CU PHYS221 Practice Test 3 Solution - vC

# Find p know side 020 m l 015 m t i 50 ºc t f 650 ºc

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Find: P Know: Side = 0.20 m L = 0.15 m T i = 5.0 ºC T f = −65.0 ºC k ice = 1.6 J/s-m-ºC Sketch : Equation(s) : P = kAΔT/L 5 Version C 0.20 m 0.20 m 0.15 m

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PHYS221 Practice Test 3 Solve : This is a heat transfer through conduction problem The temperature difference across the ice block is: ΔT = (T f – T i ) = (−65.0 ºC) – (5.0 ºC) = −70.0 ºC We know the length of the ice block, “L = 0.15 m”, that the power passes through. The surface area of the square ice block is: A = l x w = (0.20 m)(0.20 m) = 0.04 m 2 To find the rate of heat transferred (power) through the ice block, use the equation “ P = kA(ΔT)/L” with the known values “k ice = 1.6 J/s-m-ºC”, and “L = 0.15 m” with the calculated “A = 0.04 m 2 ” and “ΔT = −70.0 ºC” P = kAΔT = (1.6 J/s-m-ºC)(0.04 m 2 )(−70.0 ºC) = −29.9 W L (0.15 m) 7. A small 0.50 kg cart 1 moving at an unknown initial velocity on an air track collides elastically with a larger 0.65 kg cart 2 moving at 0.85 m/s in the opposite direction as shown below. After the collision, cart 1 is moving to the left at 0.35 m/s and cart 2 is moving to the left at 0.45 m/s. What was the initial velocity of cart 1? Find : v 1-i Know: m 1 = 0.50 kg v 1-i = ? v 1-f = −0.35 m/s x m 2 = 0.65 kg v 2-i = −0.85 m/s x v 2-f = −0.45 m/s x Sketch: Equation(s): p initial = ∑ p final p = m v Solve: This is a conservation of momentum problem for a one dimension elastic collision If we define to the right as the direction of + x , the initial velocity is: “ v 2-i = −0.85 m/s x ” and the two final velocities are “ v 1-f = −0.35 m/s x ” and “ v 2-f = −0.45 m/s x Write conservation of momentum equation in the x direction for the elastic collision m 1 v 1-i + m 2 v 2-i = m 1 v 1-f + m 2 v 2-f Solve for “ v 1-i ”: 6 Version C v 2 1 a) +1.34 m/s x b) +0.87 m/s x c) +0.17 m/s x d) −2.04 m/s x 1 2 v 2 1 1 2 v 2 1 1 2 Initial Final
PHYS221 Practice Test 3 m 1 v 1-i = m 1 v 1-f + m 2 v 2-f − m 2 v 2-i v 1-i = {m 1 v 1-f + m 2 v 2-f − m 2 v 2-i } m 1 Substitute known values “m 1 = 0.50 kg”, “ v 1-f = −0.35 m/s x ”, “m 2 = 0.65 kg”, “ v 2-f = −0.45 m/s x ” and “ v 2-i = −0.85 m/s x ” to solve for “ v 1-i ”: v 1-i = {(0.50 kg)(−0.35 m/s) + (0.65 kg)(−0.45 m/s) – (0.65 kg)(−0.85 m/s) } (0.50 kg) v 1-i = +0.17 m/s x

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