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3 years ago

14

If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the

change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions? Please express your answers with 4 decimal place. _________________________ m

1 answer:

gtnhenbr [62]3 years ago

5 0

Answer:

$L_f = 0.0196$

Explanation:

given,

mass of the car,m = 0.2 Kg

height to width of ramp, y/x = 12/75

initial displacement, L_i = 2.25 m

change in momentum, Δp = 0.58 kg.m/s

distance of change in direction of the ramp, L_f = ?

Using equation

$\Delta P = m \sqrt{\dfrac{2gy}{x}} (\sqrt{L_i}+\sqrt{L_f})$

inserting all the values

$0.58 = 0.2\times \sqrt{\dfrac{2\times 9.8\times 12}{75}} (\sqrt{2.25}+\sqrt{L_f})$

$1.5 + \sqrt{L_f} = 1.64$

$\sqrt{L_f}=0.14$

$L_f = 0.0196$

The car will cost back up to the distance of 0.0196 m

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