Question

how fast must a 2.7 - g ping - pong ball move in order to have the same kinetic energy as a 145 g baseball moving at 38.0 m/s?
physical constants

Explanation:

Step1: Write kinetic - energy formula

The kinetic - energy formula is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the velocity. Let $m_1 = 145\ g=0.145\ kg$, $v_1 = 38.0\ m/s$, $m_2 = 2.7\ g = 0.0027\ kg$, and $v_2$ be the velocity of the ping - pong ball we want to find. Since $K_1 = K_2$, we have $\frac{1}{2}m_1v_1^{2}=\frac{1}{2}m_2v_2^{2}$.

Step2: Solve for $v_2$

Canceling out the $\frac{1}{2}$ on both sides of the equation $\frac{1}{2}m_1v_1^{2}=\frac{1}{2}m_2v_2^{2}$, we get $m_1v_1^{2}=m_2v_2^{2}$. Then $v_2^{2}=\frac{m_1v_1^{2}}{m_2}$. Substitute $m_1 = 0.145\ kg$, $v_1 = 38.0\ m/s$, and $m_2 = 0.0027\ kg$ into the equation:

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Take the square - root of both sides: $v_2=\sqrt{77548.15}\approx278\ m/s$.

Answer:

$278$