Question

If k.e. of a body increases by 0.1%, the percentage increase in its momentum will be

Answer 1

Answer:

The percentage increase in momentum is 0.05%.

Explanation:

Initially, we have K1=mv212
Let's denote the increase of K.E as δ

K2=mv222=(1+δ)⋅K1=(1+δ)mv212
We have:
mv222=(1+δ)mv212
Divide both sides by m2
v22=(1+δ)v21
v2=√1+δ⋅v1
Consequently, the new momentum is:
M2=m⋅v2=m⋅√1+δ⋅v1=√1+δ⋅m⋅v1=√1+δ⋅M1

Now, since δ=0.001, so √1+δ≈1.0005, therefore the increase in momentum is0.05%.

Answer 2

$K.E=p^{2}/2m$

So,$p=\sqrt{K.E*2m}$    ----------> 1

So,if K.E is increased by 0.1%

then,

$p_{1} =\sqrt{1.001*K.E*2m}$

$p_{1} =1.0005*\sqrt{K.E*2m}$     ---------> 2

COMPARING 1 AND 2,

$p_{1} =1.0005*p$

$p_{1}=p+0.0005p$

$p_{1}=p+0.05p/100$

Hence,Increase in Momentum Will Be 0.05%.