During a goal-line stand, a 112-kg fullback moving eastward with a speed of 6 m/s collides head-on with a 120-kg lineman moving westward with a speed of 8 m/s.
The two players collide and stick together, moving at the same velocity after the
collision. Determine the post-collision velocity of the two players.

-1.24 m/s

Explanation:

Total momentum before collision = total momentum after collision

Total momentum before collision = (mass of full back * velocity of fullback) + (mass of lineman * velocity of line man).

Mass of full back = 112 kg, mass of line bag = 120 kg, velocity of full back  6 m/s (east), velocity of line back = -8 m/s (west). Hence:

Total momentum before collision = (112 * 6) + (120 * -8) = 672 - 960 = -288 kgm/s

The total momentum after collision = (mass of full back + mass of line back) * velocity after collision.

Let velocity after collision be v, hence:

The total momentum after collision = (112 + 120)v = 232v

Total momentum before collision = total momentum after collision

-288 = 232v

v = -288 / 232

v = -1.24 m/s

Therefore after collision, the two players would move at a velocity 1.24 m/s west (the same direction as the lineman).

i think its c if i'm wrong

answer///terminator or twilight zone;