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

Question

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

in progress 0
RuslanHeatt 4 years 2021-08-24T16:49:22+00:00 1 Answers 9 views 0

Answers ( )

  1. thanhcong
    0
    2021-08-24T16:51:06+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )