Question

In a football game, a tackle running at a constant speed of 5 m/s tackles a stationary receiver. The two fly off horizontally at 3 m/s, before they hit the ground and begin to slow down. If the mass of the tackle is 125 kg

What is the speed of the system's center of mass before the collision?

Answer 1

a) Let
m1 = mass of tackle,
m2 = mass of receiver,
u1 = velocity of tackle before collision,
u2 = velocity of receiver before collision,
v1 = velocity of tackle after collision,
v2 = velocity of receiver after collision
P1 = momentum of tackle+receiver before collision,
P2 = momentum of tackle+reciver after collision

m1 = 120 kg
u1 = 5 m/s
u2 = 0
v1 = 4 m/s
v2 = 4 m/s

P1 = m1u1 + m2u2 = 120*5 + 0 = 600 kg-m/s
P2 = m1v1 + m2v2 = 120*4 + m2*4 = 480 + m2*4
By conservation of momentum, P1 = P2
Or 600 = 480 + m2*4
Or m2*4 = 600 - 480
Or m2*4 = 120
Or m2 = 120/4
Or m2 = 30 kg

Ans: 30 kg

b) Kinetic energy before collision = 1/2 * m1 * u1^2 + 1/2 * m2 * u2^2
= 1/2 * 120 * 5^2 + 0
= 1/2 * 120 * 25
= 1500 J

Kinetic energy after collision = 1/2 * m1*v1^2 + 1/2*m2*v2^2
= 1/2*120*4^2 + 1/2*30*4^2
= 60*16 + 15*16
= 960 + 240
= 1200 J

Ans: KEsys, i = 1500 J
KEsys, f = 1200 J