February 4th, 2016, 08:09 PM  #1 
Member Joined: Jun 2013 Posts: 42 Thanks: 0  please explain this momentum problem
A 40gram bullet is fired with a speed of 300m/s into a ballistic pendulum of mass 5kg suspended from a chord 1m long. Compute the vertical height through which the pendulum rises. Solution: m1v1+m2v2 = m1v1'+m2v2' 400(300)+5000(0)=40v1+5000v1 v1=2.38m/s v2=0 v2^2=v1^2 + 2gS S=h 0 = 2.38^2  2(9.81)(h) h=0.289m My Question: this is the solution of the book. I am just wondering why he used S =h instead of S = d? Isn't "S" suppose to be the actual distance that the object travelled? Please clarify, thanks in advance. Last edited by skipjack; February 5th, 2016 at 04:42 AM. 
February 4th, 2016, 08:12 PM  #2 
Member Joined: Jun 2013 Posts: 42 Thanks: 0 
Please consider this diagram

February 4th, 2016, 08:25 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
That should be an equation describing conservation of energy (kinetic energy and potential energy), not the distance calculated from Newton's Third Law.

February 4th, 2016, 08:28 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
In the original post, the book is misquoted.

February 5th, 2016, 02:13 AM  #5 
Member Joined: Jun 2013 Posts: 42 Thanks: 0 
So the book is wrong? I knew it... so what's the right way of solving for h? I guess it's something like this: v2^2=v1^2 + 2gS S=d 0 = 2.38^2  2(9.81)(d) d=0.289m. After I find the length d, I will solve for h using trigonometry? Thanks for reply... in advance.. Last edited by skipjack; February 5th, 2016 at 02:38 AM. 
February 5th, 2016, 02:37 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
I'm not sure whether the book is wrong, but I doubt that what you posted here is precisely what the book gave, as the arithmetic doesn't work.

February 5th, 2016, 02:43 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
I don't think the book is wrong. They've just decided (slightly oddly) to have the vertical coordinate axis so that $s=0$ at $v=v_2$, the top of the swing. Then $\text{(kinetic energy at bottom)} + \text{(potential energy at bottom)} = \text{(kinetic energy at top)}+\text{(potential energy at top)}$ Last edited by skipjack; February 5th, 2016 at 04:33 AM. 
February 5th, 2016, 04:38 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
That doesn't explain why the initial arithmetic is incorrect.

February 5th, 2016, 05:01 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
Oh, yes, that additional zero on the 40 gram bullet. But the value for $v_1$ is correct. The reuse of variable names is a bit confusing too. 
February 5th, 2016, 05:11 AM  #10 
Math Team Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 
Using $v_2$ as the speed of the bullet/block combination after the collision ... $\frac{1}{2}(M+m)v_2^2=(M+m)gh$ $h=\dfrac{v_2^2}{2g} \approx 0.3 \text{ m}$ The book's solution is correct ... horizontal displacement is not considered since there is no change in gravitational potential energy in that dimension. Last edited by skipjack; February 6th, 2016 at 12:28 AM. 

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