Question

Two balls collide head-on on a table. The first ball has a mass of 0.50 kg and an initial velocity of 12.0 m/s. The second ball has a mass of 0.75 kg and an initial velocity of -16.0 m/s. After the collision, the first ball travels at a velocity of -21.6 m/s. What is the velocity of the second ball after the collision? Assume a perfectly elastic collision and no friction between the balls and the table.

Answer 1

\[\frac{1}{2}m_av_a^2+\frac{1}{2}m_bv_b^2=\frac{1}{2}m_av`_a^2+\frac{1}{2}m_bv`_b^2\] Now solve for v of the second ball. \[\frac{1}{2}(.5kg)(12\frac{m}{s})^2+\frac{1}{2}(.75kg)(-16\frac{m}{s})^2=\frac{1}{2}(.5kg)(-21\frac{m}{s})^2+\frac{1}{2}(.75kg)v`_b^2\]\[\frac{2(\frac{1}{2}(.5kg)(12\frac{m}{s})^2+\frac{1}{2}(.75kg)(-16\frac{m}{s})^2-\frac{1}{2}(.5kg)(-21\frac{m}{s})^2)}{.75kg}=v`_b^2\]\[\sqrt{\frac{2(\frac{1}{2}(.5kg)(12\frac{m}{s})^2+\frac{1}{2}(.75kg)(-16\frac{m}{s})^2-\frac{1}{2}(.5kg)(-21\frac{m}{s})^2)}{.75kg}}=v`_b\]\[\sqrt{\frac{(.5kg)(12\frac{m}{s})^2+(.75kg)(-16\frac{m}{s})^2-((.5kg)(-21\frac{m}{s})^2)}{.75kg}}=v`_b\]\[\sqrt{\frac{72\frac{kg*m^2}{s^2}-192\frac{kg*m^2}{s^2}+220.5\frac{kg*m^2}{s^2}}{.75kg}}=v`_b\]\[\sqrt{72\frac{m^2}{s^2}-192\frac{m^2}{s^2}+220.5\frac{m^2}{s^2}}=v`_b \rightarrow \sqrt{100.5\frac{m^2}{s^2}}=v`_b\]\[v`_b=10.03\frac{m}{s}\]