JEE Main 18 March 2021 Shift 2 Solved Paper

The mass m1 is moving with speed u1 initially and mass m2 is at rest. After the collision, the mass m1 and m2 move with speed v in opposite directions.Using the law of conservation of linear momentum,
m1U1+m2U2=m1v1+m2v2
⇒‌‌m1u1+m2(0)=m1(−v)+m2v
⇒‌‌m1u1=(−m1+m2)v...(i)
Since, the collision is elastic because they move with same speed after the collision. Hence, coefficient of restitution, e=1
∴‌‌e=‌

v2−v1

u1−u2

⇒1=‌

v−(−v)

u1−0

u1=2v...(ii)
Putting the above value in Eq. (i), we get
m7(2v)=(−m1+m2)v
⇒‌‌3m1=m2⇒‌

m2

m1

=‌

3

1

⇒‌‌m7:m1=3:1