A stone of mass 2 kg moving with a velocity of 4 m/s collides to another stone of mass 4 kg which is at rest. If after the collision they move with the same velocity, then what would be the combined kinetic energy of two stones after the collision?
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a) By momentum conservation,
2(4)−4(2)=2(−2)+4(ν
2
)
⟹ν
2
=1m/s.
b) e=
Velocityofapproach
Velocityofseparation
=
4−(−2)
1−9−2)
=
6
3
=0.5
c) At the maximum deformed state, by conservation of momentum, common velocity is ν=0
J
D
=m
2
(ν−u
2
)=4[(0−(−2)]=8Ns
d) Potential energy at the maximum deformed state,
U= loss in kinetic energy during defirmation
or U=(
2
1
m
1
u
1
2
+
2
1
m
2
u
2
2
)−
2
1
(m
1
+m
2
)ν
2
=(
2
1
2(4)
2
+
2
1
4(2)
2
)−
2
1
(2+4)(0)
2
=24J
e) J
R
=m
2
(ν
2
−ν)=4(1−0)=4Ns
also J
R
=eJ
D
v
=0.5×8=4Ns thank you bro
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