Question

A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The energy released in the process of explosion is

• A. 3/2 mv²
• B. 3 mv²
• C. 2 mv²
• D. 1/2 mv²

Answer 1

Answer: (A)

3/2 mv²

Answer 2

Here momentum of third fragment is

p³ = $\sqrt { \mathrm { p } _ { 1 } ^ { 2 } + \mathrm { p } _ { 2 } ^ { 2 } }$

or p³ = $\sqrt { ( \mathrm { mv } ) ^ { 2 } + ( \mathrm { mv } ) ^ { 2 } }$

= $\sqrt { 2 } \mathrm { mv }$

Final K.E. of the system

= $\frac { \mathrm { p } _ { 1 } ^ { 2 } } { 2 \mathrm {~m} } + \frac { \mathrm { p } _ { 2 } ^ { 2 } } { 2 \mathrm {~m} } + \frac { \mathrm { p } _ { 3 } ^ { 2 } } { 2 ( 2 \mathrm {~m} ) }$

= $\frac { 3 } { 2 }$mv² + $\frac { 3 } { 2 }$mv² + $\frac { 3 } { 2 }$ mv

= $\frac { 3 } { 2 }$mv²

Since initial K.E. = 0 therefore energy released = $\frac { 3 } { 2 }$mv².