Question

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 250 g, remains at rest. (a) What is the mass of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is 1.8 m/s?

Answer 1

Answer:

m2  = 83.3 g

Explanation:

by conservation of momentum principle we have

$m_1v_{i1} + m_2v_{i2} = m_2v_{f2}$

as both sphere has same speed so $v_{i2} = v_{i1}$

$m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}$

from conservation of kinetic energy principle we have

$\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2$

$v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}$

$v_{f1} = v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}$

$\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}$

substituting this value in above equation to get m2 value

$m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}$

solving for m2 we  get

$m2 = \frac{m_1}{3}$

m_1 = 250 g

      $=\frac{250}{3}$

  m2  = 83.3 g