Question

On a very muddy football field, a 109-kg linebacker tackles an 82-kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 9.0 m/s north and the halfback is sliding with a velocity of 6.3 m/s east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

magnitude ------m/s

direction --------° north of east

Answer 1

p1 = m1v1 = 109*9.0 = 981
p2 = m2v2 = 82*6.3 = 516.6

  P = combined momentum of the two players together = ?(981² + 516.6²) = 1108.709 kg-m/s
Combined mass of the 2 players = M1 + M2 = M = 191 kg
V = velocity of the combination = 1108.709 / 191 = 5.804 m/s - Correct

Let the direction of P make an angle ? with the N direction (i.e. E of N)
P Cos? = 109*9 kg-m/s = 981 kg-m/s ; P sin? = 82*6.3 kg-m/s = 516.6 kg-m/s
tan? = 516.6/ 981 = 0.526

=> ? = tan^-1(0.526) = 27.74 deg