Question

A body 'p' has a mass of 20 kg and is moving with a velocity of 5 metre per second . Another body 'q' has a mass of 5 kg and is moving with a velocity of 20 metre per second . Calculate the ratio of their momentum. And kinetic energy of p in si unit

Answer 1

Answer:

Given :

Mass of p = 20 kg

Mass of q = 5 kg

Velocity of p = 5 m/s

Velocity of q = 20 m/s

To find:

1. Ratio of momentum

2. Kinetic Energy of p in SI units

Calculation:

1. Momentum of p = m1 × v1

=> Momentum of p = 20 × 5 = 100 kg m/s

Momentum of q = m2 × v2

=> momentum of q = 5 × 20 = 100 kg m/s

Therefore ratio = 100:100

=> Ratio = 1:1

2. Kinetic energy of p

= ½ mv²

= ½ × 20 × 5²

= 10 × 25

= 250 Joules .

Answer 2

Question :-

Given above ↑

Answer :-

→ ratio of their momentum = 1:1

→ kinetic energy of p is 250 joule .

To find :-

♪ The ratio of the momentum of both .

♪ Kinetic energy of body "p" .

Formula used :-

$\star \sf \boxed{ \sf momentum \: = mass \times \: velocity} \\ \\ \star \sf \: \boxed{ \sf \: kinetc \: energy \: = \frac{1}{2} \: m \: {v}^{2} }$

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Explanation :-

Given that :

♠ Mass of body "p"( m ) = 20 kg

♣ velocity of body "p" ( v ) = 5 m/s

♠ mass of body "q" ( M ) = 5 kg

♣ velocity of body "q"( u ) = 20 m/s

Now ,

let the momentum of body "p" is a -

$\to \sf \: momentum \: of \: p \: (a)= m \: v \\ \\ \to \sf a = 20 \times 5 \\ \\ \to \boxed{ \sf \: a \: = 100}$

hence momentum of body "p" is { a = 100 (kg-m)/s }

Now let the momentum of body "q" is b -

$\to \sf b = M \times u \\ \\ \to \sf b \: = 5 \times 20 \\ \\ \to \sf \boxed{ \sf b \: = 100}$

hence momentum of body "q" is

100 (kg-m)/s .

hence ratio of both momentum is

→ a : b = 1 : 1

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Now ,

♠ kinetic energy of body "p" = ?

$\to \sf k_p \: = \frac{1}{2} \: m \: {v}^{2} \\ \\ \to \sf k_p \: = \frac{1}{2} \times 20 \times {5}^{2} \\ \\ \to \boxed{\sf k_p \: = 250 \: joule \: }$

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