Unformatted text preview: fnal mass oF liquid water= initial mass o± liq Δ m= 0.5kg  0.071kg= 0.429kg Problem 2) Car A will have a larger compression on the spring, since both cars have the same speed but di±±erent masses and mass A> mass B => Δ KE ( car A) > Δ KE ( car B), where as we know Δ KE = 1/2 m Δ v ² . And considering the system is closed, whichever car has greater change in KE will have the greater change in PE as well, so Δ PE (car A) > Δ PE (car B). With Δ PEs = 1/2 k Δ x ² , which means that Δ x ( car A) > Δ x (car B). Ice Water Water Eb m (liq) Eth T Eth T Ti= 27°C Ti= 10° C mi= 0.5kg T±= 0°C T±= 0° C m±= 0.429kg...
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 Winter '08
 PARDINI

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