Question

Two small bodies of masses m' and 2m' are placed in a fixed smooth horizontal circular hollow tube of mean radius 'r' as shown. The mass m' is moving with speed u' and the mass 2m' is stationary. After their first collision, the time elapsed before next collision is: (coefficient of restitution e

Answer 1

Answer:

Let the velocity of the particle of mass

m

m, after the collision is

u

1

u1 and the velocity of the particle of mass

2

m

2m is

u

2

u2.

From conservation of momentum, it can be written as,

m

u

=

m

u

1

+

2

m

u

2

mu=mu1+2mu2

u

=

u

1

+

2

u

2

u=u1+2u2

Since, the collision is elastic, the kinetic energy is conserved in the collision and it can be written as,

1

2

m

u

2

=

1

2

m

u

1

2

+

1

2

2

m

u

2

2

12mu2=12mu12+122mu22

u

2

=

u

1

2

+

2

u

2

2

u2=u12+2u22

u

2

−

u

1

2

=

2

u

2

2

u2−u12=2u22

(

u

−

u

1

)

(

u

+

u

1

)

=

2

u

2

2

(u−u1)(u+u1)=2u22

(

u

+

u

1

)

=

u

2

(u+u1)=u2

u

=

u

2

−

u

1

u=u2−u1

The time taken is given as,

t

=

2

π

r

u

2

−

u

1

t=2πru2−u1

t

=

2

π

r

u

t=2πru

Thus, the time elapsed for the next collision is

2

π

r

u

2πru.