question_answer

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 4.50 g each and speeds of\[{{v}_{1}}\]and \[{{v}_{2}}\]. The third bullet has a mass of 2.50 g and a speed of 575 m/s. Find the unknown speeds.

A)  200 m/s each

B)  145 m/s and 256 m/s

C)  536 m/s and 320 m/s

D)  None of the above

Correct Answer: D

Solution :

Three guns are fired towards the centre of circle as shown in figure. Since, total final momentum is zero, and no external force is acting on the system. So total initial momentum should also be zero. So,         \[\overrightarrow{{{P}_{1}}}+\overrightarrow{{{P}_{2}}}+\overrightarrow{{{P}_{3}}}=0\] Three vectors, which are at an angle of \[120{}^\circ \] leads to zero resultant if and only, if they have same magnitude. So,\[4.50{{v}_{1}}=2.5\times 575=4.5{{v}_{2}}\] After solving, we will get \[{{v}_{1}}\]and \[{{v}_{2}}\]be 320 m/s.