Question

A stationary bomb explodes into three pieces. One piece of 2 Kg mass moves with a velocity of $8m{s}^{-1}$ at right angles to the other pieces of mass 1 kg moving with a velocity of $12m{s}^{-1}$. If the mass of the third piece is 0.5 kg, then its velocity is

• A. $10m{s}^{-1}$
• B. $20m{s}^{-1}$
• C. $30m{s}^{-1}$
• D. $40m{s}^{-1}$

Answer 1

The correct option is D $40m{s}^{-1}$

Stationary so initial $momentum$ will be $zero.$

momentum of $2kg$ is $P_1=2\times 8=16kgm/s$

that if $1kg$ is $P_2=1\times 12=12kgm/s$

both being at $right$ angle so their resultant will be $P=\sqrt{P_1^2+P_2^2}=\sqrt{256+144}=20kgm/s$

the third should have $same$ momentum but in $opposite$ direction to make the $net$ momentum $zero.$

Now its velocity should be $V=\frac{P}{m}=\frac{20}{0.5}=40m/s$