Ask Question
28 July, 07:39

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).

+5
Answers (2)
  1. 28 July, 07:40
    0
    = 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
  2. 28 July, 07:52
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers