Question

`A force ¯ {F}=(4t+3t^{2}) N acts an an abjoct moving in xyplane. The magnitude of change in linear momentum of the object in time interval t=0 ta 1= 4 s is`​

Answer 1

Given

• F = 4t + 3t² N
• t = 0 to 4 = 4 sec

To Find

• Magnitude of change in momentum

Solution

☯ F = dp/dt

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✭ According to the Question :

⇒ F = dp/dt

⇒ dp = F dt

⇒ ∫dp = ∫F dt

⇒ ∆P = ∫4t+3t²

⇒ ∆P = [4t²/2 + 3t³/3]

⇒ ∆P = 2t² + t³

⇒ ∆P = 2(4)² + (4)³

⇒ ∆P = 2×16 + 64

⇒ ∆P = 96 N s

∴ The change in momentum is equal to 96 N s

Answer 2

❂ To calculate:

Mag of change in  linear momentum from t = 0 to t = 4

❂ Solⁿ:

→ Given that, F = 3t² + 4t

Now, by Newton's 2nd law

F =  dP / dt

⇒ dP = F dt

⇒ ∫dP = ∫F  dt

⇒ ∫dP = ∫ (3t² + 4t) dt

⇒ ∫dP = ∫ 3t² dt + ∫ 4t dt

⇒ ∫dP = t³ + 2t²

→ Now, put limits

₀ˣ∫ dP = [ t³ + 2t² ]₀⁴

⇒ x - 0 = [ 4³ + 2(4)² ] -  [ 0³ + 2(0)² ]

⇒ x = 64 + 32

⇒ x = 96 kgm/s

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❂ Final Answer:

96 kgm/s