Q:

Dear Sir,
My question is as follows:
# A projectile is fired at an angle of 60 degrees to the horizontal from a point A with a velocity of v and reaches a point B. What is the change in momentum between A and B?
In the above question, the horizontal component of velocity remains the same throughout the flight. Only the vertical component changes. Is it enough to take the vertical component into consideration? In that case the change in momentum is (mv)sin60-(-mvsin60)=2mvsin60. But the answer in my book says it is 2mv. I require your kind assistance to solve this problem because I want make myself so perfect that no question remains unanswered.
Thanking you,

- Amitabh (age 21)

India

- Amitabh (age 21)

India

A:

Amitabh-

I guess here we have to assume that points A and B are both at the same height. Then the short answer is that you're right and the book is wrong. Your reasoning is completely correct.

It would probably be best, since momentum is a vector, to write its change as a vector, e.g. 2mv*sin(60deg) down, or (in standard axes) (0, -2mv*sin(60deg)).

(Even without our assumption, the book is wrong, unless point B is lower than point A by a special amount.)

Mike W.

I guess here we have to assume that points A and B are both at the same height. Then the short answer is that you're right and the book is wrong. Your reasoning is completely correct.

It would probably be best, since momentum is a vector, to write its change as a vector, e.g. 2mv*sin(60deg) down, or (in standard axes) (0, -2mv*sin(60deg)).

(Even without our assumption, the book is wrong, unless point B is lower than point A by a special amount.)

Mike W.

*(published on 10/22/2007)*