Question

the change in momentum of a particle when the kinetic energy is increasd by 0.1% will be

Answer 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

Hope this will help you buddy.... ✌