Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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76 2 Particle Dynamics
Displacement r
12
= r
1
− r
2
= 7
ˆ
k −(20
ˆ
i +15
ˆ
j) = (−20
ˆ
i −15
ˆ
j −7
ˆ
k) cm
Work done W = F · r
12
= (5
ˆ
i −3
ˆ
j +
ˆ
k) · (−0.20
ˆ
i −0.15
ˆ
j +0.07
ˆ
k)
=−0.48 J.
2.23
(i) U(x) = 5x
2
− 4x
3
F(x) =−
dU
dx
=−(10x − 12x
2
) = 12x
2
− 10x
(ii) For equilibrium F(x) = 0
x(12x −10) = 0orx = 5/6mor0
dF
dx
= 24x −10
dF
dx
|
x=0
= (24x −10)
|
x=0
=−10
The position x = 0 is stable:
dF
dx
x=
5
6
= (24x −10)
x=
5
6
=+10
The position x = 5/6 is unstable.
2.24 Let the body travel a distance s on the incline and come down through a
height h.
Potential energy lost = mgh = mgs sin θ .
Work down against friction W = fs = μmg cos θ ·s.
By problem μ mg cos θs =
70
100
mgs sin θ
∴ μ = 0.7tanθ = 0.7 tan 30
◦
= 0.404
2.25 At the bottom of the ramp the kinetic energy K available is equal to the loss
of potential energy, mgh:
K = mgh
On the flat track the entire kinetic energy is used up in the work done against
friction
W = fd = μmgd
∴ μ mgd = mgh
μ =
h
d
2.3 Solutions 77
2.26
(i) Work done by the spring W
s
=
1
2
kx
2
=
1
2
×20 ×10
3
×(0.12)
2
= 144 J
(ii) Work done by friction W
f
=
1
2
mv
2
− W
s
=
1
2
× 50 × 3
2
− 144 = 81 J
(iii) W
f
= μmgs
∴ μ =
W
f
mgs
=
81
50 × 9.8 × (0.60 + 0.12)
= 0.2296
(iv)Ifv
1
is the velocity of the crate as it passes position A after rebonding
1
2
mv
2
1
= W
s
− μ mgs
1
2
× 50v
2
1
= 144 −0.2296 × 50 × 9.8 × (0.60 + .012) = 63
∴ v
1
= 1.587 m/s
2.3.4 Collisions
2.27 In the CMS the velocity of m
1
will be v
∗
1
= v − v
c
and that of m
2
will be
v
∗
2
=−v
c
,Fig.2.27. By definition in the CMS t otal momentum is zero:
Fig. 2.27
m
1
v
∗
1
+ m
2
v
∗
2
= 0
∴ m
1
(v − v
c
) − m
2
v
c
= 0
∴ v
c
= v
∗
2
=
m
1
v
m
1
+ m
2
(1)
∴ v
∗
1
= v −v
c
=
m
2
v
m
1
+ m
2
(2)
Note that as the collision is elastic, the velocities of m
1
and m
2
after the colli-
sion in the CMS remain unchanged. The lab velocity v
1
of m
1
is obtained by
the vectorial addition of v
∗
1
and v
∗
c
. From the triangle ABC, Fig. 2.28.After
collision
v
2
1
= v
∗2
1
+ v
2
c
− 2v
∗
1
v
c
cos(180
◦
− θ) (3)
78 2 Particle Dynamics
Fig. 2.28
Using (1) and (2) and simplifying
v
1
=
v
m
1
+ m
2
m
2
1
+ m
2
2
+ 2m
1
m
2
cos θ (4)
2.28 (a) Let the initial momentum of one object be p. After scattering let the
momenta be p
1
and p
2
, with the angle θ between them, Fig. 2.29.
p = p
1
+ p
2
(momentum conservation)
∴ ( p · p) = p
2
= ( p
1
+ p
2
) · ( p
1
+ p
2
)
= p
1
· p
1
+ p
2
· p
2
+ p
1
· p
2
+ p
2
· p
1
= p
2
1
+ p
2
2
+ 2 p
1
+ p
2
(1)
p
2
2m
=
p
2
1
2m
+
p
2
2
2m
(energy conservation) or p
2
= p
2
1
+ p
2
2
(2)
Combining (1) and (2), 2 p
1
· p
2
= 0
∴ p
1
and p
2
are orthogonal
Fig. 2.29a
Fig. 2.29b
2.3 Solutions 79
Fig. 2.30
(b) Suppose one of the objects (say 2) is scattered in the backward direction
then the momenta would appear as in Fig. 2.30, and because by (a) the angle
between p
1
and p
2
has to be a right angle, both the objects will be scattered on
the same side of the incident direction (x-axis). In that case, the y-component
of momentum cannot be conserved as initially
p
y
= 0. Both the objects
cannot be scattered in the backward direction. In that case the x-component
of momentum cannot be conserved.
2.29 We work out in the CM system. The total kinetic energy available in the
CMS is
= K
∗
=
1
2
μv
2
(1)
where μ =
mM
m + M
(2)
is the reduced mass.
