Could I see the solution to #6? I am stumped on part c!

Jillian, I just sent it to your email.  John

we keep getting 4sqrt(gR) instead of 8sqrt(gR) for 5a. we tried using energy and momentum and keep getting 4sqrt(gR) any tips or suggestions?

Ben, I assumed that for the block/bullet system that the velocity at the top of the circle would be zero - then it would fall down at the top. so using cons of energy: 1/2 (4m) v bottom squared = 4mg times 2R as the height of the circle is 2R, not R. I get v bottom = sqrt (4gR)

Then use cons of momentum where m v zero = 4m times v bottom. and v zero would be 8 sqrt (gR).

What do you think?

John

Hello John,

Thank you for taking the time to answer, we agree with your reasoning. an energy to momentum problem was really fun.

we are now looking at part b. it seems its a force to energy to momentum problem (also fun). we got v=sqrt(gr) for the speed needed at the top of the curve and v=2sqrt(gr) needed at the bottom of the curve to reach the top. So sqrt(gr) + 2sqrt(gr) = 3sqrt(gr)

plugging that into the momentum equation got v = 12sqrt(gr). please advise, thank you.

Ben, these are fun problems!  You are correct about the v at the top of the curve. And that comes from assuming N = 0. That means mg is enough to keep it moving in a circle without the normal force from the track.

Then use conservation of energy to find the velocity of the bullet - but be careful. The velocity of the bullet/block system is not the velocity of the bullet - it is equal to v zero/4, where v zero is the velocity of the bullet.

The equation is then (and it's hard without a math editor!):

1/2 times 4m times (v zero/4)^2 = 4mg(2R) + 1/2 times 4m times (sqrt(gR))^2

That gives 4 sqrt(5gR).

John

Could I get the solutions to number 6 parts c and d. I have gotten a and b with no problem but I am stuck on how to approach c and d.

Thanks!

I am also struggling with your answer to 6c.  Maybe there is something that I am misunderstanding about the question.  I have used conservation of energy to solve it and I end of with a quadratic equation.  But I am getting 1.5 m instead of 2.5.

Kim, I apologize for the late response. I'm emailing you my solutions for #6 and #10 - #6 didn't have a numerical solution, but #10 did, so I'm guessing that's the problem where you got 1.5 m instead of 2.5 m.  John

Kim, my email to you with the problems bounced back - do you have another email I can send it to?  John

Hi John, could I also have the solution for 6c please? Thank you

Caroline, I'll email the whole solution to you.  John