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3 years ago

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A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 k

g and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.

1 answer:

12345 [234]3 years ago

7 0

Answer:

(a) vf = 1.51 m/s

(b) vf = 36.22 m/s

Explanation:

The rate of change of momentum is equal to the force:

$F = \frac{mv_f-mv_i}{t}$

$Ft = m(v_f-v_i)$

where,

F = Force = 1035 N

t = time = 0.175 s

vi = initial speed = 0 m /s

vf = final speed = ?

(a)

m = mass of body = 120 kg

Therefore,

$(1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\$

<u>vf = 1.51 m/s</u>

<u></u>

(b)

m = mass of head = 5 kg

Therefore,

$(1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\$

<u>vf = 36.22 m/s</u>

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