Question 289: Centre of Mass and Rotational Motion

Question

Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)
A
: (K.E.)
B
will be $${{A \over 1}}$$, so the value of A will be ________.

Accepted Answer

Given that, $${{{M_1}} \over {{M_2}}} = {1 \over 2}$$
Also, p
1
= p
2
= p
$$ \Rightarrow $$ M
1
V
1
= M
2
V
2
= p
Also, we know that
$$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$$ & $${K_2} = {{{p^2}} \over {2{M_2}}}$$
$$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} 	imes {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$$
$$ \Rightarrow {A \over 1} = {2 \over 1}$$
$$ \Rightarrow $$ $$ 	herefore $$ A = 2

Centre of Mass and Rotational Motion

Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.) A : (K.E.) B will be ${{A \over 1}}$, so the value of A will be ________.

Integer / Numeric Answer Type