A man holding a rock sits on a sled that is sliding across a frozen lake (neglig

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 96.0 kg. The mass of the rock is 0.330 kg and the man can throw it with a speed of 16.0 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving).

________m/s

Determine the speed of the sled if the man throws the rock directly backward.

_________m/s

Explanation / Answer

M = mass of sled, man, and rock = 96 kg

V = velocity of the combination = 0.460 m/s

m = mass of rock = 0.33 kg

v = velocity of rock relative to ground when thrown forward = 0.460 + 16 = 16.460 m/s

m' = mass of sled and man = 96 - 0.33 = 96.67 kg

using conservation of momentum

MV = mv + m' v'

96 (0.460) = (0.33) (16.460) + (96.67) v'

v' = 0.401 m/s

when thrown backward :

M = mass of sled, man, and rock = 96 kg

V = velocity of the combination = 0.460 m/s

m = mass of rock = 0.33 kg

v = velocity of rock relative to ground when thrown backward = 0.460 - 16 = - 15.54 m/s

m' = mass of sled and man = 96 - 0.33 = 96.67 kg

using conservation of momentum

MV = mv + m' v'

96 (0.460) = (0.33) (- 15.54) + (96.67) v'

v' = 0.51 m/s

v' = 0.401 m/s

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