Two solids A and B of mass $1 k g$ and $2 k g$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(K . E .)_{A}$ : $(K . E .)_{B}$ will be $\frac{A}{1}$ . So the value of $A$ will be .

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Solution

Given that, $\frac{M_{1}}{M_{2}} = \frac{1}{2}$

We know that $K = \frac{p^{2}}{2 M}$
$\Rightarrow \frac{K_{1}}{K_{2}} = \frac{p^{2}}{2 M_{1}} \times \frac{2 M_{2}}{p^{2}} = \frac{2}{1}$

$\Rightarrow \frac{K_{1}}{K_{2}} = \frac{M_{2}}{M_{1}} = \frac{2}{1}$

$\Rightarrow \frac{A}{1} = \frac{2}{1}$

Therefore , $A$ =2.

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Two solids A and B of mass 1kg and 2kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be A1. So the value of A will be .

Given that, M1M2=12 We know that K=p22M ⇒K1K2=p22M1×2M2p2=21 ⇒K1K2=M2M1=21 ⇒A1=21 Therefore , A =2.

Two solids A and B of mass $1 k g$ and $2 k g$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(K . E .)_{A}$ : $(K . E .)_{B}$ will be $\frac{A}{1}$ . So the value of $A$ will be .