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A 75-kg man is riding in a 30-kg cart at 2.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is:

Respuesta :

Answer:

Explanation:

We shall apply law of conservation of momentum .

M and m be mass of man and boat . V and v be velocity of man and boat .

MV = mv

75 V = 30 v

2 .0 m/s is relative velocity of man wrt boat .

Given v + V = 2

V = 2 - v

75 ( 2 - v )  = 30 v

150 - 75 v = 30 v

150 = 105 v

v = 1.43 m /s

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