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:$2.32\times 10^{-3} kg$

Explanation:

Given

mass of first and second bullet $m_1=m_2=3.90\times 10^{-3} kg$

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

$3.90\times 10^{-3}\times 368+3.90\times 10^{-3}\times 368\times \cos (120)+m_3\times 618\times \cos (240)=0$

$3.90\times 10^{-3}\times 368+3.90\times 10^{-3}\times 368\times (-0.5)+m_3\times 618\times (-0.5)=0$

$0.5\times 3.90\times 10^{-3}\times 368=m_3\times 618\times 0.5$

$m_3=3.90\times 10^{-3}\times \frac{368}{618} kg$

$m_3=2.32\times 10^{-3} kg$