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A 75.0 kg man is standing at rest on ice while holding a 4.00kg ball. If the man throws the ball at a velocity of 3.50m/s forward what will his resulting velocity be

Respuesta :

abidemiokin

Explanation

According to law of conservation  of momentum

m1u1 + m2u2 = (m1+m2)v

m1 is mass of the man 75kg

u1 is the velocity of the man 0m/s

m2 is the mass of the ball = 4kg

u2 is the velocity of the ball = 3.5m/s

We are to find v

Substitute

75(0)+4(3.5) = (75+4)v

0 + 14 = 79v

v = 14/79

v = 0.177m/s

Hence his resulting velocity will be 0.177m/s

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