Answer:
-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).
