Answer :
(a) The mass of the bullet is [tex]m_b = 2.00g=0.002 kg[/tex] and its speed is [tex]v_b = 1.5 \cdot 10^3 m/s[/tex], so its momentum is
[tex]p=m_b v_b = (0.002 kg)(1.5 \cdot 10^3 m/s)=3 kg m/s[/tex]
The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
[tex]v= \frac{p}{m}= \frac{3 kg m/s}{0.148 kg}=20.3 m/s [/tex]
(b) The kinetic energy of the bullet is:
[tex]K_b = \frac{1}{2} m_b v_b^2= \frac{1}{2}(0.002 kg)(1.5 \cdot 10^3 m/s)^2=2250 J [/tex]
while the kinetic energy of the ball is:
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(0.148 kg)(20.3 m/s)^2=30.5 J [/tex]
So, the bullet has greater kinetic energy than the ball.
[tex]p=m_b v_b = (0.002 kg)(1.5 \cdot 10^3 m/s)=3 kg m/s[/tex]
The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
[tex]v= \frac{p}{m}= \frac{3 kg m/s}{0.148 kg}=20.3 m/s [/tex]
(b) The kinetic energy of the bullet is:
[tex]K_b = \frac{1}{2} m_b v_b^2= \frac{1}{2}(0.002 kg)(1.5 \cdot 10^3 m/s)^2=2250 J [/tex]
while the kinetic energy of the ball is:
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(0.148 kg)(20.3 m/s)^2=30.5 J [/tex]
So, the bullet has greater kinetic energy than the ball.