March 18th, 2017, 09:37 PM  #1 
Member Joined: Mar 2017 From: Tasmania Posts: 35 Thanks: 2  conservation of momentum
I've done 3 and 4 just want a clarification that its right Q3) Maxim, of mass 70kg, is running at 3.5ms1 when he jumps on a skateboard, of mass 2.2kg, travelling at 0.8 ms1 in the same direction. i. Calculate their new combined velocity. Q4) Maxim is on his skateboard again travelling at 6ms1 in the opposite direction to which his hat is pointed. He then leaps off the skateboard. If the skateboard’s velocity after the leap was 0.5ms1, find Maxim’s velocity post leap. (Mass of skateboard = 3kg, mass of Mr VT = 70kg). I need help with this question Q5) Maxim is on his skateboard again, travelling at 4ms1 to the right. He then leaps off the skateboard. If the distance between Maxim and the skateboard was increasing at a rate of 4 metres per second, find Maxim’s velocity post leap. (Mass of skateboard = 3kg, mass of Mr vT = 70kg). (There are two solutions to this problem, explain what both mean in this situation). 
March 19th, 2017, 04:50 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,429 Thanks: 1196 
Post your calculations for Q3 & Q4 if you want verification. Q5. $\vec{p_0}=\vec{p_f}$ $(M+m)v_0 = Mv_{1f}+mv_{2f}$ Case 1: $v_{2f}v_{1f} = 4$ if the skateboard is ahead of Maxim post separation Case 2: $v_{1f}v_{2f} = 4$ if Maxim is ahead of the skateboard post separation Solve the system of equations for each case Last edited by skeeter; March 19th, 2017 at 05:21 AM. 
March 19th, 2017, 12:35 PM  #3 
Member Joined: Mar 2017 From: Tasmania Posts: 35 Thanks: 2 
I get most of it by could you go into a bit more detail into how you got the first equation like how did u derive v0 Also for question 3 I got 3.2ms^1 and for question 4 I got 6.29ms^1 Last edited by Posher; March 19th, 2017 at 12:52 PM. 
March 19th, 2017, 01:58 PM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,429 Thanks: 1196  Quote:
Quote:
Q5. Quote:
$M = 70 \, kg$, $m = 3 \, kg$, $v_0 = 4 \, m/s$ Quote:
Quote:
case 1 ... $v_{2f}  v_{1f} = 4 \implies v_{2f} = v_{1f}+4$ substitute for $v_{2f}$ in the total momentum conservation equation given inthe previous post ... $(M+m)v_0 = Mv_{1f} + m(v_{1f}+4)$ $(M+m)v_0 = Mv_{1f} + mv_{1f}+ 4m$ $(M+m)v_0  4m = v_{1f}(M+m)$ $\dfrac{(M+m)v_0  4m}{M+m} = v_{1f}$ sub in your given values to calculate $v_{1f}$, then add 4 to get $v_{2f}$ ... same idea for case 2.  
March 19th, 2017, 09:07 PM  #5 
Member Joined: Mar 2017 From: Tasmania Posts: 35 Thanks: 2 
skeeter I'm pretty sure even though you got slightly different answers to me in q3 and q4 I still was in the ballpark of the questions just rounding differences also thanks for question 5. Also may I ask what your background knowledge is to be able to answer all of my questions Last edited by Posher; March 19th, 2017 at 09:11 PM. 
March 20th, 2017, 07:43 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,429 Thanks: 1196  Quote:
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