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Grade 11Mechanics

A body of mass M= 9.8kg with a small disc of mass 0.2kg placed on its horizontal surface ab, rests on a smooth horizontal plane. The disc can move freely along the smooth groove abc of mass M. To what height (relative to the initial position) will the disc rise after separating from the body of mass M when the initial velocity v=5m/s is given to it in the horizontal direction.

Profile image of yashasvi
6 Years agoGrade 11
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2 Answers

Profile image of Vikas TU
6 Years ago
Dear Student 
to find the height,
h=Mv^2/2g(M-m)
h=9.8*5*5/2*9.8(9.8-0.2)
h=25/2*9.6=1.302
Hope this hep
Profile image of Rohit Upadhyay
5 Years ago
Let ‘V’ be the velocity with which the mody is moving
loss of kinetic energy :-
    ½ mu – 1/2(m+M)v2
½ *0.2*(5)- ½ (9.8+2)v2
= 2.5-5v2
ACCORDING TO THE LAW OF CONSERVATION OF MOMENTUM
mu=(m+M)v
= V=(m+M)/ mu
V=0.2*5/ 10
V= 1/10
V=0.1m/s
 
LOSS OF ENERGY=2.5-5(0.1)2
                               =2.45J
 
POTENTIAL ENERGY GAINED BY THE DISK = mgh
                                                                            =0.2*9.8*h
 ACCORDING TO THE LAW OF CONSERVATION OF ENERGY
                                                    2.45= 0.2*9.8*h
                                                   h= 1.25m