Respuesta :
Answer:
Explanation:
It is the case of perfectly elastic collision . So we shall apply formula of velocity after collision as follows .
Let m₁ and m₂ be the mass colliding with velocity u₁ and u₂ and their velocity become v₁ and v₂ after collision .
[tex]v_1=\frac{(m_1-m_2)u_1 }{m_1+m_2)} +\frac{2 m_2u_2}{(m_1+m_2)}[/tex]
Putting the values
[tex]v_1=\frac{ (.30-.80).40 }{( .30+.80)} +\frac{2\times .80\times(-.15) }{(.30+.80 )}[/tex]
= - 0.4 m /s
So direction of .30 kg mass will be reversed .
[tex]v_2=\frac{ ( m_2-m_1) u_2 }{( m_1+m_2)} +\frac{2 m_1u_1}{(m_1+m_2)}[/tex]
putting the values
[tex]v_2=\frac{ ( .80-.30)(-.15) }{( .30+.80)} +\frac{2 \times.30\times.40}{(.30+.80)}[/tex]
= .15 m /s
The direction of .80 kg will become from left to right ie its direction will be reversed .
b ) Minimum amount of kinetic energy will be at the position when they move with common velocity
common velocity
v = .3 x .4 - .8 x .15 / (.3 + .8)
= 0
c )
Missing energy is stored as elastic potential energy in the spring .
