IIT JEE , AIEEE Forum - Physics , Mechanics

sachin_gupta1991

Q. There is a rectangular plate of mass M kg of dimensions (a x b). The plate is held in horizontal position by striking n small balls each of mass m per unit area per unit time. These are striking in a half region of the plate s.t. area being striked= b x a

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2

The balls are colliding elastically with vel. v. What is v?

n=100,M=3 kg,m=0.01 kg,b=2 m , a= 1m,g=10 m/s²

feynmann

Balls striking per unit area per unit time = n/ 2 ;

Now , the total momentum change of a single ball = 2mv ( mv - ( - mv) )

hence the moment of force applied at the region of plate between x & x + dx from the axis is = x * 2mv * n/2 * a dx

= nmva x dx

So the total moment of force applied due to impact = nmva x dx ( from 0 to b )

= nmvab^2/2

this must balance the torque due to gravity = Mg b/2

so we get , Mg b/2 = nmvab^2 /2

or, v = Mg/ ( nmab ) = 15 m/s

waterdemon

Solution:

Torque about Hinge Side is given by:

a x (nb/2)(2mv) * (3b/4) = Mg(b/2)
v = 2Mg/3abnm

Therefore,
Substituting the values we have:

v = 2*3*10/3*2*100*0.01
v = 60/600*0.01
v = 60/6
v = 10 m/s.

Hope you find it useful.
Cheers !!!!!!!!!!!!!

feynmann

Plz explain the factor 3b/4 . ( It seems to be wrong )

karthik2007

Even I have a hunch about that... probably they are considering the n ball system to be concentrated at the center of the "half rectangle"

rohith291991

*edited*

karthik2007

@feymann - how did you get that n/2 as the no of balls striking per unit area?

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