Question

A body of mass 2kg makes an elastic collision with another body at rest and continuous to make in the original direction with one fourth of its original speed find the mass of ii body

Answer 1

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  

$v_{cm} = \frac{m_{1}v_{1i} + m_{2}v_{2i}}{m_{1}+m_{2}}$  

$v_{cm} = \frac{(2)(4)}{(2 + 1.2)}$  

vcm = 2.5 m/s  

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  

$v_{cm} = \frac{m_{1}v_{1i} + m_{2}v_{2i}}{m_{1}+m_{2}}$  

$v_{cm} = \frac{(2)(4)}{(2 + 1.2)}$  

vcm = 2.5 m/s