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2 years ago

6

Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the

speed of Betty and her dog after the catch is about 0.8 m/s

1 answer:

MA_775_DIABLO [31]2 years ago

7 0

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

$Mv_{1o}+mv_{2o}=(M+m)v$ (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

$v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s$

The speed fo both Betty and her dog is 2.18m/s

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