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

#### Chemist116

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
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 $\dfrac{m}{M+m}$

Chemist116

#### Chemist116

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!.