VITEEE 2009 Solved Paper

Mass of the first body m1=5 kg and for elastic collision coefficient of restitution, e = 1.

Let initially body m1 moves with velocity v after collision velocity becomes (

u

10

).
Let after collision velocity of M block becomes (v2).
By conservation of momentum
m1u1+m2u2=m1v1+m2v2
or ‌‌5u+M×0=5×

u

10

+Mv2
or ‌‌5u=

u

2

+Mv2...(i)
Since, v1−v2=−e(u1−u2)
or ‌‌

u

10

−v2=−1(u)
or ‌‌

u

10

+u=v2
or ‌‌

11u

10

=v2...(ii)
Substituting value of v2 in Eq.(i) from Eq.(ii)
5u=

u

2

+M(

11u

10

)
or ‌‌5−

1

2

=M(

11

10

)
⇒ M=

9×10

2×11

=

45

11

=4.09‌kg