EXAMINATION SOLUTIONS
MECHANICS 1
MATH 11009
(Paper Code MATH-11009)
May-June 2015, 1 hour and 30 minutes
A.1. a) A.B = 2 ∗ 3 + 3 ∗ −1 + 2 ∗ 1 = 5,
|A| = |B| =
√
14 so θ = cos−1 (5/14).
√
b) A × B = 7î − ĵ − 11k̂ so unit vector would be (7î − ĵ − 11k̂)/ 171.
A.2. a) need to differentiate v as follows
˙
a = dv/dt = r̈r̂ + ṙ r̂˙ + ṙ θ̇θ̂ + r θ̈θ̂ + r θ̇θ̂
which is, using the relations given,
a = r̈r̂ + ṙ θ̇θ̂ + ṙ θ̇θ̂ + r θ̈θ̂ − r θ̇2 r̂
or
a = (r̈ − r θ̇2 )r̂ + (2ṙ θ̇ + r θ̈)θ̂
b) constant radius so ṙ = 0 and θ̇ = ω
a = −Rω 2 r̂
so centripetal acceleration is Rω 2 directed radially inwards.
A.3. (8 total marks possible)
Z
P2
F.dr =
P1
Z
0
1
2 dx
t.t
dt
dt +
Z
0
1
t2
dy
dt
dt
2
where the path is parametrised by x = t and y = t for t ∈ [0, 1].
Z 1
Z P2
Z 1
3
2
F.dr =
t .1 + t .2t dt =
3t3 dt = [3t4 /4]10 = 3/4
P1
0
0
The force is not conservative as (xy)y 6= (x2 )x so the path does matter.
A.4. a) Separating variables,
1
k
dṡ = − dt
ṡ
m
and integrating ODE, we get
log(ṡ/v) = −
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k
t
m
or ṡ = ve−kt/m . Clearly ṡ > 0 for all finite t and ṡ → 0 as t → ∞.
b) total distance travelled (as t → ∞) is
∞
Z t=∞
Z ∞
Z ∞
mv
mv −kt/m
−kt/m
=
e
ds =
ṡdt =
ve
dt = −
k
k
t=0
0
0
0
A.5. a) multiply ODE by ẋ and integrate to get
mẋ2 + kx2 = constant
b) Initial conditions ẋ(0) = 0 and x(0) = l give constant so
mẋ2 + kx2 = kl2
p
Therefore maximum speed = l k/m. Largest (magnitude of ) acceleration will be at
largest x from ODE, so maximum |ẍ| = kl/m
R0
Rl
c) Work done = l −kxdx = 0 kxdx = [0.5kx2 ]l0 = 0.5kl2 which is just the gain in kinetic
p
energy = 0.5m(l k/m)2 .
B.1. (30 marks possible)
a) Apply Newton’s laws
ÿ = 0
&
z̈ = −g
integrate once (using initial velocity)
ẏ = v0 cos β
&
ż = v0 sin β − gt
and again (using initial position)
y(t) = v0 t cos β
&
z(t) = v0 t sin β − 1/2gt2
which give position of particle at time t. (4 marks for y and 4 marks for z)
b) h = z cos α − y sin α = v0 t(cos α sin β − cos β sin α) − 1/2gt2 cos α using formulae from
a). Simplify to
h(t) = v0 t sin(β − α) − 1/2gt2 cos α
Maximum h occurs at
ḣ(t) = v0 sin(β − α) − gt cos α = 0
or
tmaxheight = v0 sin(β − α)/(g cos α)
(5 marks) Substituting this into expression for h gives maximum height
hmax =
v02 sin2 (β − α)
.
2g cos α
(3 marks)
c) Projectile hits plane when h(t) = 0 i.e.
v0 t sin(β − α) − 1/2gt2 cos α) = t(v0 sin(β − α) − 1/2gt cos α) = 0
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Hence flight time is
tf light = 2v0 sin(β − α)/(g cos α)
which is twice tmaxheight regardless of α. (3 marks) Substitute this into expression for r(t)
to get maximum distance.
r(t) = y cos α + z sin α = v0 t cos β cos α + v0 t sin β sin α − 1/2gt2 sin α
So
r(t) = v0 t cos(β − α) − 1/2gt2 sin α
With tf light inserted,
rmax
4v02 sin2 (β − α)
2v02 cos(β − α) sin(β − α)
− 1/2g sin α
=
g cos α
g 2 cos2 α
2
2v0 sin(β − α)
=
cos(β − α) cos(α) − sin(α) sin(β − α)
g cos2 α
2v 2 sin(β − α) cos β
.
= 0
g cos2 α
(3 marks)
d) i) With a smooth (frictionless) plane, this question is about converting the potential
energy at point A into kinetic energy at point O. Vertical height of A above O is rmax sin α
so, if the speed at O is u we must have
1/2mu2 = mgrmax sin α
so
u=
(4 marks)
2v0 p
(sin α sin(β − α) cos β.
cos α
ii) with friction, the revised work equation is
1/2mu2 = (mg sin α − F )rmax
to take account of work done against friction. Then
u=
2v0 p
(sin α − F/mg) sin(β − α) cos β.
cos α
(4 marks)
B.2. (30 marks possible)
a) L = r × mṙ = mr 2 θ̇ẑ. (3 marks)
Newton’s 2nd law gives mr̈ = −k/r 2 r̂. Crossing with r gives
mr × r̈ =
dL
d
(mr × ṙ) =
=0
dt
dt
so angular momentum is conserved by a central force. (5 marks)
b) Equation of motion in the radial direction is
mr̈ − mr θ̇2 = −
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k
r2
(2 marks) (see A2 for a hint) Substitute θ̇ = L/mr 2 in to get
L
mr̈ − mr
mr 2
2
=−
k
r2
rearranging
mr̈ −
L2
k
+ 2 =0
3
mr
r
Multiply by ṙ and integrate gives
k
L2
−
=E
1/2mṙ +
2mr 2
r
2
where E is a constant (the total energy). So the effective potential U(r) :=
(6 marks)
L2
2mr 2
− kr .
c) Circular orbit given by radius such that dU/dr = 0 or
L2
k
=
mr 3
r2
so circular orbit has r0 = L2 /(mk) (3 marks). From angular momentum mr02 θ̇ = L so
θ̇ =
mk 2
L3
and the rotation period = 2π/θ̇ = 2πL3 /(mk 2 ). (3 marks)
d) Particle is in circular orbit so ṙ = 0 and E = U(r0 ) For escape need E ≥ U(r → ∞) = 0.
So energy required is E = U(r → ∞) − U(r0 ) or
E = −U(r0 ) = −
L2
k
mk 2
+
=
.
2mL4 /(m2 k 2 ) L2 /(mk)
2L2
(8 marks)
End of solutions.
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