posted 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.
(b) Find the momentum of the ball of dough halfway to its maximum height on the way up.
(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 = ?