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Problem-solving: Problem-solving 101 - Solutions
Dr Kathryn Boast, Hertford College
20th July 2020
Problems
1. Shelf and Brackets
A shelf of uniform density is supported by two brackets at a
1 1
distance of /8 and /4 of the total length, L, from each end
respectively.
(a) Find the ratio of the reaction forces from the brackets on
the shelf. (Express your answer in the form of a decimal
less than or equal to 1, to three significant figures.)
The shelf is not moving, so the forces must be balanced, i.e. they
must sum to zero, or be equal in opposite directions.
=⇒ mg=R +R
1 2
The shelf is also not rotating, so the moments acting on it must be
balanced. We can take moments about any point we like. Let us
take moments about the centre so that we do not have to consider
the weight of the shelf (which acts at the centre). (The moment is the
force times the distance from the point about which you are taking
1
moments.)
R ×3L =R ×L
1 8 2 4
R L 8
1 = ×
R 4 3L
2
= 2
3
=0.667 (3sf)
(b) The rectangular shelf is now replaced with a uniform right-
angled triangular shelf, as shown in the figure below. The
1
centre of mass of a triangle lies at a point /3 of the per-
pendicular distance from the base to the tip. Find the new
ratio of the forces on the two brackets. (Express your an-
swer in the form of a decimal less than 1, to three significant
figures.)
L
Taking moments about the centre of mass, a distance /3 from the
2
end of the shelf, we have
R ×2L−1L=R ×L−L
3 3 8 4 3 4
R 16 − 3 =R 1
3 24 24 4 12
13R =2R
3 4
R3 = 2
R4 13
=0.154 (3sf)
2. Energy of a bullet
Abullet of mass m, moving horizontally with speed u, meets a
block of wood of mass M travelling along the same line but in
the opposite direction with speed U, and remains embedded in
it. Show that the loss of kinetic energy is of the form 1kMm and
2
give the expression for k in terms of u, U, m and M.
The initial kinetic energy of the bullet is
1 2
Tm = 2mu
The initial kinetic energy of the wood is
T =1MU2
M 2
The kinetic energy of the final block (with the bullet embedded) is
T = 1(M+m)v2
f 2
So the change in kinetic energy is
1 2 1 2 1 2
∆KE=2(M+m)v −2mu −2MU (1)
3
By the conservation of momentum, the initial momentum must equal the
final momentum. Hence
mu−MU=(M+m)v
=⇒ v= mu−MU
M+m
(mu−MU)2
=⇒ v2 =
(M+m)2
Substituting this into equation (1) gives
(M+m)(mu−MU)2 1
∆KE= − mu2+MU2
(M+m)2 2
2 2 2 2
1 m u −2muMU+M U 2 2
= 2 M+m −mu −MU
2 2 2 2 2 2
= 1 m u +M U −2muMU−(M+m)mu −(M+m)MU
2 M+m
= 1 −2muMU−Mmu2−MmU2
2 M+m
1 Mm 2 2
=−2M+m 2uU+u +U
2
=−1(u+U) Mm
2 M+m
Hence
(u+U)2
k = M +m
3. A Jet in the Fog
Aperson is standing on top of a mountain on a foggy day. The
fog obscures their vision beyond a distance r = 500m . A fighter
jet is travelling in a straight line at constant speed on a horizontal
flight path at the same altitude as the person on the mountain
(but is not flying directly towards them).
The person uses a stopwatch to measure the time interval be-
tween when the jet first becomes visible and when the jet is
loudest (sound was emitted when it was closest to them). They
then measure the time interval between the loudest noise they
hear and the jet exiting their field of vision. Their first mea-
surement is T = 1.60s , and their second is T = 1.00s.
1 2
−1
Thespeedofsoundinairis340ms andwecanassumethetime
taken for light to travel between the plane and the mountain is
negligible.
4
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