Question

On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8m/s north and the halfback is sliding with a velocity of 7.2m/s east.

A. What is the magnitude of the velocity at which the two players move together immediately after the collision?

B.What is the direction of this velocity ?

Answer 1

A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s

B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum

Step-by-step explanation:

This problem bothers on the inelastic collision

Given data

Mass of linebacker m1= 110kg

Mass of halfbacker m2= 85kg

Velocity of linebacker v1= 8.8m/s

Velocity of halfbacker v2= 7.2m/s

Applying the principle of conservation of momentum for inelastic collision we have

m1v1 +m2v2= (m1+m2)v

Where v is the common velocity after impact

Substituting our data into the expression we have

110*8.5+85*7.2= (110+85)v

935+612=195v

1547=195v

v=1547/195

v=7.9m/s

Momentum of linebacker after impact = 110*7.9= 869Ns

Momentum of halfbacker after impact = 85*7.9= 671.5Ns

the direction after impact is due north since the linebacker has greater momentum