(a) 1. A coordinate system attached to car A is a translating system, in which
aB = a A + aB / A
Observe that
( a B ) t = 3 m/s2
2
v B 144
(a B ) n =
=
= 1.44 m/s2
100
Thus
a B = 3.33 m/s2
64.4
Either from a graphical or analytical solution,
a B / A = 5.32 m/s2
= 62.72
(b) A coordinate system attached to B is a rotating system. If a rel is the acceleration of car A as
observed from car B, a rel a B / A . It can be shown from rigid-body kinematics that a rel satisfies
&
a A = a B + rA / B + ( rA / B ) + 2 v rel + a rel
aB = 3.33 m/s2
aB/A
60
64.4
aA = 2 m/s2
2. Let T1 be the tension in the upper string and T2 tension in the lower string. Force balance on
each mass gives
&&
(1)
4mg T1 = 4mq1
&&
3mg T2 = 3mq3
(2)
&&
(3)
mg T2 = mq4
&& && &&
There are five unknowns q1 , q3 , q4 , T1 , and T2 in three equations. However,
&& &&
&& &&
&& &&
&&
&&
q3 2 q + ( q 4 q 2 ) = 0
q3 + q4 = 2q2 = 2 q1
T1 = 2T2
Upon solution,
1
&&
q1 = g = 1.40 m/s2
7
q4
q2
q1
T1
T1
4m
T2
q3
3m
T2
m
3. Let position 1 of the block be its initial position at 150 mm above the springs. Suppose
position 2 corresponds to deflection x in the two springs. Between positions 1 and 2,
U = T + Vg + Ve
where the work done by forces other than gravitational and spring forces is
U =0
T = T2 T1 = 0
Define the reference level for measuring potential energy as the level associated with the
precompressed springs before impact. Then
Vg = mgh2 mgh1 = mg ( x ) mg (0.15) = mg ( x + 0.15)
1 2 1
Ve = 2 kx2 2 kx12
2
2
1
1
= 2 k (0.075 + x ) 2 2 k (0.075) 2 = k [(0.075 + x ) 2 0.0752 ]
2
2
Thus
U = T + Vg + Ve
mg ( x + 0.15) + k [(0.075 + x ) 2 0.0752 ] = 0
5000 x 2 + 651.9 x 14.715 = 0
x = 0.0196 or 0.150
The additional deflection is x = 19.6 mm.
10 kg
Position 1
150 mm
Reference
x
Position 2
k
k = 5 kN/m