A bomb of 12 kg divides in two parts whose ratio of masses is 1 : 3. If kinetic energy of smaller part is 216 J , then momentum of bigger part in kg m / sec will be

Options

A. 36
B. 72
C. 108
D. Data is incomplete

Answer

A

Explanation

The bomb of mass 12kg divides into two masses
m and m then `m_(1)+m_(2)=12 " "` ….(i)
and `(m_(1))/(m_(2))=(1)/(3)`
by solving we get `m_(1)=3kg` and `m_(2)=9kg`
Kinetic energy of smaller part `= (1)/(2)m_(1)v_(1)^(2)=216 J`
`therefore v_(1)^(2)=(216xx2)/(3)rArr v_(1)=12m//s`
So its momentum `= m_(1)v_(1)=3xx12=36` kg.m/s
As both parts possess same momentum therefore momentum of each part is 36 kg-m/s