Question

14. A bullet of mass 10 g travelling horizontally with a velocity of
150 m s- strikes a stationary wooden block and comes to rest
in 0.03 s. Calculate the distance of penetration of the bullet
into the block. Also calculate the magnitude of the force exerted
by the wooden block on the bullet.
1 lr trouelling in a straight line with a velocity​

Answer 1

Answer:

$\bigstar{\bold{Distance\:travelled=2.25\:m}}$

$\bigstar{\bold{Force\:exerted=50\:N}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Mass of bullet (m) = 10 g = 0.01 kg
• Initial Velocity (u) = 150 m/s
• Final velocity (v) = 0 m/s
• Time taken (t) = 0.03 s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Distance travelled (s)
• Force exerted (F)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we have to find the acceleration of the bullet

→ By the first equation of motion we know that,

  v = u + at

→ Substituting the datas we get,

  0 = 150 + a × 0.03

  -150 = 0.03 a

   a = -150/0.03

   a = -5000 m/s²

→ Here acceleration is negative since it is retardation or deceleration.

→ By the second equation of motion, distance travelled is given by,

  s = ut + 1/2 × a × t²

→ Substituting the datas,

  s = 150 × 0.03 + `1/2 × - 5000 × 0.03 × 0.03

  s = 4.5 + -4.5/2

  s = 4.5 - 2.25 m

  s = 2.25 m

→ Hence the distance travelled is 2.25 m

$\boxed{\bold{Distance\:travelled=2.25\:m}}$

→ Now we have to find the force exerted.

→ By Newton's second law of motion, we know that

 F = m a

→ Substituting the datas,

  F = 0.01 × -5000

  F = - 50 N

→ Taking only the magnitude we get,

   Force exerted = 50 N

$\boxed{\bold{Force\:exerted=50\:N}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as

Answer 2

Answer:

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Explanation:

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