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Change in speed of the bullet after striking an object 
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#1
Dec1712, 03:03 PM

P: 100

1. The problem statement, all variables and given/known data
A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with signs and units! 2. Relevant equations conservation of momentum, mv1+mv2=mv1+mv2 3. The attempt at a solution the man is not moving initially, and final, i got 0.183 for the final bullet speed but thats not right, Iam confused about how to find the final for the bullet? 


#2
Dec1712, 03:49 PM

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The problem doesn't state the direction in which the bullet bounces away, so I would assume that the bullet goes back toward the location it was fired from. Did you use the fact that it's an elastic collision ? Such a problem can often be solved most easily in the center of mass reference frame. If you do, don't roundoff anything until your final answer. 


#3
Dec1712, 04:00 PM

P: 100

i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?



#4
Dec1712, 04:06 PM

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Change in speed of the bullet after striking an object



#5
Dec1712, 04:38 PM

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but i dont know either final velocity...



#6
Dec1712, 10:12 PM

P: 31

Is the answer 0.367ms^{1}.
Sol^{n}: We have two equations (one from momentum conservation and the other from kinetic energy conservation) with two variables (final velocities of the bullet and the man).Solve these two equations and get the final velocity of the man in terms of initial velocity of both and their masses. The final equation will be: V=M_{2}U_{2}/(M_{1}+M_{2}) where M_{1} is the mass of the man, M_{2} is the mass of the bullet, U_{2} is the velocity of the bullet. 


#7
Dec1812, 05:22 AM

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yes, that is the answer, why is the mass combined because the bullet bounces off.



#8
Dec1812, 10:42 AM

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Let's call the final velocity of the man, V_{2}, and the final velocity of the bullet, u_{2}. The initial velocities being V_{1} = 0 m/s and u_{1} = 190 m/s . Similarly, let M be the mass of the man, M = 31 kg. Let m be the mass of the bullet, m = 30 grams = 0.03 kg. (I took it upon myself to change some variable names.) Conservation of momentum gives [itex]\ mu_1+MV_1=mu_2+MV_2\ [/itex] where V_{1}=0, for the (initially) stationary man. [itex]\ mu_1=mu_2+MV_2\ [/itex]The other equation you need is from conservation of kinetic energy (It's an elastic collision.) [itex]\ (1/2)m{u_1}^2=(1/2)m{u_2}^2+(1/2)M{V_2}^2\ [/itex] 


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