JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 11)
A body of mass m1 moving with an unknown
velocity of $${v_1}\mathop i\limits^ \wedge $$, undergoes a collinear collision
with a body of mass m2 moving with a velocity
$${v_2}\mathop i\limits^ \wedge $$ . After collision, m1 and m2 move with
velocities of $${v_3}\mathop i\limits^ \wedge $$ and $${v_4}\mathop i\limits^ \wedge $$ , respectively.
If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :-
$${v_4} - {{{v_2}} \over 2}$$
$${v_4} - {{{v_2}} \over 4}$$
$${v_4} - {v_2}$$
$${v_4} + {v_2}$$
Explanation
Applying linear momentum conservation
$${m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i$$
m1v1 + 0.5 m1v2 = m1(0.5 v1) + 0.5 m1v4
0.5 m1v1 = 0.5 m1(v4 - v2)
v1 = v4 - v2
$${m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i$$
m1v1 + 0.5 m1v2 = m1(0.5 v1) + 0.5 m1v4
0.5 m1v1 = 0.5 m1(v4 - v2)
v1 = v4 - v2
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