Q 1 :    

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2:2:1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is            [2022]

  • v

     

  • 2v

     

  • 22v

     

  • 32v

     

(3)

Given, mass of the shell = m

Ratio of masses of the fragments is 2 : 2 : 1

Therefore, masses of the three fragments are

m1=m2, m2=m2 and m3=m4

Now fragments with equal masses i.e., m1 and m2 fly off perpendicularly with speeds v1=v2=v. Let the velocity of third fragment be v'.

Applying law of conservation of momentum, 

m1v1i^+m2v2j^+m3v=0

mv2i^+mv2j^+m4v'=0v'=2v(-i^-j^)

|v|=2v(-1)2+(-1)2=22v



Q 2 :    

An object flying in air with velocity (20i^+25j^-12k^) suddenly breaks into two pieces whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity (100i^+35j^+8k^).  The velocity of the larger piece will be:               [2019]

  • 4i^+23j^-16k^

     

  • -100i^-35j^-8k^

     

  • 20i^+15j^-80k^

     

  • -20i^-15j^-80k^

     

(1)

From the law of conservation of linear momentum mv=m1v1+m2v2

6k(20i^+25j^-12k^)=k(100i^+35j^+8k^)+5kv2

5v2=(120-100)i^+(150-35)j^+(-72-8)k^

5v2=20i^+115j^-80k^  v2=4i^+23j^-16k^