(a) The pitcher must throw the ball at 27.7 m/s
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is the object's velocity
Let's calculate the momentum of the bullet, which has a mass of
m = 2.70 g = 0.0027 kg
and a velocity of
[tex]v=1.50\cdot 10^3 m/s[/tex]
Its momentum is:
[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]
The pitcher must throw the baseball with this same momentum. The mass of the ball is
m = 0.146 kg
So the velocity of the ball must be
[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]
So, the pitcher must throw the ball at 27.7 m/s.
(b) a. The bullet has greater kinetic energy
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass of the object and v is its speed.
For the bullet, we have:
[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]
For the ball:
[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]
So, the bullet has greater kinetic energy.
