)

13. T = 20

2vg

T ' = 2

0( )2(2 )v

g =

2T

Ans. (B)

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PG Brainstormer - 1H (Solutions)

14.Range

4 = max height

22 sin cos4 ug =

2 2sin2

ug

sec = cos = 45°

also H = 2

4u

g

u = 2 gH

t = / 4

cosR

u =

2 12

122

u

g u =

22 2

gHg = 2

Hg

Ans. (B)

15.

30º30º

30º

VH1

15º15º

15º

VH2

1HV

= 1HV V

2HV

= 2HV V

1

H

VV = 3

2

H

VV =

13

1

2

VV = 3

Ans. (A)

16. All objects outside have opposite velocity v wrt train so angular velocity = vr

far objects < near objectsAns. (B)

PHYS

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PG Brainstormer - 1H (Solutions)17. Given that t2 = t1 + t3

200 = (VL + 20) 1

2t

… (1)

VL

20 m/s

t

v

t1 t2 t3200 m 400 m 600 m

600 = (VL + 20) 3

2t

… (2)

t3 = 3t1400 = VLt2 … (3)

= VL (4t1) VL t1 = 100

from (1) & (2)200 = (VL + 20) 1

2t

= 1

2LV t

+ 10 t1

10 t1 = 150

t1 = 15010 = 15 sec

t3 = 45 sect2 = 60 sec

Total time = t1 + t2 + t3 = 120 secAns. (C)

18.AV = ˆ ˆ10 10i tn [ ˆ ˆˆ n ai bj ]BV = ˆ ˆ10 3 10j tj

for COM to be in a straight line

tan = cy

cx

VV = independent of t

tan = ( ) / 2( ) / 2

Ay By

Ax Bx

V VV V =

[10 (10 3 10 )]10 10

tb t

t a = const

which will happen when a = 3

2 = and b =

12

Ans. (A)

19.(20 / )

RVm s

(5 / )MV

m s

(15 / )Vm s

VRM

VMtan =

| | M

R

V VV =

1020

= tan–1 (1/2)Ans. (A)

20. WMV

in south direction VM

45°

VM

VW

VWM

V WM'

45°

MV

is along N-Efrom the vector velocity triangle shown

VW = 2

V toward east Ans.

Ans. (D)

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PG Brainstormer - 1H (Solutions)21. If axis is vertical then for any horizontal wind, rate remain same but if axis is made vertical, rate of water filling (Av)

changes due to A (area normal to relative speed) decreases in absence of wind.Ans. (B, D)

22. A specific case is shown in figure2 m/s

t

v

t = 0t = 0.5s1 t = 1

here tan 1 = tan 2 = 4 m/s2

if t1 is < 0.5 or t1 > 0.5either tan 1 or tan 2will exceed 4 m/s2.Ans. (A, C)

23. Time for crossing the compartment t = 2.44

Such that t = 2hg =

2 3.210

= 0.64 = 0.8 sec

u = 2.40.8 = 3 m/s

and 16 = vt

v = 160.8 = 20 m/s

Ans. (A, B, C, D)

24.BAV =

B AV V

Shortest distance AC = 4 sin 45° = 2 2 m.45°

VBA

2i^

– 2j^

B (0, 4)

2j^

A

4 m

Y

X

C

time to reach point C is

t = 2 2BC

= 2 22 2

= 1 sec.

Ans. (B, C, D)

25. Speed of bob = 3 m/s

time to fall = 2hg =

2 510

= 1 sec

particle will move along line x = 2m and fall at a distancey = vt = 3mAns. (A, C)

26.RV = ˆ4i

RBV = – ˆ ˆ2 4i jBGV =

BRV +

RV

= ˆ ˆ2 4i j

VBG = 4 16 = 20 = 2 5 m/s

Crossing time t = 100

4 = 25 sec

drift = 2 × 25 = 50 mAns. (A, B, C)

PHYS

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PG Brainstormer - 1H (Solutions)

27. If collision pt is (x, y) then x = 4 cos t = 4 cos (t – T) >

y = 4 sin t – 12

gt2 = 4 sin (t – T) – 12

(t – T)2

Ans. (A, B, D)

28. Particle will have a tangential speed of cart so it will land outside the circle following a projectile motion.Ans. (B, D)

29. v = v0 + kx

dvdx = k

a = vdvdx = k (v0 + kx)

also1

00 x dx

v kx = 0t

dt

0 1

0

1 ln

v kxk v = t

t = 1

0

1 ln

vk v

Ans. (A, B, C)

30. For minimum time man must swim normal to river current so t = dv

For direct motion toward point P his absolute velocity will be –

vSG = 2 2v u

thus t = 2 2

d

v u

for u > v drift can never be zeroAns. (A, C, D)

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