Three guns are aimed at the center of a circle. They are mounted on the circle $120 °$ apart. The fire in a timed sequence such that the three bullets collide simultaneously at the center and Combine to form a stationary lump of three bullets. Two of the bullets have masses of $4.5 g$ each, and are moving with a speed $v_{1} & v_{2} .$ while the third bullet is having a mass of $2.5 g$ and moving with a speed of $575 m s^{- 1}$ . Determine the values of $v_{1} & v_{2} .$
Analyze the situation if all the three bullets are having same mass? Same momentum?

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Solution

Step 1:
Given, $v_{3} = 575 m / s$ , $m_{1} = m_{2} = 4.5 g$ , $m_{3} = 2.5 g$
Step 2:
The total final momentum is zero, and no external force is applied to the system, so the total initial momentum should be also zero
$p_{1} + p_{2} + p_{3} = 0$
Three vectors, which are at an angle of 120𝆩 lead to zero resultant if and only if they have the same magnitude.
so,
$4.5 v_{1} = 2.5575 = 4.5 v_{2}$
After solving we get $v_{1} = v_{2} = 319.4 m s^{- 1} .$
if all the bullets had the same mass or same momentum, they won't form a lump and have zero final momentum.
Hence velocity of the mass $m_{1} a n d m_{2} i s 319.4 m / s$

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