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 \underline\hspace2cm .

Center of Mass question

2021 · 24 Feb · Shift 2 · Q64

2021 · 24 Feb · Shift 2 · Q64

View written solutionFree

Correct answer: 2

Let the common linear momentum of both solids be $p$ .
Kinetic energy in terms of momentum is $K = \frac{p^2}{2m}.$
For solid $A$ of mass $1\,\text{kg}$ : $K_A = \frac{p^2}{2\cdot 1} = \frac{p^2}{2}.$
For solid $B$ of mass $2\,\text{kg}$ : $K_B = \frac{p^2}{2\cdot 2} = \frac{p^2}{4}.$
Therefore, the ratio is $K_A : K_B = \frac{p^2}{2} : \frac{p^2}{4} = 2:1.$
Given $(\text{K.E.})_A : (\text{K.E.})_B = \frac{A}{1},$ so we get $A = 2.$

Hence, the required integer answer is $2$ .

2021 · 24 Feb · Shift 2 · Q64

2021 · 24 Feb · Shift 2 · Q64

View written solutionFree

Correct answer: 2

Let the common linear momentum of both solids be $p$ .
Kinetic energy in terms of momentum is $K = \frac{p^2}{2m}.$
For solid $A$ of mass $1\,\text{kg}$ : $K_A = \frac{p^2}{2\cdot 1} = \frac{p^2}{2}.$
For solid $B$ of mass $2\,\text{kg}$ : $K_B = \frac{p^2}{2\cdot 2} = \frac{p^2}{4}.$
Therefore, the ratio is $K_A : K_B = \frac{p^2}{2} : \frac{p^2}{4} = 2:1.$
Given $(\text{K.E.})_A : (\text{K.E.})_B = \frac{A}{1},$ so we get $A = 2.$

Hence, the required integer answer is $2$ .