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
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Asked By: AMARNENI SIVA TEJA
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Asked By: SHIVANSHU KUMAR
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Joshi sir comment
Asked By: MITUL
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Joshi sir comment
Please sir post the solution of this question.   Thanks in advance
Asked By: TUSHAR GUPTA
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Plz help sir. EDIT: Sir can you also tell how to solve part E?
Asked By: DARIUS
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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. NowxCM=[m(x-y)j^-(M-m)yj^+Myj^]/2M 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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At a general point -2Tsinθ = ma    (θ is measured from horizontal) a = -2Mgθ/m = -2Mgy/ml compare with a = -ω2y ω=2Mgml            (1) Now for max displacement of small m mg(h+y) = 2Mgx and x = y2/2l on solving this quadratic eq. for y/l we get  yl-m2M=mMhlyl=mMhl  (given that mM <<hl) so x = m2Mh       (2) so vmax = xω now solve

a particle moving in a straight line with an initial velocity u is subjected to a constant acceleration,a. Show that the distance traveled by the particle in the nth second is; Sn=u+a(n-0.5)
Asked By: KAGGWA BEN
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Joshi sir comment

distance covered in n sec. = un+an2/2   (1)

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

so sn = (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 Ml2

answer given by sarika is not completely correct

a ball falls on an inclined plane of inclination @ from a height h above the point of impact and makes a perfectly elastic collisiom.where will it hit the plane again?
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 + 1/2at2

here remember that you should divide gravity in comnents too

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