Two particles 'P' and 'Q', of masses 0.3kg and 0.2kg respectively, which are free to move on a smooth horizontal table. A vertical wall is situated at the right end of the table (right of P and Q). the particles are moving directly towards the wall in the same straight lane, 'P' with speed 3.2ms-1 and 'Q' with speed 1.2ms-1,
'Q' hits the wall and rebounds from it with speed 2ms-1. the particles subsequently collide, after which they move in opposite directions, 'P' with speed 'v'ms-1 and 'Q' with speed '4v'ms-1
Find the value of 'v' ?
2007-02-16 02:45:39 · 3 answers · asked by Leonidus 2 in Science & Mathematics ➔ Physics
the law of conservation of momentum applies here. This means that all the momentum before any collision is equal to the total momentum after the collision if you assume elasticity. After Q hits the wall, the momentum is (0.3*3.2)+(0.2*-2). The momentum after this collision is therefore (0.3*-V)+(0.2+4V). If you assume that the collision between the particles is elastic, the conservation of momentum gives
(0.3*3.2)+(0.2*-2)=(0.3*-V)+(0.2+4V).
you can rearrange and so on from there
2007-02-16 03:02:06 · answer #1 · answered by thesilvernewt 2 · 0⤊ 1⤋
Firstly draw 2 diagrams of P & Q , one before and one after collision, label the values to see what is missing.
Use your formula and DO YOUR OWN DAMN HOMEWORK!
2007-02-16 10:57:58 · answer #2 · answered by Jimbobarino 4 · 0⤊ 2⤋
This site will help you in gaining your momentum
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2007-02-16 10:57:36 · answer #3 · answered by Anonymous · 0⤊ 1⤋