Question

A 0.12 kg ball of dough is thrown straight up
into the air with an initial speed of 11 m/s.
The acceleration of gravity is 9.8 m/s2
What is its momentum halfway to its maxi-
mum height?
Answer in units of kgm/s

Answer 1

Not certain if correct, but hope it provides something.

Answer 2

First, find the half of the maximum that the ball will reach.
Use the formula \[\frac{ vf -vi }{ a }=t\] to find the time.\[vf=0m/s, vi=11m/s, a=-9.8m/s ^{2}\]
\[\frac{ 0-11 }{-9.8 }=\frac{ 11 }{ 9.8}s\]
Now use this formula \[vt+\frac{ 1 }{ 2 }a ^{2}=s\]
then get \[11\times \frac{ 11 }{ 9.8 }-\frac{ 1 }{ 2 }\times9.8\times \left( \frac{ 11 }{ 9.8 } \right)^{2}=\frac{ 1 }{ 2 }\times \frac{ 11^{2} }{ 9.8}\]
And \[\frac{ 1 }{ 2 }s =\frac{ 1 }{ 4 }\times \frac{ 11^{2} }{ 9.8 }\]
Second, find the time that the ball spend to reach the half of the the maximum it will reach.\[vf ^{2}-vi ^{2}=2gs\]
\[vf ^{2}=2\times \left( -9.8 \right)\times \frac{ 1 }{ 4 }\times \frac{ 11^{2} }{ 9.8 }+11^{2}\]
\[vf ^{2}=\frac{ 1 }{ 2 }\times11^{2}\]
\[vf =11\sqrt{\frac{ 1 }{ 2 }}m/s\]
As p(momentum)=mv,
the momentum of the ball is \[0.12\times11\sqrt{\frac{ 1 }{ 2 }}=1.32\sqrt{\frac{ 1 }{ 2 }}kg \times m/s\]

Answer 3

You should be able to skip solving for any sort of time just by using the 4th equation of motion. Use
V = final velocity = 0 m/s
Vi = 11 m/s
h = max height

\[ v^2 = v_i^2 + 2ah \\ -v_i^2 = -2gh \\ h = \frac{v_i^2}{2g} \]

Then just use the same equation with a different final velocity and new height; Vi is the same, V is the unknown, and h/2 is half the max height by definition.
\[ v_{(h/2)}^2 = v_i^2 + 2a{\frac{h}{2}} \\ v_{(h/2)}^2 = v_i^2 - g \left( \frac{v_i^2}{2g} \right) \\ v_{(h/2)}^2 = \frac{v_i^2}{2} \\
v_{(h/2)} = v_i \frac{\sqrt2}{2} \]
then momentum
\[ p_{(h/2)}=mv_{(h/2)} \\ p_{(h/2)} = mv_i \frac{\sqrt2}{2} \]

Gets the same answer without the pesky time :)