Question

A 130 kg tackler moving at 2.5 m/s meets head-on (and tackles) an 82 kg halfbackmoving at 5.0 m/s. What will be their mutual speed immediately after the collision?Choose the direction of halfback to be the positive direction and direction of tackler tobe negative direction.

Answer 1

mass 1 = m1 = 130kg

velocity 1 = v1= -2.5 m/s

mass 2 = m2= 82 kg

velocity 2 = v2= 5 m/s

Conservation of momentum:

pi= pf

pi = initial momentum = m1v1 + m2 v2

pf = final momentum = mf Vf

Where:

mf= final mass = m1+m2= 130 + 82 = 212 kg

vf= final velocity (mutual speed after the collision)

Replacing:

pi=pf

m1v1 + m2v2= mfvf

(130 kg) * (-2.5 m/s )+(82kg) * (5 m/s) = 212 kg Vf

-325 kgm/s + (410 kgm/s) = 212 kg Vf

85 kgm/s = 212 kg * Vf

(85 kgm/s ) / 212 kg = Vf

Vf = 0.4 m/s