Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

[tex]\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}[/tex]

The momentum (P₂) after a collision will be

[tex]\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}[/tex]

Applying momentum conservation, we have

[tex]\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \\\\[/tex] ...1

[tex]\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta[/tex] ...2

From equations 1 and 2, we have

[tex]\rm \theta = tan \ ^{-1} \dfrac{ (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2)}{ (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2)}\\\\\\\theta = tan \ ^{-1} \dfrac{5.12*239*cos21.3+3.05*282*cos15.4}{5.12*239*sin21.3-3.05*282*sin15.4}\\\\\\\theta = 62.7^o[/tex]

From equation 1, we have

[tex]\rm u = \dfrac{(m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) }{ (m_1 + m_2) \ cos\ \theta } \\\\\\u = \dfrac{5.12*239*cos21.3 - 3.05*282*cos15.4}{(5.12+3.05)cos62.2}\\\\\\u = 83.0325 m/s[/tex]

Then the change in kinetic energy, we have

[tex]\rm \Delta KE = \dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2-\dfrac{1}{2}(m_1+m_2)u^2\\\\\\\Delta KE = \dfrac{1}{2} * 5.12 * 239^2 + \dfrac{1}{2}*3.05*282^2 - \dfrac{1}{2}(5.12+3.05)*83.032^2\\\\\\\Delta KE = 239.34 \ J[/tex]

The fraction of kinetic energy lost will be

[tex]\rm Energy \ lost = \dfrac{239.34}{267.5} = 0.895[/tex]

More about the collision link is given below.

https://brainly.com/question/13876829