Q. Three guns are aimed at the centre of a circle. They are mounted on the circle, $120{^\circ}$ apart. They fire in a timed sequence, such that the three bullets collide at the centre and mash into a stationary lump. Two of the bullets have identical masses of 4.5 g and speed of $v_{1\ \ }$and $v_{2}$. The third bullet has a mass of 2.50 g and a speed of ${575\ ms}^{- 1}$ Find the unknown speeds.
A) $200\ {ms}^{- 1}$ each
B) $145\ {ms}^{- 1}\ $and $256\ {ms}^{- 1}$
C) $536\ {ms}^{- 1\ }and\ 320\ {ms}^{- 1}$
D) None of the above
Explanation
(d) Three guns are fired towards the centre of circle as shown in figure.
[image:5_image165.emf]
Since, total final momentum is zero, and no external force is acting on the system, so total initial momentum should be also zero.
So, $\mathbf{p}_{1} + \mathbf{p}_{2} + \mathbf{p}_{3} = 0$
Three vectors, which are at an angle of $120{^\circ}$ leads to zero resultant if and only if they have same magnitude.
So, ${4.5v}_{1} = 2.5 \times 575 = {4.5v}_{2}$
After solving, we will get $v_{1}\ and\ v_{2}$ come out be ${320\ ms}^{- 1}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.