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

= 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