Question

A block of mass m, moving with speed u undergoes
head-on elastic collision with another mass m, at
rest. The fraction of energy transferred by m, to my
is
m, - m₂)²
m, + m)
(1)
(2)
m, + m2)
(m, - m
2mm
4mm
(3
(4) (m, + m2)
(m, + m2)​

Answer 1

Answer:

mv=mv

1

+nmv

2

,(Conservationofmeomentum)

v−0=v

2

−v

1

i.e.,v

1

=v

2

−v(e=1)

mv=−mv+mv

2

(1+n)

v

2

=

1+n

2v

K.E.∝mv

2

The fraction of the incident energy, transferred to the heavier ball is=

mv

2

nm(

1+n

2v

)

2

=

(1+n)

2

4n