(a) [tex]9.1 \cdot 10^{-21} kg m/s[/tex]
The magnitude of the linear momentum of an object is given by
[tex]p=mv[/tex]
where
m is the object's mass
v is its speed
In this case, we have
[tex]m=1.67\cdot 10^{-27} kg[/tex] (mass of the proton)
[tex]v=5.45\cdot 10^6 m/s[/tex] (speed of the proton)
So, the momentum is
[tex]p=(1.67\cdot 10^{-27} kg)(5.45\cdot 10^6 m/s)=9.1 \cdot 10^{-21} kg m/s[/tex]
b) 7.0 kg m/s
In this case, we have
m = 16.0 g = 0.016 kg (mass of the bullet)
v = 435 m/s (speed of the bullet)
By applying the same formula, the linear momentum is
[tex]p=(0.016 kg)(435 m/s)=7.0 kg m/s[/tex]
c) 797.5 kg m/s
In this case, we have
m = 72.5 kg (mass of the sprinter)
v = 11.0 m/s (speed of the sprinter)
By applying the same formula, the linear momentum is
[tex]p=(72.5 kg)(11.0 m/s)=797.5 kg m/s[/tex]
d) [tex]1.8\cdot 10^{29} kg m/s[/tex]
In this case, we have
[tex]5.98\cdot 10^{24} kg[/tex] (mass of the Earth)
[tex]v=2.98\cdot 10^4 m/s[/tex] (speed of the Earth)
By applying the same formula, the linear momentum is
[tex]p=(5.98\cdot 10^{24} kg)(2.98\cdot 10^4 m/s)=1.8\cdot 10^{29} kg m/s[/tex]
