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Unformatted text preview: Problem 2
In the ”rest” frame the velocity of the ﬁrst car (m1 = 2000kg) is u1x =
20m/s, that of the second car (m2 = 1500kg) is u2x = 0. So the total
momentum before the collision is
px = m1 u1x + m2 u2x = 4 × 104 kg m/s
which must be conserved and is the same after the collision. The total
mass of the wreck (the two cars stuck together) after the collision is M =
m1 + m2 = 3500kg. Thus, the velocity of the wreck after the collision is
ux = 4 × 104 kg m/s
px
=
≈ 11.43m/s
M
3500kg Now let us transform to the ”moving” frame with velocity v = 10m/s.
Applying the Galilean transformation law (1.2) for velocities we get that the
velocity of the ﬁrst car before the collision in this frame is
u′ x = u1x − v = 10m/s
1
the velocity of the second car is
u′ x = u2x − v = −10m/s
2
and that of the wreck is
u′ = ux − v ≈ 1.43m/s
x
The total momentum before the collision in the ”moving” frame is thus
′
′
p′
x,initial = m1 u1x + m2 u2x = 5000kg m/s while after the collision we have
p′ inal = M u′ = 5000kg m/s
x,f
x
Hence p′ inal = p′
x,f
x,initial and the momentum is conserved in the ”moving” frame. Note that the ﬁnal result may be slightly oﬀ because of the
rounding of the numerical results above. 1 ...
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This note was uploaded on 07/20/2011 for the course PHYS 2D 2D taught by Professor Sinha during the Fall '10 term at UCSD.
 Fall '10
 SINHA

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