6 - Centre of mass and momentum conservation Questions Answers

Plz help sir
Asked By: DARIUS
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Asked By: DARIUS
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Joshi sir comment

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
is this question helpfull: 1 3 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 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
is this question helpfull: 4 2 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 + 1/2at2

here remember that you should divide gravity in comnents too

a gun is mounted on a railroad car .the mass of the car,the gun,the shells, and the operator is 50m where m is the mass of one shell.if velocity of the shell w.r.t. The gun(in its state before firing) is 200m/s,what is the recoil of the car after the second shot?neglect friction.
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

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