A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.460 m/s. The total mass of

Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.460 m/s. The total mass of the sled, man, and rock is 95.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 15.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving).

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Ngọc Hoa 3 years 2021-08-27T01:02:52+00:00 2 Answers 0 views 0

Answers ( )

  1. huyenthanh
    0
    2021-08-27T01:04:09+00:00
    Reply

    Answer:

    0.41 m/s.

    Explanation:

    Mt = 95 kg

    Mr = 0.31 kg

    Vr = 15.5m/s

    Ut = 0.46 m/s

    Mass of the man and sled = (95 – 0.31) kg

    = 94.69 kg

    Using conservation of momentum equation,

    Momentum before the throw = momentum after the throw

    95 × 0.46 = 0.31 × 15.5 + 94.69 × V2

    43.7 = 4.805 + 94.69 V2

    V2 = 0.41 m/s.

  2. sori
    0
    2021-08-27T01:04:40+00:00
    Reply

    Answer:

    = 0.417 m/s

    Explanation:

    Momentum before throwing the rock: m*V = 95.0 kg * 0.460 m/s

    = 44.27 N*s  

    A) man throws the rock forward

    mass of rock m1 = 0.310 kg

    V1 = 15.5 m/s, in the same direction of the sled with the man

    sled and man:

    m2 = 95 kg – 0.310 kg = 94.69 kg

    v2 = ?

    Conservation of momentum:

    momentum before throw = momentum after throw

    44.27N*s = 0.310kg * 15.5m/s + 94.69kg*v2

    ⇒ v2 = [44.27 N*s – 0.310 * 15.5N*s ] / 94.69 kg

    = 0.417 m/s

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