If the compression of the spring is x then the spring energy would be
1
2
kx
2
.
Equating the total kinetic energy available in the CM–system to the spring
energy
1
2
μv
2
=
1
2
kx
2
x = v
μ
k
= v
mM
k(m + M)
2.30 After the first collision (head-on) with the sphere 2 on the right-hand side,
sphere 1 moves with a velocity
v
1
=
2m
2
u
2
+ u
1
(m
1
− m
2
)
m
1
+ m
2
=
0 + u
1
(m − 4m)
m +4m
=−0.6u
1
(1)
and the sphere 2 moves with a velocity
v
2
=
2m
1
u
1
+ u
2
(m
2
− m
1
)
m
1
+ m
2
=
2mu
1
+ 0
m +4m
= 0.4u
1
(2)
80 2 Particle Dynamics
where u and v with appropriate subscripts refer to the initial and final
velocities.
The negative sign shows that the ball 1 moves toward left after the collision
and hits ball 3. After the second collision with ball 3, ball 2 acquires a velocity
v
2
and moves toward right
v
1
=
2m
3
u
3
+ υ
1
(m
1
− m
3
)
m
1
+ m
3
=
0 − u
1
(m
1
− 4m)
m +4m
= 0.36u
1
(3)
But v
1
<v
2
. Therefore ball 1 will not undergo the third collision with ball 2.
Thus in all there will be only two collisions.
2.31
Momentum conservation gives m
1
u
1
+ m
2
u
2
= m
2
v
2
(1)
Conservation of kinetic energy in elastic collision gives
1
2
m
1
u
2
1
+
1
2
m
2
u
2
2
=
1
2
m
2
v
2
2
(2)
By the problem
1
2
m
1
u
2
1
=
1
2
m
2
u
2
2
(3)
u
2
= αu
1
(4)
∴ α
2
=
m
1
m
2
(5)
Using (3) in (2)
m
2
u
2
2
=
1
2
m
2
v
2
2
∴ v
2
=
√
2u
2
(6)
Using (4) and (6) in (1)
m
1
u
1
+ m
2
αu
1
=
√
2m
2
u
2
=
√
2m
2
αu
1
or m
1
+ m
2
α =
√
2αm
2
Dividing by m
2
and using (5) and rearranging
α
α −
√
2 − 1
= 0
since α = 0,α =
√
2 − 1
∴ α =
u
2
u
1
=
√
2 − 1
∴
u
1
u
2
=
1
√
2 − 1
=
√
2 + 1
2.3 Solutions 81
From (5)
m
1
m
2
= α
2
=
√
2 − 1
2
= 3 −2
√
2
2.32
(i) If the common velocity of the merged bodies is v then momentum con-
servation gives
(m
A
+ m
B
)v = m
A
v
A
+ m
B
v
B
∴ v =
m
A
v
A
+ m
B
v
B
m
A
+ m
B
(ii) v =
3
2
m
B
(5
ˆ
i +3
ˆ
j) + m
B
(−
ˆ
i +4
ˆ
j)
3
2
m
B
+ m
B
= 2.6
ˆ
i +3.4
ˆ
j
(iii) p
A
= m
A
(υ − υ
A
) = m
A
2.6
ˆ
i +3.4
ˆ
j −(5
ˆ
i +3
ˆ
j)
= m
A
−2.4
ˆ
i +0.4
ˆ
j
p
A
= 1200
(−2.4)
2
+ (0.4)
2
= 2920 N m
F
A
=
p
A
t
=
2920
0.2
= 14,600 N
p
B
= m
B
(
υ −
υ
B
) = m
B
2.6
ˆ
i +3.4
ˆ
j −(−
ˆ
i +4
ˆ
j)
= m
B
3.6
ˆ
i −0.6
ˆ
j
m
B
=
2
3
m
A
=
2
3
× 1200 = 800 kg
p
B
= 800
(3.6)
2
+ (−0.6)
2
= 2920 N m
F
B
=
p
B
t
=
2920
0.2
= 14,600 N
(iv) K
=
1
2
(m
A
+ m
B
)υ
2
=
1
2
(1200 + 800)
(2.6)
2
+ (3.4)
2
= 18, 320 J
2.33 Let the velocity of the particle moving below the x-axis be v’. Momentum
conservation along x- and y-axis gives
mv
0
= mv cos θ +mv
cos β (1)
0 = mv sin θ − mv
sin β (2)
82 2 Particle Dynamics
Cancelling m and reorganizing (1) and (2)
v
cos β = v
0
− v cos θ (3)
v
sin β = v sin θ (4)
Dividing (4) by (3)
tan β =
v sin θ
v
o
− v cos θ
(5)
If the collision is elastic, kinetic energy must be conserved:
1
2
mv
2
0
=
1
2
mv
2
+
1
2
mv
2
(6)
or v
2
0
= v
2
+ v
2
(7)
Squaring (3) and (4) and adding
v
2
= v
2
+ v
2
0
− 2v
0
v cos θ (8)
Eliminating v
2
between (7) and (8) and simplifying
v = v
0
cos θ
2.34 The momenta of β and
14
N are indicated in both magnitude and direction
in Fig. 2.31. The resultant R of these momenta is given from the diagonal
AC of the rectangle (parallelogram law). The momentum of u is obtained by
protruding CA to E such that AE = AC:
Fig. 2.31 Decay of
14
Cat
rest
2.3 Solutions 83
R =
AB
2
+ BC
2
=
p
2
+ (4p/3)
2
=
5p
3
tan θ =
BC
AB
=
4p/3
p
=
4
3
∴ θ = 53
◦
Thus the neutrino is emitted with momentum 5p/3 at an angle (180 −53
◦
)or
127
◦
with respect to the β particle.
2.35 Denoting the angles with (*) for the CM system transformation of angles for
CMStoLSisgivenby
tan θ =
sin θ
∗
cos θ
∗
+
m
M
(1)
But θ
∗
= π − φ
∗
= π − 2φ
∴ sin θ
∗
= sin(π − 2φ) = sin 2φ
cos θ
∗
= cos(π − 2φ) =−cos 2φ
(i) becomes
tan θ =
sin 2φ
m
M
− cos 2φ
Furthermore
sin θ
cos θ
=
sin 2φ
m
M
− cos 2φ
Cross-multiplying and rearranging
m
M
sin θ = sin θ cos 2φ + cos θ sin 2φ = sin(θ +2φ)
∴
m
M
=
sin(2φ + θ)
sin θ
2.36 In the lab system let M be projected at an angle φ with velocity v.Inthe
CMS the velocity v
∗
for the struck nucleus will be numerically equal to v
c
,
the centre of mass velocity. Therefore, M is projected at an angle 2φ with
velocity v
∗
=
mv
M + m
. The CM system velocity v
c
=
mv
M + m
. The velocities
v
∗
and v
c
must be combined vectorially to yield v,Fig.2.32. Since v
c
= v
∗
the velocity triangle ABC is an isosceles triangle. If BD is perpendicular on
AC, then
84 2 Particle Dynamics
AC = 2AD = 2AB cos φ
∴ v = 2v
∗
cos φ =
2mu cos φ
M + m
Fig. 2.32
2.37
(a) Kinetic energy of A before collision K
A
=
1
2
(2m)u
2
= mu
2
. Since B
is initially stationary, its kinetic energy K
B
= 0. Hence before collision,
total kinetic energy K
0
= mu
2
+ 0 = mu
2
.
Let A and B move with velocity υ
A
and υ
B
, respectively, after the colli-
sion, Fig. 2.33. Total kinetic energy after the collision,
Fig. 2.33
K
= K
A
+ K
B
=
1
2
(2m)v
2
A
+
1
2
(10m)v
2
B
= mv
2
A
+ 5mv
2
B
If an energy Q is lost in the collision process, conservation of total energy
gives
mu
2
= mv
2
A
+ 5mv
2
B
+ Q (1)
Applying momentum conservation along the incident direction and per-
pendicular to it
2mu = 10mv
B
cos 37
◦
= 8mv
B
(2)
2mv
B
= 10mv
B
sin 37
◦
= 6mv
B
(3)
From (2) and (3) we find
2.3 Solutions 85
v
A
=
3u
4
; v
B
=
u
4
(b) From (1), Q = m(u
2
− v
2
A
− 5v
2
B
)
= m
u
2
−
9u
2
16
−
5u
2
16
=
mu
2
8
∴
Q
k
A
=
mu
2/8
mu
2
=
1
8
Since Q is positive, energy is lost in the collision process.
2.38
(a) In the elastic collision (head-on) of a particle of mass m
1
and kinetic
energy K
1
with a particle of mass m
2
initially at rest, the fraction of
kinetic energy imparted to m
2
is
K
2
K
0
=
4m
1
m
2
(m
1
+ m
2
)
2
=
4 × 1 × 12
(1 + 12)
2
=
48
169
(b)
1
2
m
2
v
2
2
1
2
m
1
v
2
1
=
12
1
v
2
2
v
2
0
=
48
169
∴ v
2
=
2
13
v
0
K
1
= K
0
− K
2
= K
0
−
48
169
K
0
=
121
169
K
0
∴
1
2
m
1
v
2
1
=
121
169
×
1
2
m
1
v
2
0
∴ v
1
=−
11
13
v
0
Negative sign is introduced because neutron being lighter then the carbon
nucleus will bounce back.
2.39 Let the heavy body of mass M with momentum P
0
collide elastically with a
very light body of mass m be initially at rest. After the collision both the bodies
will be moving in the direction of incidence, the heavier one with velocity v
H
and the lighter one with velocity v
L
.
Momentum conservation gives
p
0
= p
L
+ p
H
(1)
Energy conservation gives
p
2
0
2M
=
p
2
L
2m
+
p
2
H
2M
(2)