Question

A pitcher claims he can throw a 0.148-kg baseball with as much momentum as a 2.00-g bullet moving with a speed of 1.50 ✕ 103 m/s. (a) what must the baseball's speed be if the pitcher's claim is valid? m/s (b) which has greater kinetic energy, the ball or the bullet? the bullet has greater kinetic energy. the ball has greater kinetic energy. both have the same kinetic energy

Answer 1

(a) The mass of the bullet is $m_b = 2.00g=0.002 kg$ and its speed is $v_b = 1.5 \cdot 10^3 m/s$, so its momentum is
$p=m_b v_b = (0.002 kg)(1.5 \cdot 10^3 m/s)=3 kg m/s$

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:
$v= \frac{p}{m}= \frac{3 kg m/s}{0.148 kg}=20.3 m/s$

(b) The kinetic energy of the bullet is:
$K_b = \frac{1}{2} m_b v_b^2= \frac{1}{2}(0.002 kg)(1.5 \cdot 10^3 m/s)^2=2250 J$

while the kinetic energy of the ball is:
$K= \frac{1}{2}mv^2= \frac{1}{2}(0.148 kg)(20.3 m/s)^2=30.5 J$

So, the bullet has greater kinetic energy than the ball.