11. Consider two bodies m,, m, undergoes head on
collision in one dimension. Find an expression for
velocity of two bodies after collision?
Answers
Answer:
Let two bodies of masses m
1
and m
2
moving with velocities u
1
and u
2
along the same straight line.
And consider the two bodies collide and after collision v
1
and v
2
be the velocities of two masses.
Before collision
Momentum of mass m
1
=m
1
u
1
Momentum of mass m
2
=m
2
u
2
Total momentum before collision
p
1
=m
1
u
1
+m
2
u
2
Kinetic energy of mass m
1
=
2
1
m
1
u
1
2
Kinetic energy of mass m
2
=
2
1
m
2
u
2
2
Thus,
Total kinetic energy before collision is
K.E=
2
1
m
1
u
1
2
+
2
1
m
2
u
2
2
After collision
Momentum of mass m
1
=m
1
v
1
Momentum of mass m
2
=m
2
v
2
Total momentum before collision
P
f
=m
1
v
1
+m
2
v
2
Kinetic energy of mass m
1
=
2
1
m
1
v
1
2
Kinetic energy of mass m
2
=
2
1
m
2
v
2
2
Total kinetic energy after collision
K
f
=
2
1
m
1
v
1
2
+
2
1
m
2
v
2
2
So, according to the law of conservation of momentum
m
1
u
1
+m
2
u
2
=m
1
v
1
+m
2
v
2
⇒m
1
(u
1
−v
1
)=m
2
(v
2
−u
2
)
............(1)
And according to the law of conservation of kinetic energy
2
1
m
1
u
1
2
+
2
1
m
2
u
2
2
=
2
1
m
1
v
1
2
+
2
1
m
2
v
2
2
⇒m
1
(u
1
2
−v
1
2
)=m
2
(v
2
2
−v
1
2
)
.............(2)
Now dividing the equation
(u
1
+v
1
)=(v
2
+u
2
)
⇒u
1
−u
2
=v
2
−v
1
Therefore, this is, relative velocity of approach is equal to relative velocity of separation.