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A 5.00-kg howler monkey is swinging due east on a vine. It overtakes and grabs onto a 6.00-kg monkey also moving east on a second vine. The first monkey is moving at 10.0 m/s at the instant it grabs the second, which is moving at 7.00 m/s. 1) After they join on the same vine, what is their common speed

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asuwafohjames

Answer:

8.36 m/s

Explanation:

From the law of conservation of momentum,

Total momentum of the monkeys before grab = Total momentum after grab

mu+m'u' = V(m+m')................... Equation 1

Where m = mass of the first money, m' = mass of the second monkey, u = initial velocity of the first monkey, u' = initial velocity of the second monkey, V = Common velocity of both monkeys.

make V the subject of the equation

V = [mu+m'u']/(m+m')................ Equation 2

Given: m = 5 kg, m' = 6 kg, u = 10 m/s, u' = 7 m/s

Substitute into equation 2

V = [5×10+6×7]/(5+6)

V = [50+42]/11

V = 92/11

V = 8.36 m/s

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