Question

A professional boxer hits his opponent with a 1345-N horizontal blow that lasts for 0.122 s.

a) Calculate the impulse (in kg · m/s) imparted by this blow.

b) What is the opponent's final velocity (in m/s), if his mass is 113 kg and he is motionless in midair when struck near his center of mass?

c) Calculate the recoil velocity (in m/s) of the opponent's 11.4 kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.

Answer 1

Part A.

We know that relation between Average force and change in momentum is given by:

F_avg = dP/dt

Also Impulse = change in momentum,

J = dP = m*dV = m*(Vf -Vi)

J = F_avg*dt

F_avg = Average force applied = 1345 N

dt = impact time = 0.122 sec

So,

J = Impulse = 1345*0.122

J = 164.1 kg.m/sec

Part B.

Since J = dP = m*(Vf - Vi)

Vi = Initial velocity of opponent = 0 m/s

Vf = final velocity = ?

m = mass of opponent = 113 kg

So,

Vf = J/m + Vi

Vf = 164.1/113 + 0

Vf = 1.45 m/s

Part C.

Using momentum conservation before and after blow

Pi = Pf

Pi = 0, Since initial opponent was stationary

So, Pf = 0

Pf = final momentum of opponent's center of mass + momentum of opponent's head

Pf = 0

m*Vf + m_head*V_recoil = 0

V_recoil = recoil speed of head = m*Vf/m_head

m = mass of the body = 113 kg

m_head = mass of the head = 11.4 kg

So,

V_recoil = 113*1.45/11.4

V_recoil = 14.4 m/s

[It can simply be found if we hit blow on the head, then rest of the body remain stationary, So in that case, Vf = J/m_head + Vi

Vf = 164.1/11.4 = 14.4 m/s]

Let me know if you've any query.