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PG Brainstormer - 5H (Solutions)

Section - ASingle Choice Correct (Q. No. 1 to 10)

The Questions given in this section are MCQs with four answer options (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Choose the correct answer and fill the answer in the OMR sheet provided at the end.

PG Brain Stormer - 5 HKINEMATICS, NEWTON’S LAWS OF MOTION, FRICTION

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

Time Allowed 45 MinMaximum Marks 55

1. Sol. Ans. (B)At equilibrium, net force along tangent will be zero.

ma cos = mg sin

a = g tan a = dyg dx

a = g 2kx

x = 2agK

2. Sol. Ans. (C)

2xdxv tdt ; 2 2 4y

dy dxv xdt dt

ˆ ˆ ˆ ˆ2 4 /x yv v i v j i j m s .

3. Sol. Ans. (A)

14 2

(8 2) (8 2)x tdx t dx t dt

dt

x = 4t2 – 2t + 2

4 2

2 2y tdy dy dt

dt y = 2t

x = y2 – y + 2

4. Sol. Ans. (C)Let the particles move perpendicular to each other at time t.

Hence ˆ ˆ ˆ ˆ(4 ) ( 3 ) 0i gt j i gt j

–12 + g2t2 = 0

t = 12 3100 5

t

Hence distance d is

d = 2

2 2 21 1(4 3 )2 2

t t h gt h gt

= 7 37 5t m

5. Sol. Ans. (B)Area under a-t graph gives change in velocity

1 4 5 5 4 30 /2v m s

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PG Brainstormer - 5H (Solutions)6. Sol. Ans. (C)

ucos = 32

u , = 30°

2 sin 2ug

= 2 2sin2

P ug

32g =

(1/ 4)2

Pg 4 3P

7. Sol. Ans. (B)As net force on block A is zero

aA = 0and aB 0.

8. Sol. Ans. (A)When both child and cart move together, common acceleration will be

a = 12040 = 3 m/s2

If friction on cart is f, we usef = 10 a = 30 N.

For the friction to be limiting valueF = µN

30 = µ (30 g)

µ = 110 = 0.1.

9. Sol. Ans. (A)As shown in figure

dx = dy – 2dy sin

dx = dy (1 – 2 sin 37º)

dy

dx

y

dx = dy 615

= – 5dy

vA = dxdt

vA = – 5Bv

vB = dydt

10. Sol. Ans. (B)Equation of motion for ball

2T – 1.5 mg = 1.5 ma … (1)Equation of motion for rod

mg – T = 2 ma … (2)(1) + 2 x (2) 0.5 mg = 5.5 ma

a = 11g

Time for ball to cross the length of rod is

t = 23

la =

2 1 113 g = 11

15 sec.

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PG Brainstormer - 5H (Solutions)

Section - BMore than one Choice Correct (Q. No. 11 to 15)

The Questions given in this section are MCQs with four answer options (A), (B), (C) and (D) out of which ONE or MOREthan ONE answers may be correct. Choose all the correct answer options and fill the answer in the OMR sheet providedat the end. You will get marks in a question only if all the answer options of that question are chosen by you.

11. Sol. Ans. (B, D)As A is thrown upward it will take more time but as at the top of the tower speed of both stones in downwarddirection is same so they will reach ground with the same speed.

12. Sol. Ans. (B, C)CB : v = x – 60

dvdx = 1

a = x – 60at x = 35

a = – 25 m/s2

BA : v = –xdvdx = –1

a = xat x = 20 m

a = 20 m/s2 along +x axis.

13. Sol. Ans. (A, B, C)As both blocks are moving toward right the direction of friction can be shown in diagram.

F

F

F

F

F

2

elastic

2

0

1

>

14. Sol. Ans. (A, C, D)As whole system is at rest we use k1x1 = k2x2 = 6 N and Fnet on both m and M = 0

15. Sol. Ans. (A, B, C)If a particle is projected with velocity v0 at an angle with the plane then maximum displacement of the particlenormal to the plane is given by :

h = 2

0( sin )2 cosv

g

where g cos is the component of gravitational acceleration normal to that plane. Since, in the both the cases theparticle is projected at an angle with the plane and plane ahs inclination with horizontal, therefore, h1 and h2 both

will be equal to each other and their magnitudes will be equal to 2 20 sin / 2cos .v

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PG Brainstormer - 5H (Solutions)

The time of flight of each particle will be equal to 0 sin2 cosvg

, where v0 sin is normal component of projection. If

we consider motion of particles along the plane then the range of first particle will be :

R1 = 20

1cos – sin Τ2v T g

while range of other particle will be :

R2 = 20

1cos sin Τ2v T g

where value of T is written above.Hence, (R2 – R1) will be equal to g sin T2.The tangential velocity of the first particle at the point shown is figure is :

1tv = 1

0Τ1cos – sin

2 2v g

while 2tv = 2

0Τ1cos sin

2 2v g

2tv 1t

v

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