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 3.90 x 10⁻³ kg and are fired with a speed of 368 m/s. The third bullet is fired with a speed of 618 m/s and we wish to determine the mass of this bullet.

Answer 1

Answer:
`2.32* 10^(-3) kg`

Step-by-step explanation:

Given

mass of first and second bullet
`m_1=m_2=3.90* 10^(-3) kg`

Velocity of two bullets
`v_1=v_2=368 m/s`

velocity of third bullet
`v_3=618 m/s`

angles between guns is
`120^(\circ)`

Suppose First gun is at
`0^(\circ)` and second is at
`120^(\circ)` and third is at
`240^(\circ)`

therefore

conserving momentum in x-direction

`m_1v_1\cos 0+m_2v_2\cos 120+m_3v_3\cos 240=0`

as three bullets club together to become lump

`3.90* 10^(-3)* 368+3.90* 10^(-3)* 368* \cos (120)+m_3* 618* \cos (240)=0`

`3.90* 10^(-3)* 368+3.90* 10^(-3)* 368* (-0.5)+m_3* 618* (-0.5)=0`

`0.5* 3.90* 10^(-3)* 368=m_3* 618* 0.5`

`m_3=3.90* 10^(-3)* (368)/(618) kg`

`m_3=2.32* 10^(-3) kg`