33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Work, Energy and Power
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45
K.E. of 12 kg = 1 2 1
mv = × 12 × 81 = 486 J
2 2
20. (a) : Kinetic energy = p 2
2m
E
\ 1 p1 2 / 2m1
E1
m2
=
E2
p m E m
2 2 ⇒ =
/ 2 2 2 1
or E 1 < E 2 [as m 1 > m 2 ]
21. (c) : Ratio of their kinetic energy is given as
KE1
1 2 11 2
= ( / ) mv
KE 2 ( 1/ 2)
mv 2 2 2
v 2 = 2gs
(zero initial velocity)
which is same for both
KE1
m1
2 1
∴ = = = ⋅
KE2
m2
4 2
22. (a) : m 1 v 1 = m 2 v 2
(conservation of linear momentum)
E1
(/ 1 2)
mv 11 2 mv 1 2 1 2 m2
m2
=
E2
mv mv m m
2 2 2 =
1 2
2 2 2 2 ⋅ = .
(/ )
1 1
23. (a) : Let m be the mass of the body and v 1 and v 2 be
the initial and final velocities of the body respectively.
\ Initial kinetic energy = 1 mv 2
2 1
Final kinetic energy = 1 mv 2
2 2
Initial kinetic energy is increased 300% to get the final
kinetic energy.
1 1 ⎛ ⎞
∴ mv = +
2 2 ⎝
⎜ 1 300
2 2 100 ⎠
⎟ mv1 2
⇒ v 2 = 2v 1 or v 2 /v 1 = 2
Initial momentum = p 1 = mv 1
Final momentum = p 2 = mv 2
p2
mv2
v2
⎛ ⎞
∴ = = = 2 or, ∴ p2 = 2p1 = +
p1
mv1
v
⎝
⎜1 100
⎠
⎟ p1
1
100
So momentum has increased 100%.
24. (b) : Kinetic energy of the ball = K and angle of
projection (q) = 45°.
Velocity of the ball at the highest point = v cosq
v
= v cos 45° =
2
Therefore kinetic energy of the ball
= × ⎛ 2
1
⎝ ⎜ v ⎞
2 ⎠
⎟
2
= 1 2
m
4 = K
mv
2
2
K.E.
25. (d) : 1
K.E. = p
⇒ K.E.
= m2
m
= 4
2
2 m1
1
m
or 1 1
=
m2
4
26. (a) : Mass of first body = m;
Mass of second body = 4m and KE 1 = KE 2 .
Linear momentum of a body
p= 2mE ∝ m
Therefore p 1 m m
p
= 1
2 m
= 1 1
2 4m
= 4
= 2
or p 1 : p 2 = 1 : 2.
27. (b) : v 2 = u 2 + 2as or v 2 – u 2 = 2as
or v 2 – (0) 2 F
2
=
2
2 × × =
m s v Fs
or
m
and KE . . = 1 2
mv = 1
m × 2 Fs
2 2 m
= Fs.
Thus K.E. is independent of m or directly proportional
to m 0 .
28. (c) :
K1
p1 2 M2 2 =
K2
p2 2 ×
M1 2
Here K 1 = K 2
\
p1
M1
1
= = = 1 or, p 1 : p 2 = 1 : 3
p M 9 3
2
2
29. (c) : On the diametrically opposite points, the
velocities have same magnitude but opposite directions.
Therefore change in momentum is
Mv – (–Mv) = 2Mv
30. (d) : Given : F = 20 + 10y
work done, W = ∫ F.
dy
1
⎡ 10
= ∫( 20 + 10y)
dy = 20y+
⎣⎢
2
0
31. (d) :
1
2 ⎤
y 20 10 25
⎦⎥ = + 0 2
= J
Work done = Area under (F-d) graph
= Area of rectangle ABCD + Area of triangle DCE
1
= 2× ( 7− 3) + × 2× ( 12−
7)= 8 + 5 = 13 J
2
t
2 ds t d 2 s
32. (d) : s = 3
; dt
= 2
3
; 2
dt
= 2
3
m/s2
Work done, W Fds m d 2
= ∫ = ∫
s ds
2
dt
= ∫m d 2
s 2
2
ds
= ∫ × × = ∫
dt dt dt t
3 2 2 4 dt tdt
2
3 3 3
0
0
2
2
4 t 4
= = × =
3 2 3 2 8
J.
3
0
33. (b) : Work done = area under F-x curve
= area of trapezium
1
9
= × + × = × 3
( 6 3) 3 = 13. 5 J
2
2