33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Laws of Motion
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33
( mv 11) 2 + ( mv 2 2)
2 2 2
= ( 21) + ( 21)
= 21 2 kg m/s
According to law of conservation of linear momentum
m 3 v 3 = 21 2 or 3v 3 = 21 2
or v 3 = 7 2 m/s
31. (d) : As per triangle law,
F1+ F2+ F3 = 0 i.e., net force on the
particle is zero. So, acceleration is
also zero.
Hence velocity of the particle will
remain constant.
Y
32. (c) : Taking x-components,
4 N 1 N
the total should be zero.
1 × cos 60° + 2cos 60°
30°
+ x – 4cos 60° = 0
60° 60°
\ x = 0.5 N
x
X
33. (d) : Coefficient of sliding
friction has no dimension.
2 N
f
f = m s N ⇒ µ s =
N
34. (d) : Let m s and m k
be the coefficients of
static and kinetic friction
between the box and the
plank respectively.
When the angle of inclination q reaches 30°, the block
just slides,
1
∴ µ s = tanθ= tan 30°= = 06 .
3
If a is the acceleration produced in the block, then
ma = mgsinq – f k
(where f k is force of kinetic friction)
= mgsinq – m k N (as f k = m k N)
= mgsinq – m k mgcosq (as N = mgcosq)
a = g(sinq – m k cosq)
As g = 10 m s –2 and q = 30°
\ a = (10 m s –2 )(sin30° – m k cos30°) ...(i)
If s is the distance travelled by the block in time t, then
s= 1 at
2 s
(as u = 0) or a = 2
2
2
t
But s = 4.0 m and t = 4.0 s (given)
240 (. m)
1 −2
∴ a = = ms
2
(. 40s)
2
Substituting this value of a in eqn. (i), we get
1 −2 −2
⎛
ms 10 ms
1 3 ⎞
= ( ) −
2
⎝
⎜ µ
2 k
2 ⎠
⎟
1
10 1 3 3 1 1 9
= − µ k or µ k = − = =09 .
10 10
09 .
µ k = = 05 .
3
35. (a)
36. (c) : Force of friction on massm 2 = mm 2 g
Force of friction on mass m 3 = mm 3 g
Let a be common acceleration
of the system.
mg 1 −µ mg 2 −µ
mg 3
∴ a =
m + m + m
1 2 3
Here, m 1 = m 2 = m 3 = m
mg −µ mg −µ mg mg − 2µ mg g( 1−
2µ
)
∴ a =
=
=
m+ m+
m 3m
3
Hence, the downward acceleration of mass m 1 is g( 1− 2µ
) .
3
37. (a) :
For upper half smooth plane
Acceleration of the block, a = gsinq
Here, u = 0 ( block starts from rest)
L
a = g sinq, s = 2
Using, v 2 – u 2 = 2as, we have
2
L
v − 0= 2× gsinθ
×
2
v= gLsinθ
...(i)
For lower half rough plane
Acceleration of the block, a′ = g sinq – mg cosq
where m is the coefficient of friction between the block
and lower half of the plane
Here, u= v= gLsinθ
v = 0 (block comes to rest)
L
a = a′ = g sinq – mgcosq, s = 2
Again, using v 2 – u 2 = 2as, we have
L
2
0− ( gLsin θ) = 2× ( gsinθ − µ gcos θ ) ×
– gLsinq = (g sinq – mgcosq)L
2
–sinq = sinq – mcosq
m cos q = 2sinq or m = 2tanq
38. (a) : Force of friction, f = mmg
f µ mg
−2
∴ a = = = µ g = 05 . × 10 = 5 ms
m m
Using v 2 – u 2 = 2aS, 0 2 – 2 2 = 2(–5) × S ⇒ S = 0.4 m
39. (c) : Pseudo force or fictitious force, F fic = ma
Force of friction, f = mN = mma
The block of mass m will
not fall as long as
f ≥ mg ; mma ≥ mg
α ≥ g
µ