Question

(d) None
133. Two bodies 'A' and 'B' have velocities in the
ratio of 2 : 1 mass of body A is half of the mass of
body 'B' the kinetic energies of 'A' and 'B' are in
the ratio of ​

Answer 1

Given :

• Two bodies 'A' and 'B' have velocities in the

ratio of 2 : 1.

• Mass of body A is half of the mass of body 'B'.

To calculate :

• The ratio of kinetic energies of A and B.

Calculation :

As velocities of 'A' and 'B' are in the ratio of 2 : 1. So,

Let ,

• Velocity of body A = 2v

• Velocity of body B = 1v or v

Also, let us assume ,

• Mass of body B = m

So, as per the given question mass of body A is half of the mass of body 'B'.

$\longrightarrow$ Mass of Body A = $\sf { \dfrac{1}{2} m }$

Now, let's calculate the ratio of the kinetic energies of 'A' and 'B' .

$\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$

Kinetic Energy of Body A :

We know that,

$\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}}$

• K.E = Kinetic Energy

• m = mass

• v = velocity

Here,

✰ Mass of A = $\sf { \dfrac{1}{2}m}$

✰ Velocity of A = 2v

Substituting values,

$\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{2} \times \dfrac{1}{2}m \times {(2v)}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: A)} = \dfrac{1}{\cancel{4}}m \times {\cancel{4}v}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: A)} = 1m \times {v}^{2} }$

$\longrightarrow \boxed{\sf {K.E_{(Body \: A)} = m{v}^{2}} }$

Kinetic Energy of Body B :

We know that,

$\bigstar \: \boxed{\sf {K.E = \dfrac{1}{2}m{v}^{2}}}$

• K.E = Kinetic Energy

• m = mass

• v = velocity

Here,

✰ Mass of B = m

✰ Velocity of B = v

Substituting values,

$\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {(v)}^{2} }$

$\longrightarrow \sf {K.E_{(Body \: B)} = \dfrac{1}{2} \times m \times {v}^{2} }$

$\longrightarrow \boxed{ \sf {K.E_{(Body \: B)} = \dfrac{1}{2}m{v}^{2}} }$

Calculating Ratio :

$\longrightarrow \sf { Ratio = \dfrac{Kinetic \: Energy_{(Body \: A)} }{Kinetic \: Energy_{(Body \: B)}} }$

$\longrightarrow \sf { Ratio = \dfrac{ \cancel{m{v}^{2}} }{ \cfrac{1}{2} \cancel{m {v}^{2}}} }$

$\longrightarrow \sf { Ratio = \dfrac{1 }{ \cfrac{1}{2}} }$

$\longrightarrow \sf { Ratio = 1 \times \dfrac{2}{1}}$

$\longrightarrow \sf { Ratio = \dfrac{2}{1}}$

$\longrightarrow \boxed {\pmb{ \rm \red { Ratio = 2:1}}}$

Therefore, the kinetic energies of 'A' and 'B' are in

the ratio of 2 : 1.

Answer 2

TIP: In case of such situations, your answer is always the given ratio of masses.

But we need some explanations:

(")(ಠ_ಠ)(")

We know, KE = ½ mv²

Given that, ratio of masses of A and B = 2:1.

So, Ratio of KE = (KE of A)/(KE of B)

⇒ Ratio of KE = (½ Mv²)/(½ mv²) = M/m

⇒ Ratio = 2/1 or 2:1.

Thus, ratio of KE in this case of 2:1.

More:

There is an increase in mass of a body in motion.