Question IV

IV) A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the struck body? (15 pts). What is the speed of the center-of-mass of the system if the initial speed of the 2.0 kg body was 4.0 m/s? (5 pts).

Solution:
The initial velocity of the second mass is zero. From our equation for the velocity of an object in an elastic collision,
$v_{1f} = (\frac{m_1 - m_2}{m_1 + m_2}) v_{1i} + (\frac{2 m_2}{m_1 +m_2}) v_{2i}$
we can solve for m₂.
$v_{1f} = \frac{1}{4}v_{1i} ~=~ \frac{m_{1} - m_{2}}{m_{1}+m_{2}}$
4(m₁ - m₂) = m₁ + m₂
(8 - 4m₂) = 2 + m₂
or m₂ = 1.2 kg
 
The velocity of the center of mass of the system is given by
$v_{cm} = \frac{m_{1}v_{1i} + m_{2}v_{2i}}{m_{1}+m_{2}}$
$v_{cm} = \frac{(2)(4)}{(2 + 1.2)}$
v_cm = 2.5 m/s