# pr2ch1 - Problem 2 In the ”rest” frame the velocity of...

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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.

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