Impulse question about boxing

[21385]

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This is from the NOVA book.

When a boxer is hit, he or she is often advised to "ride the punch", that is, to move his or her head backwards during contact with the opponent's fist. Which would be a reasonable explanation for this advice?
I am stuck between two choices. C is the correct one.
a) The impulse received by the head will be less during the collision.
c) Increasing the time of collision will decrease the force of contact.

Let me just say that I understand why C is correct. However, I also think that A should also be correct because instead of the punch just immediately stopping when it hits the head, it goes forward with the head for a while. So, instead of going from mv1 to 0, it is going from mv1 to mv2, where v2 is smaller than v1, both in the same direction. Since the impulse received by the head is equal to the impulse lost by the punch, wouldn't the impulse be smaller and consequently make choice A also correct?

Can someone explain why I am wrong (assuming the book is correct)?

Thanks a lot!

 

[lorenzomicron]

strictly speaking, p=m*deltav
however, the impulse is not technically decreased. speed still goes from v to 0.so while the impulse is the same, it is spread out in time, decreasing the force.

 

[venacontracta]

This is from the NOVA book.

When a boxer is hit, he or she is often advised to "ride the punch", that is, to move his or her head backwards during contact with the opponent's fist. Which would be a reasonable explanation for this advice?
I am stuck between two choices. C is the correct one.
a) The impulse received by the head will be less during the collision.
c) Increasing the time of collision will decrease the force of contact.

Let me just say that I understand why C is correct. However, I also think that A should also be correct because instead of the punch just immediately stopping when it hits the head, it goes forward with the head for a while. So, instead of going from mv1 to 0, it is going from mv1 to mv2, where v2 is smaller than v1, both in the same direction. Since the impulse received by the head is equal to the impulse lost by the punch, wouldn't the impulse be smaller and consequently make choice A also correct?

Can someone explain why I am wrong (assuming the book is correct)?

Thanks a lot!

The fist is moving at v1 the moment before it contacts the head.

The fist is still going to go to v = 0 eventually. It's not like when you throw a punch you immediately pull your arm back if the guy starts to ride the punch. No matter what he does, you still fully extend your arm and you don't start pulling your fist back towards you until AFTER it has reached v = 0. In other words, you don't pull your fist back suddenly just because it is going at some arbitrary speed v2.

So, the fist is still going to go from v = v1 to v = 0 no matter what the guy taking the punch does. Thus, regardless of whether the guy rides the punch or not, the impulse to his head will be exactly the same. All riding the punch does is ensure that it takes longer for the fist to go from v = v1 to v =0, and thus decrease the force delivered to the head.

 

[Rayhaan]

^ Yup.

This is sort of like when you are in a car accident (pray you never are!). If you are with your friend, and your friend decided not to wear his seat belt he gets thrown out of the car and sustains major trauma! Why though? How come nothing happens to you? You both have the same impulse since your starting and final momentums would be the same, but since you wear a seat belt it increase the time of the impact. So if time goes up, force must go down since Impulse = Force x Time. So you get a lesser force. That's why the boxer rides out the punch! He increase the collision time so the force he receives is lower.

 

[21385]

Thanks everyone! Makes perfect sense