# 9 - Centre of mass and momentum conservation Questions Answers

**Asked By: DARIUS**

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**Joshi sir comment**

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Initially there is no external force. Centre of mass of the system accelerates when insect moves with acceleration.

Now suppose displacement of insect at t = t is x with respect to rod, and y is the acceleration of rod in downward direction with respect to ground. Thus acceleration of counterweight is y with respect to ground. Now$\overrightarrow{{x}_{CM}}=\left[m\right(x-y)\hat{j}-(M-m)y\hat{j}+My\hat{j}]/2M\phantom{\rule{0ex}{0ex}}$Now solve for x = l,

For dynamics, take friction upward in free body diagram of insect and downward in the free body diagram of rod.

Now solve

Inform for any issue.

**Asked By: DARIUS**

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**Joshi sir comment**

$Atageneralpoint\phantom{\rule{0ex}{0ex}}-2T\mathrm{sin}\theta =ma(\theta ismeasuredfromhorizontal)\phantom{\rule{0ex}{0ex}}\Rightarrow a=-2Mg\theta /m=-2Mgy/ml\phantom{\rule{0ex}{0ex}}comparewitha=-{\omega}^{2}y\phantom{\rule{0ex}{0ex}}\omega =\sqrt{\frac{2Mg}{ml}}\left(1\right)\phantom{\rule{0ex}{0ex}}Nowformaxdisplacementofsmallm\phantom{\rule{0ex}{0ex}}mg(h+y)=2Mgxandx={y}^{2}/2l\phantom{\rule{0ex}{0ex}}onsolvingthisquadraticeq.fory/lweget\phantom{\rule{0ex}{0ex}}\frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}}(giventhat\frac{m}{M}\frac{h}{l})\phantom{\rule{0ex}{0ex}}sox=\frac{m}{2M}h\left(2\right)\phantom{\rule{0ex}{0ex}}so{v}_{max}=x\omega \phantom{\rule{0ex}{0ex}}nowsolve$

**Asked By: KAGGWA BEN**

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**Joshi sir comment**

distance covered in n sec. = un+an^{2}/2 (1)

distance covered in n-1 sec. = u(n-1)+a(n-1)^{2}/2 (2)

so s_{n} = (1) - (2)

put equation (1) and (2) and get the answer

the moment of inertia of a thin uniform rod about an axis through its end and perpendicular to its length is **ML²/3 **. The moment of inertia of the rod about an axis parallel to given axis and at a distance L from the center of the rod is

**Asked By: AVANTIKA PURI**

**Solved By: SARIKA SHARMA**

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**Joshi sir comment**

for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml^{2}

answer given by sarika is not completely correct

**Asked By: GAURAV MAHATE**

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**Joshi sir comment**

velocity of impact = √2gh

so components of velocity after rebound = √2ghcosθ perpendicular to the plane and √2ghsinθ parallel to the plane for the collision is perfectly elestic

now with perpendicular component take net displacement 0 and calculate time

then find displacement along the plane by using s = ut + 1/2at^{2}

here remember that you should divide gravity in comnents too

**Asked By: GAURAV MAHATE**

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**Joshi sir comment**

in first firing

m200 = 49mv so velocity of gun = 200/49

now apply momentum conservation again by taking 200/49 as the velocity of gun