Question

show that
(1/1+2ω)-(1/1+ω)+(1/2+ω)=0​

Answer 1

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➥ This question says that there is a bullet its mass is 30 grams and it's moving with initial velocity of 150N and hit wood as it's target. Now this question says that the bullet penetrates 100 cm in the target. Now at last we have to find the magnitude of initial velocity means we have to find the acceleration and final velocity

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➥ Mass of the bullet = 30 g = 00.3 kg

➥ Bullet is moving with initial velocity of 150N and hit wood as it's target.

➥ The bullet penetrates 100 cm = 1 m in the target.

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➥ Acceleration

➥ Final velocity

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➥ Acceleration = -5000 m/s²

➥ Final velocity = 100 m/s

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{Using \; concepts}}}}}$

➥ Newton's second law of motion formula

➥ Using Newton's third law of motion

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➥ Newton's second law of motion => a = -F / m

Note : We use a = -F / m as the formula of Newton's second law of motion here because this question says that the bullet penetrates means it's retards.

➥ Using Newton's third law of motion => v² - u² = 2as

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{These \: denotes}}}}}$

➥ v denotes final velocity

➥ u denotes initial velocity

➥ a denotes acceleration

➥ s denotes displacement

➥ m denotes mass

➥ F denotes force

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{Full \: solution}}}}}$

~ Finding acceleration by using Newton's second law of motion formula (atq)

➨ a = -F / m

➨ a = -150 / 0.03

Cancelling the above digits afterwards multiplying the left terms we get –

➨ a = -5000 m/s²

~ Finding final velocity by using Newton's third law of motion formula

➨ v² - u² = 2as

➨ 0² - u² = 2(-5000)(1)

Note : We write v as 0 because at that time the object is at rest ( bullet penetrates and it stop ) and when an object is at rest then it's velocity be 0 by itself.

➨ 0² - u² = 2 × -5000 × 1

➨ 0² - u² = 2 × -5000

➨ 0 - u² = -10000

➨ -u² = -10000

➨ u² = 10000

➨ u = √10000

➨ u = 100 m/s

Henceforth, initial velocity magnitude is 100 m/s