Work Energy and Power - Result Question 48

51. On a frictionless surface a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed

v3. The second block’s speed after the collision is :

[2015 RS]

(a) 34v

(b) 32v

(c) 32v

(d) 223v

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Answer:

Correct Answer: 51. (d)

Solution:

  1. (d) Here, M1=M2 and u2=0

u1=V,V1=V3;V2=?

From figure, along x-axis,

M1u1+M2u2=M1V1cosθ+M2V2cosϕ (i) Along y-axis

0=M1V1sinθM2Vssinϕ

By law of conservation of kinetic energy,

12M1u12+12M2u22=12M1V12+12M2V22

Putting M1=M2 and u2=0 in equation (i), (ii) and (iii) we get,

θ+ϕ=π2=90

and u12=V12+V22

V2=(V3)2+V22[u1=V. and .V1=V3]

or, V2(V3)2=V22

V2V29=V22

or V22=89V2V2=223V



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