Question Number 223712 by fantastic last updated on 02/Aug/25
Commented by fantastic last updated on 02/Aug/25
$${a}\:{bulet}\:{of}\:{mass}\:{m}\:{and}\:{velocity}\:{v} \\ $$$${hits}\:{a}\:{wooden}\:{block}\:{of}\:{mass}\:{M}\:{which}\:{is}\:{tied}\:{with}\:{a}\: \\ $$$${rope}\:{of}\:{length}\:{l}.\left({the}\:{bullet}\:{sticks}\:{with}\:{the}\:{block}\right) \\ $$$${Find}\:{theta} \\ $$
Answered by mahdipoor last updated on 02/Aug/25
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\left({m}+{M}\right){gh} \\ $$$$\Rightarrow{h}=\frac{{mv}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right){g}} \\ $$$${l}−{l}.{cos}\theta={h}\Rightarrow\theta={arccos}\left(\frac{{l}−{h}}{{l}}\right) \\ $$
Commented by fantastic last updated on 03/Aug/25
![sir I got θ=cos^(−1) [1−((m^2 v^2 )/(2(M+m)gl))]](https://www.tinkutara.com/question/Q223715.png)
$${sir}\:{I}\:{got}\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{1}−\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right){gl}}\right] \\ $$
Answered by fantastic last updated on 03/Aug/25
Commented by fantastic last updated on 03/Aug/25
![Let V be the speed after they attached from conservation of momentum we can say mv=(M+m)V or V=((mv)/(M+m)) So E_k_(in point A) =(1/2)mv^2 ⇒(1/2)(M+m)×((m^2 v^2 )/((M+m)(M+m)))=((m^2 v^2 )/(2(M+m))) OB=l OC=lcos θ So CA=l−lcos θ=l(1−cos θ) so E_p_(in point B) =mgh⇒(M+m)gl(1−cos θ) B is the highest point it goes A is the lowest point So E_k_(in point A) =E_p_(in point B) ((m^2 v^2 )/(2(M+m)))=(M+m)gl(1−cos θ) or 1−cos θ=((m^2 v^2 )/(2(M+m)^2 gl)) ......(i) or cos θ= 1−((m^2 v^2 )/(2(M+m)gl)) So θ=cos^(−1) [1−((m^2 v^2 )/(2(M+m)^2 gl))]✓ we can also write ...(i) as 2sin^2 (θ/2)=(1/2).((m^2 v^2 )/((M+m)^2 gl)) or sin^2 (θ/2)=((m^2 v^2 )/(4(M+m)^2 gl)) or sin (θ/2)=(√((m^2 v^2 )/(4(M+m)^2 gl)))=((mv)/(2(M+m)(√(gl)))) so (θ/2)=sin^(−1) [((mv)/(2(M+m)(√(gl))))] or θ=2sin^(−1) [((mv)/(2(M+m)(√(gl))))]✓](https://www.tinkutara.com/question/Q223718.png)
$${Let}\:{V}\:{be}\:{the}\:{speed}\:{after}\:{they}\:{attached} \\ $$$${from}\:{conservation}\:{of}\:{momentum}\:{we}\:{can}\:{say} \\ $$$${mv}=\left({M}+{m}\right){V} \\ $$$${or}\:{V}=\frac{{mv}}{{M}+{m}} \\ $$$${So}\:{E}_{{k}_{{in}\:{point}\:{A}} } =\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\cancel{\left({M}+{m}\right)}×\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\left({M}+{m}\right)\cancel{\left({M}+{m}\right)}}=\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right)} \\ $$$${OB}={l}\: \\ $$$${OC}={l}\mathrm{cos}\:\theta \\ $$$${So}\:{CA}={l}−{l}\mathrm{cos}\:\theta={l}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${so}\: \\ $$$${E}_{{p}_{{in}\:{point}\:{B}} } ={mgh}\Rightarrow\left({M}+{m}\right){gl}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${B}\:{is}\:{the}\:{highest}\:{point}\:{it}\:{goes} \\ $$$${A}\:{is}\:{the}\:{lowest}\:{point} \\ $$$${So} \\ $$$${E}_{{k}_{{in}\:{point}\:{A}} } ={E}_{{p}_{{in}\:{point}\:{B}} } \\ $$$$\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right)}=\left({M}+{m}\right){gl}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${or}\:\mathrm{1}−\mathrm{cos}\:\theta=\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right)^{\mathrm{2}} {gl}}\:\:\:\:\:\:\:……\left(\mathrm{i}\right) \\ $$$${or}\:\mathrm{cos}\:\theta=\:\mathrm{1}−\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right){gl}} \\ $$$${So}\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{1}−\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{2}\left({M}+{m}\right)^{\mathrm{2}} {gl}}\right]\checkmark \\ $$$${we}\:{can}\:{also}\:{write}\:…\left(\mathrm{i}\right)\:{as} \\ $$$$\mathrm{2sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\left({M}+{m}\right)^{\mathrm{2}} {gl}} \\ $$$${or}\:\mathrm{sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{4}\left({M}+{m}\right)^{\mathrm{2}} {gl}} \\ $$$${or}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{{m}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{4}\left({M}+{m}\right)^{\mathrm{2}} {gl}}}=\frac{{mv}}{\mathrm{2}\left({M}+{m}\right)\sqrt{{gl}}} \\ $$$${so}\:\frac{\theta}{\mathrm{2}}=\mathrm{sin}^{−\mathrm{1}} \left[\frac{{mv}}{\mathrm{2}\left({M}+{m}\right)\sqrt{{gl}}}\right] \\ $$$${or}\:\theta=\mathrm{2sin}^{−\mathrm{1}} \left[\frac{{mv}}{\mathrm{2}\left({M}+{m}\right)\sqrt{{gl}}}\right]\checkmark \\ $$
Answered by mr W last updated on 03/Aug/25
![V=velocity of wooden block (with bullet inside) after hit mv=(m+M)V ⇒V=((mv)/(m+M))=(v/(1+(M/m))) V^2 =2g(1−cos θ)l cos θ=1−(V^2 /(2gl))=1−(v^2 /(2gl(1+(M/m))^2 )) ⇒θ=cos^(−1) [1−(v^2 /(2gl(1+(M/m))^2 ))] assume l is large enough such that (v^2 /(2gl(1+(M/m))^2 ))≤2, i.e. l≥(v^2 /(4g(1+(M/m))^2 ))](https://www.tinkutara.com/question/Q223717.png)
$${V}={velocity}\:{of}\:{wooden}\:{block} \\ $$$$\:\:\:\:\:\:\:\left({with}\:{bullet}\:{inside}\right)\:{after}\:{hit} \\ $$$${mv}=\left({m}+{M}\right){V} \\ $$$$\Rightarrow{V}=\frac{{mv}}{{m}+{M}}=\frac{{v}}{\mathrm{1}+\frac{{M}}{{m}}} \\ $$$${V}^{\mathrm{2}} =\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){l} \\ $$$$\mathrm{cos}\:\theta=\mathrm{1}−\frac{{V}^{\mathrm{2}} }{\mathrm{2}{gl}}=\mathrm{1}−\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gl}\left(\mathrm{1}+\frac{{M}}{{m}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{1}−\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gl}\left(\mathrm{1}+\frac{{M}}{{m}}\right)^{\mathrm{2}} }\right] \\ $$$$ \\ $$$${assume}\:{l}\:{is}\:{large}\:{enough}\:{such}\:{that} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gl}\left(\mathrm{1}+\frac{{M}}{{m}}\right)^{\mathrm{2}} }\leqslant\mathrm{2},\:{i}.{e}.\:{l}\geqslant\frac{{v}^{\mathrm{2}} }{\mathrm{4}{g}\left(\mathrm{1}+\frac{{M}}{{m}}\right)^{\mathrm{2}} } \\ $$
Commented by fantastic last updated on 03/Aug/25
$${Sir}\:{you}\:{are}\:{right} \\ $$$${I}\:\:{also}\:{expressed}\:{it}\:{with}\:\mathrm{sin}^{−\mathrm{1}} \\ $$