Solution To Problem 0001 | Higher Secondary Physics

Problem 0001

A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular direction with speed v. What willl be the speed of the third (lighter) fragment?

Answer:

From conservation of momentum

$m{v_3} + 2\sqrt 2 vm=0$

${v_3} = -2\sqrt 2 v$
 

Hence, Speed of third fragment is $2\sqrt 2 v$