JEE Main 31-Aug-2021 Shift 1 Solved Paper

Given, mass of body 1 = M
Mass of body 2 = m
Initial speed of body 1,u1=v0
Initial speed of body 2,u1=0
Final speed of body 1 and 2=v1 and v2
Angle made by body 1 and 2 after collision with respect to initial direction =θ1,θ2 By using law of conservation of momentum, Along X-axis
Mv0+m.0=Mv1‌cos‌θ1+2‌cos‌θ2
If θ1=θ2=θ
∴Mv0=Mv1‌cos‌θ+mv2‌cos‌θ ...(i)
Along Y-axis,
Mv1‌sin‌θ1=mv2‌sin‌θ2 [since, θ1=θ2=θ]
v2=

Mv1

m

...(ii)
Substituting the value of v2 in Eq. (i), we get
Mv0=Mv1‌cos‌θ+m(

Mv1

m

)‌cos‌θ
=2Mv1‌cos‌θ
v1=

v0

2‌cos‌θ

...(iii)
By using law of conservation of energy along X-axis

1

2

Mv02=

1

2

Mv12+

1

2

mv22
⇒Mv02=Mv12+m(

M

m

v1)2 [from Eq. (ii)]
⇒Mv02=Mv12+

M2v12

m

=Mv12m(m+M)
⇒v02=(

v0

2‌cos‌θ

)2(

m+M

m

) [From Eq. (iii)]
⇒v02=

v02

4cos2θ

(1+

M

m

)
⇒4cos2θ=1+

M

m

⇒

M

m

=4cos2θ−1 [ For largest possible value of

M

m

,θ=0 ]
=4cos20°−1=4−1=3