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3 years ago

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A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in

the toboggan. If KE1 is the original kinetic energy of the toboggan and KE2 is the kinetic energy after the parcel has been added, what is the ratio KE2/KE1

1 answer:

timama [110]3 years ago

5 0

Answer:

$KE2/KE1=0.71$

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

$V2 = \frac{m1}{m1+m2}*V1$

The kinetic energies are:

$KE1=1/2*m1*V1^2$

$KE2 = 1/2*(m1+m2)*V2^2$

$KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2$

Simplifying:

$KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2$

The ratio will be:

$KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}$

$KE2/KE1=0.71$

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Knowing this, let's begin with the calculaations:

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