Question 173: Centre of Mass and Rotational Motion

Question

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 :-

Accepted Answer

Applying linear momentum conservation
$${m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i$$
m
1
v
1
+ 0.5 m
1
v
2
=  m
1
(0.5 v
1
) + 0.5 m
1
v
4
0.5 m
1
v
1
= 0.5 m
1
(v
4
- v
2
)
v
1
= v
4
- v
2

Answer

$${v_4} - {{{v_2}} \over 2}$$

Answer

$${v_4} - {{{v_2}} \over 4}$$

Answer

$${v_4} - {v_2}$$

Answer

$${v_4} + {v_2}$$

Centre of Mass and Rotational Motion

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