Question

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 ?

Answer 1

Please mark it as the brainliest answer...

Answer 2

$\star{\underline{\underline{\sf\purple{ Solution:-}}}}$

Given:

• Mass of the bullet (m) = 10g (or 0.01 kg)
• Initial velocity of the bullet (u) = 150 m/s
• Terminal velocity of the bullet (v) = 0 m/s
• Time period (t) = 0.03 s

$\\ \star{\underline{\underline{\sf\purple{To\:Find:-}}}}$

• The distance of penetration ?
• Magnitude of the force - ?

$\\ \star{\underline{\underline{\sf\purple{ Formula:-}}}}$

According to the first Motion Equation,

$\bullet{\boxed{\sf{ V = u + at } }}$

Acceleration (a) must be;

$\implies{\sf{ a = \dfrac{v-u}{t} }} \\ \\ \implies{\sf{ \dfrac{0-150}{0.03} \: ms^{-2} }}$

$\implies{\sf{ -5000 \:ms^{-2} }}$

Acceleration (a) of the bullet is -5000 m/s².

Now,

According to the third Motion Equation,

$\bullet{\boxed{\sf{ v^2 = u^2 + 2as }}}$

The distance of penetration (s) can be defined;

$\hookrightarrow{\boxed{\sf{ S = \dfrac{v^2- u^2}{2a} }}}$

$\hookrightarrow{\sf { \dfrac{0^2 - (150)^2 }{2(-5000)} \: metres }}$

$\hookrightarrow{\sf{ 2.25\: metres} }$

According to the second law of motion,

$\bullet{\boxed{\sf{ F = m \times a }}}$

After substituting the values,

$\sf{ Force\: Exerted(f) = 0.01kg \times (-5000 ms^{-2}) }$

$\hookrightarrow \sf{ -50\: N}$

Hence,

Wooden block exerts a force of magnitude 50 N on the bullet in the direction.