SATHEE: Laws Of Motion Question 220

Laws Of Motion Question 220

Question: The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates into the wood for a distance of 12 cm is [CBSE PMT 2000; DPMT 2003]

Options:

A) $2.2 \times 10^{3} N$

B) $3.2 \times 10^{3} N$

C) $4.2 \times 10^{3} N$

D) $5.2 \times 10^{3} N$

Show Answer

Answer:

Correct Answer: D

Solution:

$u = 250 m / s$ , $v = 0$ , $s = 0.12 m e t r e$

$F = m a = m (\frac{u^{2} - v^{2}}{2 s}) = \frac{20 \times 10^{- 3} \times {(250)}^{2}}{2 \times 0.12}$

$∴$ $F = 5.2 \times 10^{3} N$

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Laws Of Motion Question 220

Question: The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates into the wood for a distance of 12 cm is [CBSE PMT 2000; DPMT 2003]

Options:

Answer:

Solution:

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