a ball is moving with speed u undergoes a head on elastic collision with ball of mass nm initially at rest the fraction of incident energy transferred to initial ball
Answers
Answer:
Given : m
1
=m m
2
=nm
Let the velocities of A and B after the collision be v
1
and v
2
respectively.
initial velocity of A before collision is v and that of B is zero.
Initial kinetic energy of the system E=
2
1
mv
2
Using v
2
=
m
1
+m
2
(m
2
−em
1
)u
2
+(1+e)m
1
u
1
v
2
=
m+nm
(nm−1×m)(0)+(1+1)m(v)
⟹v
2
=
(1+n)
2v
Thus Kinetic energy of B after the collision E
B
′
=
2
1
(nm)v
2
2
E
B
′
=
2
1
(nm)
(1+n)
2
4v
2
Fraction of total kinetic energy retained by B
E
E
B
′
=
(n+1)
2
4n
Step-by-step explanation: