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) A person with mass 50.0 kg, standing still, throws an object with mass 12.0
kg. If the velocity of the person after throwing is +2.0 m/s, what is the
velocity at which he threw the object? Assume the surface to be frictionless.

Respuesta :

The velocity of the object is -8.3 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the person+object system must be conserved.

Therefore, we have:

[tex]p_i = 0[/tex] --> the total momentum at the beginning is zero, because they are both at rest

[tex]p_f=mv+MV[/tex] --> this is the total momentum ,where

m = 12.0 kg is the mass of the object

v is the final velocity of the object

M = 50.0 kg is the mass of the person

V = +2.0 m/s is the final velocity of the person

Since the total momentum is conserved,

[tex]p_i = p_f[/tex]

And therefore we have:

[tex]0=mv+MV\\v=-\frac{MV}{m}=-\frac{(50.0)(+2.0)}{12.0}=-8.3 m/s[/tex]

where the negative sign means the direction of the object is opposite to that of the person.

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