Solution To Problem 0001 | Higher Secondary Physics

Submitted by pradipta pramanik on Mon, 07/04/2011 - 17:48

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:

Image removed.

From conservation of momentum

[tex]m{v_3} + 2\sqrt 2 vm=0[/tex]

[tex]{v_3} = -2\sqrt 2 v[/tex]
 

 

 

 

Hence, Speed of third fragment is [tex]2\sqrt 2 v[/tex]

 

 

 

 

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