Question

A bomb at rest explored into three fragments of equal mass . two fragments fly off at right angle to each other with velocities 9 ms^-1 and 12 ms^-1 respectively calculate the speed of third fragment​

Answer 1

Explanation:

momentum of bomb before collision=0

after collision it split into three fragments of equal masses

using law of conservation of linear momentum

0=9m+12m+vm

0=9+12+v

since the other two fragments are at right angles

v=-√9²+12²

v=-15m/s

The minus sign indicates third fragment moved in opposite direction