the change in momentum of a particle when the kinetic energy is increased by 0.1% will be
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1
Answer:
The increase in momentum is
=
0.05
Explanation:
The kinetic energy is
E
=
1
2
m
v
2
The momentum is is
p
=
m
v
v
=
√
2
E
m
Therefore,
p
=
m
v
=
m
√
2
E
m
=
√
2
m
E
Taking logs
ln
p
=
1
2
(
ln
2
+
ln
m
+
ln
E
)
Differentiating
Δ
p
p
=
1
2
(
Δ
m
m
)
+
1
2
(
Δ
E
E
)
But,
Δ
E
E
=
0.1
Therefore,
Δ
p
p
=
1
2
⋅
0.1
=
0.05
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