Asked by britt
                A 0.15 kg ball of dough is thrown straight up into the air with an initial velocity of 13 m/s.
(a) Find the momentum of the of dough at its maximum height.
kg·m/s upward
(b) Find the momentum of the ball of dough halfway to its maximum height on the way up.
kg·m/s upward
            
        (a) Find the momentum of the of dough at its maximum height.
kg·m/s upward
(b) Find the momentum of the ball of dough halfway to its maximum height on the way up.
kg·m/s upward
Answers
                    Answered by
            drwls
            
    (a) At its maximum height, the vertical component of velocity is zero. Since there is no horizontal component, the magnitude of velocity (speed) and momentum are also zero.
(b) Half way up, the speed is 1/sqrt2 = 70.7% of the original value. (Conservation of energy should tell you why).
Compute the corresponding momentum,
M*V=(0.15)(0.707)*13 = ?
 
    
(b) Half way up, the speed is 1/sqrt2 = 70.7% of the original value. (Conservation of energy should tell you why).
Compute the corresponding momentum,
M*V=(0.15)(0.707)*13 = ?
                    Answered by
            bobbobobob
            
    1.379
    
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