All Solutions
Page 245: Section Review
a)
We can use conservation of energy:
$$
mglsinphi=frac{mv^2}{2}rightarrow v=sqrt{2glsinphi}=sqrt{2cdot9.81cdot1cdotsin30^o}
$$
$$
boxed{v=3.13,,rm m/s}
$$
b)
We can use conservation of momentum:
$$
begin{align*}
m_1v&=(m_1+m_2)v’\
frac{F_{g1}}{g}v&=(frac{F_{g1}}{g}+frac{F_{g2}}{g})v’\
F_{g1}v&=(F_{g1}+F_{g2})v’\
v’&=frac{F_{g1}v}{F_{g1}+F_{g2}}\
&=frac{24.5cdot3.13}{24.5+36.8}
end{align*}
$$
$$
boxed{v’=1.25,,rm m/s}
$$
b) $v’=1.25,,rm m/s$
When the athlete is currently at the top of the pole, over the crossbar, the point of the contact of the pole to the ground reacts has opposite force that is equal. The force that reacts **from the ground to the pole** has a vertical component and that component of force changes vertical momentum over some time.
In the second case in the system we have ball earth an you, now to have conserved momentum we do not have to move because standing on the ground we transfer momentum to the earth.
