How to find the angle generated as a function of the speed of a bullet when it collisions to mass hanging to a ceiling?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

A bullet of mass $m$ collisions to a bob hanging vertically from a ceiling whose mass is $M$. As a result of this impact the bob with the bullet inside travels an arc (whose have a radius $R$) and then oscillates. Given this condition find the initial angle traveled by the mass M as a function of the speed (indicated as $v$) of the bullet. Assume the acceleration due gravity is $g$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\cos^{-1}\left(1-\frac{1}{2gR}\frac{m}{m+M}v^2\right)\\
2.&\sin^{-1}\left(1-\frac{1}{2gR}\frac{m}{m+M}v^2\right)\\
3.&\cos^{-1}\left(1-\frac{1}{2gR}\frac{m}{M}v^2\right)\\
4.&\sin^{-1}\left(1-\frac{1}{2gR}\frac{m}{M}v^2\right)\\
5.&\tan^{-1}\left(1-\frac{1}{2gR}\frac{m}{m+M}v^2\right)\\
\end{array}$

How exactly should I tackle this question?. Can someone guide me here?. The only thing which I can recall when a bullet strikes a bob is in an inelastic collision which would be given by:

$p_i=p_f$

But I dont know exactly how can I relate it with the arc?. The thing here is that the bob starts swinging or oscillating. I don't know how to translate this into an equation.

Typically this would be given by

$mv=(m+M)u$

$u=\frac{m}{m+M}v$

Then as it oscillates from the bob will be as:

$E_k=E_u$

$\frac{1}{2}(m+M)u^2=(m+M)gR(1-\cos\omega)$

Hence combining both expressions would be:

$\frac{1}{2}(m+M)\left(\frac{m}{m+M}v\right)^2=(m+M)gR(1-\cos\omega)$

$\frac{1}{2}\frac{m^2v^2}{m+M}=(m+M)gR(1-\cos\omega)$

$\frac{1}{2}m^2v^2=(m+M)^2gR(1-\cos\omega)$

$1-\cos\omega=\frac{1}{2}\frac{m^2v^2}{(m+M)^2gR}$

$\omega=\cos^{-1}\left(1-\frac{1}{2}\frac{m^2v^2}{(m+M)^2gR}\right)$

But this doesn't appear in any of the alternatives. What could it be wrong in my approach. Can someone help me here?.
 

skeeter

Math Team
Jul 2011
3,353
1,842
Texas
inelastic collision ...

$mv_0 = (M+m)v_f \implies v_f = \dfrac{mv_0}{M+m}$

post collision conservation of energy ...

$\dfrac{1}{2}(M+m)v_f^2 = \dfrac{1}{2}(M+m)\left( \dfrac{mv_0}{M+m} \right)^2 = \dfrac{1}{2}\dfrac{m^2v_0^2}{M+m} = (M+m)gh \implies$

$h = \dfrac{1}{2g} \left(\dfrac{mv_0}{M+m}\right)^2$

$\cos{\theta} = \dfrac{R-h}{R} = 1 - \dfrac{h}{R} = 1 - \dfrac{1}{2gR} \left(\dfrac{mv_0}{M+m}\right)^2$

I agree with your solution ... looks like the first answer choice omitted squaring the quantity $\dfrac{m}{M+m}$
 
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Reactions: Chemist116
Jun 2017
399
6
Lima, Peru
inelastic collision ...

$mv_0 = (M+m)v_f \implies v_f = \dfrac{mv_0}{M+m}$

post collision conservation of energy ...

$\dfrac{1}{2}(M+m)v_f^2 = \dfrac{1}{2}(M+m)\left( \dfrac{mv_0}{M+m} \right)^2 = \dfrac{1}{2}\dfrac{m^2v_0^2}{M+m} = (M+m)gh \implies$

$h = \dfrac{1}{2g} \left(\dfrac{mv_0}{M+m}\right)^2$

$\cos{\theta} = \dfrac{R-h}{R} = 1 - \dfrac{h}{R} = 1 - \dfrac{1}{2gR} \left(\dfrac{mv_0}{M+m}\right)^2$

I agree with your solution ... looks like the first answer choice omitted squaring the quantity $\dfrac{m}{M+m}$
Hopefully at least some of the questions I posted on this forum was right. :)
Btw I also thought about that. But since all of your solutions are correct as my book says. I'll take it for this as the right one!.