A stationary body of mass 3kg explodes into three equal places. Two of the places fly off in two mutually perpendicular directions, one with a velocity of `3 hat i ms^-1` and the other with a velcotiy of `4 hat j ms^-1` , If the explosion occurs in `10^-4 s`, the average force acting on the third place in newton is

Options

A. `(3 hat i+4 hat j) times 10^-4`
B. `(3 hat i-4 hat j) times 10^-4`
C. `(3hat i+4 hat j) times 10^4`
D. `-(3 hat i+4 hat j) times 10^4`

Answer

D

Explanation

By law of conservation of linear momentum
`m_v vecv_1+m_2 vec v_2+m_3 vec v_3=0`
Here : `m_1=m_2=m_3=1kg`
`vecv_1=3 hati, vec v_2=4 hat j`
`therefore 3 hat i+4 hatj + vec v_3=0`
The average force acting on the third piece is
`F=(mvec v_3)/t`
`=(1 times (3 hat i+4 hatj))/10^-4 N`
`=-(3 hat i+4 hat j) times 10^-4 N`