SATHEE: Laws Of Motion Question 297

Laws Of Motion Question 297

Question: . A bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 \times 10^{5} t$ where. F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Options:

A) 1.8 N-s

B) zero

C) 9 N-s

D) 0.9 N-s

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Given $F = 600 - (2 \times 10^{5} t)$

The force is zero at time t, given by 0 $= 600 - 2 \times 10^{5} t$

$\Rightarrow t = \frac{600}{2 \times 10^{5}} = 3 \times 10^{- 3} seconds$

$∴ I m p u l s e = \int_{0}^{t} F d t = \int_{0}^{3 \times 10^{- 3}} (600 - 2 \times 10^{5} t) d t$

$= [600 t - \frac{2 \times 10^{5} t^{2}}{2}]_{0}^{3 \times 10^{- 3}}$

$= 600 \times 3 \times 10^{- 3} - 10^{5} {(3 \times 10^{- 3})}^{2}$

$= 1.8 - 0.9 = 0.9 N s$

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Laws Of Motion Question 297

Question: . A bullet is fired from a gun. The force on the bullet is given by F=600−2×105t where. F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Options:

Answer:

Solution:

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Question: . A bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 \times 10^{5} t$ where. F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?