Question

A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.
a. find the momentum of the ball of dough at its maximum height.
b. find the momentum of the ball of dough halfway to its maximum height on the way up

Answer 1

Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh => h = v^2/2g h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) => V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s

Answer 2

(a) The momentum of ball at the maximum height is zero.

(b) The momentum of ball at halfway of maximum height is 1.06 kg-m/s.

Given data:

The mass of ball is, m = 0.10 kg.

The initial speed of ball is, u = 15 m/s.

(a)

At maximum height (h) the ball will stop, so final speed of ball is, v = 0 m/s. Then the momentum at the maximum height is,

$p=m \times v\\p=m \times 0\\p=0$

Thus, the momentum of ball at the maximum height is zero.

(b)

Apply the third kinematic equation of motion to obtain the maximum height (h) achieved by the ball as,

$v^{2}=u^{2}+2(-g)h\\0^{2}=15^{2}+2(-9.8) \times h\\19.6h=225\\h=11.47 \;\rm m$

For halfway height,

h' = h/2

h'=11.47/2

h'=5.74 m

Again apply third kinematic equation of motion to obtain the speed (v') at halfway height as,

$v'^{2}=u^{2}+2(-g)h'\\v'^{2}=15^{2}+(2(-9.8)\times 5.74)\\v'=\sqrt{15^{2}-112.50} \\v'=10.60 \;\rm m/s$

Then momentum (p') is calculated as,

$p'=m \times v'\\p'=0.10 \times 10.60\\p'=1.06 \;\rm kg-m/s$

Thus, the momentum of ball at halfway of maximum height is 1.06 kg-m/s.

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