# 11 - Centre of mass and momentum conservation Questions Answers

In a gravity free space a small ball is placed at the centre of a box of the same mass. The ball and the box both are moving with velocity u towards a fixed wall. Assume all collisions are perfectly elastic.

A) How many collisions are possible?

B) find velocities of the box and the ball and the position of the ball relative to the box after all possible collisions.

**Asked By: DIPYAMAN**

**read solutions ( 2 ) | submit your answer**

**Asked By: DARIUS**

**submit your answer**

**Joshi sir comment**

Updated:

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**

**submit your answer**

**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**

**submit your answer**

**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**

**read solutions ( 1 ) | submit your answer**

**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**

**read solutions ( 1 ) | submit your answer**

**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