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A professional boxer hits his opponent with a 1250-N horizontal blow that lasts for 0.16 s. a) what is the opponents final speed, in meters per second, if his mass is 102.5 kg and he was at rest when struck near his center of mass? B) now assume that just the opponent’s head, with a mass of 4.64 kg, was hit in this manner. Calculate the final speed of his head, in meters per second, assuming the head does not initially transfer significant momentum to the boxer’s body.

Answer :

We know that the impulse if related to the momentum as:

[tex]I=\Delta p[/tex]

and related to the average force and the time of the collision by:

[tex]I=Ft[/tex]

Which means that:

[tex]Ft=\Delta p[/tex]

a)

In this case the force is 1250 N, the time is 0.16 s, the initial velocity is 0 and the mass is 102.5 kg, then we have:

[tex]\begin{gathered} (1250)(0.16)=(102.5)(v_f-0) \\ v_f=\frac{(1250)(0.16)}{102.5} \\ v_f=1.951219512 \end{gathered}[/tex]

Therefore the velocity is 1.951219512 m/s.

b)

In this case the force is 1250 N, the time is 0.16 s, the initial velocity is 0 and the mass is 4.64 kg, then we have:

[tex]\begin{gathered} (1250)(0.16)=(102.5)(v_f-0) \\ v_f=\frac{(1250)(0.16)}{4.64} \\ v_f=43.10344828 \end{gathered}[/tex]

Therefore the velocity is 43.10344828 m/s.

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