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A person stands in a stationary canoe and throws a 5.16 kg stone with a velocity of 8.05 m/s at an angle of 31.0° above thehorizontal. The person and canoe have a combined mass of 105 kg.Ignoring air resistance and effects of the water, find thehorizontal recoil velocity (magnitude and direction) of the canoe.
magnitude m/s
direction ---Select--- opposite the horizontal component of the velocityof the stoneat right angles to the horizontal component ofthe velocity of the stonealong the horizontal component of the velocity ofthe stone

Respuesta :

Answer: -0.33909m/s

Explanation:

Mass of stone(Ms)=5.16kg

Velocity of the stone(Vfs)=8.05m/s

Angle=31.0 degrees

The person and the canoe(Mc)=105kg

The conversation of momentum law applied in the horizontal direction gives:

McVfc+MsVfscos(31.0)=0

Making Vfc subject of the formula :

Vfc=-MsVfscos(31.0)/Mc

=-(5.16kg)(8.00m/s)cos(31.0)/105kg

= -0.33909m/s

The magnitude of the velocity is 0.33909m/s. The minus sign indicates the direction opposite the horizontal component of the velocity of the Stone.

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