A few days ago
redblitz528

momentum question?

Momentum question?

A 0.20 kg ball is thrown straight up into the air with an initial speed of 11 m/s. Find the momentum of the ball at the following locations.

(a) at its maximum height

______ kg·m/s

(b) halfway to its maximum height

______ kg·m/s

I know that part a is supposed to be 0 but I dont really understand why. The final velocity would be 0, I know for the first part and momentum is velocity times mass and it would equal to 0. But I multiplied final velocity by mass for the 2nd part also but I did not get the right answer so I am confused. This is how I solved the 2nd part:

v = vo + at

0 = (11) + (-9.8)(t)

t= 1.122

So the time it would take to get to maximum height is 1.122 seconds.

delta y = vo t + 1/2 at^2

delta y = (11)(1.122) + .5 (-9.8)(1.122)^2

delta y = 6.1735

I divided the max height by 2 to get the height halfway to max and got 3.086.

v^2 = vo^2 + 2a (delta y)

v^2 = (11)^2 + 2 (-9.8) (3.086)

v^2 = 121 – 60.49

v^2 = 60.6144

v = 7.78 m/s

can someone please help?? Thank you!

Top 2 Answers
A few days ago
norcekri

Favorite Answer

Since the ball is thrown straight up, the maximum height is reached at that instant when the velocity is 0. Since the ball is not moving then, its momentum (1/2mv^2) is 0.

For the second part, I believe you’ve worked out the problem correctly. The velocity halfway up should be sqrt(2)/2 of the initial velocity, which is roughly 7.78. I think that perhaps your problem is using the wrong momentum equation. I believe you’ll find it’s 1/2mv^2, the integral of the one you were using. Is the correct answer about 7.78 times the one you got?

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4 years ago
tamala
Angular momentum is likewise a scalar. even with the shown fact that: angles themselves at the instant are not vectors. you are able to coach that angles a, b and c, in the event that they are orthogonal to a minimum of one yet another, provide diverse reulting angles for a+b+c vs. a+c+b, as an occasion.
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