We can find the momentum of the rock by using De Broglie's relationship:
[tex]p= \frac{h}{\lambda} [/tex]
where
p is the momentum
h is the Planck constant
[tex]\lambda[/tex] is the De Broglie's wavelength
By using [tex]\lambda=3.32 \cdot 10^{-34} m[/tex], we find
[tex]p= \frac{6.6 \cdot 10^{-34} Js}{3.32 \cdot 10^{-34} m}=1.99 kg m/s [/tex]
The momentum of the rock is
[tex]p=mv[/tex]
where [tex]m=50 g=0.05 kg[/tex] is the mass and v is its velocity. Rearranging the equation, we find the speed of the rock:
[tex]v= \frac{p}{m}= \frac{1.99 kg m/s}{0.05 kg}=39.8 m/s [/tex]
