Three particles P, Q and R of masses $0.1$ , $0.2$ and $0.5$ respectively lie on a smooth horizontal table , P is projected towards Q with speed $5$ ,after the collision P rebounds with speed $1$ , then Q collides with R , after the collision R moves with speed $v$ .
Given that there is no subsequent collision between P and Q , find the greatest value of $v$
My turn
For the collision between P and Q
$0.1 \times 5 + 0 = 0.1\times -1 + 0.2 v_{Q}$
$v_{Q} = 3$
For the collision between Q and R
$3 \times 0.2 + 0 = 0.2 \times v_{1} + 0.5 \times v$
Since there is no subsequent collision between P and Q , then $v_{1}$ (the speed of Q after is collides R) $-1\geq v_{1} $
Then we get
$v \geq 1.6$ which means the least value of v is 1.6 not the greatest
What is the mistake ? Thanks in advance
Actually the condition for no further collision is $$v_1\geq-1$$, so your inequality is reversed and you will arrive at the correct answer.
If you're still not sure about this, consider what would happen if, as you said, $v_1$ takes a value less than $-1$ such as $-2$...?