Question

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump. The angle between the guns is 120°. Two of the bullets have a mass of 4.30 10-3 kg and are fired with a speed of 257 m/s. The third bullet is fired with a speed of 559 m/s and we wish to determine the mass of this bullet in kg

Answer 1

The arrangement of the problem can be expressed as the figure below

The bullets are fired towards the center of the circle with following velocities

V_{​​​​​​1} =V_{​​​​​​2} =257m/s

V_{​​​​​​3} =559m/s

The masses are,

m_{​​​​​​1} =m_{​​​​​​2} =4.30×10^-3 kg

to find the mass of the third bullet(m_{​​​​​​3} ) we apply the conservation of linear momentum.

But for that,we have to resolve the momentum in x and y direction separately. The overall momentum after the collision is zero. So the sum of momenta in x and y direction must be separately zero.

So from the figure above,we can write the equation for conservation of momentum in x direction as

.

So the mass of the third bullet is 1.97×10^-3 Kg.

I would like to point out that even if we simply divide the momentum of second bullet bu the velocity of the third bullet, without resolving into the components at all,we still get the right answer. This happens because of the unique symmetry of the problem where momenta make an angle of 120° with each other.

I took the longer approach to explain the composition of the solution in detail. I hope this helps you in understanding the concept better