Question

1. A bullet of mass 10 g moving with velocity 350 ms comes to rest after penetrating 40 cm in a wooden block in resistive
force
​

Answer 1

Answer:

$\bigstar{\bold{Resistive\:force=-1531.25\:N}}$

Explanation:

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

• Mass of the bullet (m) = 10 g = 0.01 kg
• Initial velocity (u) = 350 m/s
• Final velocity (v) = 0 m/s
• Distance travelled (s) = 40 cm = 0.4 m

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

• Resistive force

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

➠ Here we have to find the resistive force offered by the block.

➠ First we have to find the acceleration of the bullet.

➠ By the third equation of motion we know that

    v² - u² = 2as

    where v = final velocity

    u = initial velocity

    a = acceleration

    s = distance travelled

➠ Substituting the data,

     0² - 350² = 2 × a × 0.4

     a = -122500/0.8  

     a = 153125 m/s²

➠ Hence acceleration of the bullet is -153125 m/s²

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

➠ Now we have to find the resistive force

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

   F = m a

    where m = mass,

    a = acceleration

➠ Substitute the data,

    F = 0.01 × -153125

    F = -1531.25 N

➠ Here force is negative since it is resistive force and acts in the opposite direction.

➠ Hence the resistive force is -1531.25 N

    $\boxed{\bold{Resistive\:force=-1531.25\:N}}$

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

➠ The three equations of motion are:

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

Answer 2

Answer:

\bigstar{\bold{Resistive\:force=-1531.25\:N}}★Resistiveforce=−1531.25N

Explanation:

\Large{\underline{\underline{\sf{Given:}}}}

Given:

Mass of the bullet (m) = 10 g = 0.01 kg

Initial velocity (u) = 350 m/s

Final velocity (v) = 0 m/s

Distance travelled (s) = 40 cm = 0.4 m

\Large{\underline{\underline{\sf{To\:Find:}}}}

ToFind:

Resistive force

\Large{\underline{\underline{\sf{Solution:}}}}

Solution:

➠ Here we have to find the resistive force offered by the block.

➠ First we have to find the acceleration of the bullet.

➠ By the third equation of motion we know that

v² - u² = 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance travelled

➠ Substituting the data,

0² - 350² = 2 × a × 0.4

a = -122500/0.8

a = 153125 m/s²

➠ Hence acceleration of the bullet is -153125 m/s²

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

➠ Now we have to find the resistive force

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

F = m a

where m = mass,

a = acceleration

➠ Substitute the data,

F = 0.01 × -153125

F = -1531.25 N

➠ Here force is negative since it is resistive force and acts in the opposite direction.

➠ Hence the resistive force is -1531.25 N

\boxed{\bold{Resistive\:force=-1531.25\:N}}

Resistiveforce=−1531.25N

\Large{\underline{\underline{\sf{Notes:}}}}

Notes:

➠ The three equations of motion are:

v = u + at

s = ut + 1/2 × a × t²

v² - u² = 2as