Answer to Question #285846 in Mechanics | Relativity for Grace

Question

A body of mass 0.120 kg and a velocity of 0.50 m/s hits another identical body at rest, and the two recoils forming 90 degrees between them. If the final velocity of one of the balls are directed 30 degrees above horizontal, find their speeds after collision.

Answer 1

by momentum conservation:

$m_1v_1=m_1v_1'+m_2v_2$

where m₁, m₂ are masses,

v₁, v₁' are speeds of first body before and after collision

v₂ is speed of second body after collision

then:

for x-direction:

$v_1'cos30\degree +v_2cos60\degree=v_1=0.5$

for y-direction:

$v_1'sin30\degree =v_2sin60\degree$

$v_1'\sqrt 3+v_2=1$

$v_1'=v_2\sqrt 3$

speeds after collision:

$v_2=1/4=0.25$ m/s

$v_1'=0.25\sqrt 3=0.43$ m/s