Question

A 30g bullet initially travelling 500m/s penetrates 12cm into a wooden block . what average force does if exert on the block?​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

S = 12 cm = 0.12 m.

u = 500 m/s.

v = 0 m/s.

m = 30g = 0.03Kg.

$\huge{\bold{\underline{Explanation:-}}}$

Applying third kinematic equation.

$\large{\bold{v^2 - u^2 = 2as}}$

Substituting the values.

$\large{ \implies 0^2 - (500)^2 = 2 \times a \times 0.12}$

$\large{ \implies - 250000 = 0.24 \times a}$

$\large{ \implies a = \frac{ - 250000}{0.24}}$

$\large{\boxed{\boxed{ a = - 10,41,666.67 \: m/s^2}}}$

Now,applying Newton's Second law.

$\large{\bold{F = ma}}$

$\large{ \implies F = 0.03 \times - 10,41,666.67}$

$\large{ \implies F = - 31,249.99 \: N}$

Taking magnitude.

$\large{ \implies F = 31,249.99 \: N}$

$\huge{\boxed{\boxed{F = 31.249 \: KN}}}$

So,the Force exerted is 31.249 KiloNewton (or) 31,249.99N.

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Initial Velocity,u = 500m/s

• Displacement,s = 12cm = 0.12m

• Mass,m = 0.03Kg

When the bullet comes in contact with the wooden block it stops » v = 0m/s

Using the relation,

$\huge{ \mathsf{v {}^{2} - u {}^{2} = 2as }}$

Putting the values,we get:

$\sf{0 {}^{2} - 500 {}^{2} = 2a(0.12) } \\ \\ \rightarrow \: \sf{0.24a = - 250000} \\ \\ \rightarrow \: \underline{ \boxed{ \sf{a = - 1041666.6 \: ms {}^{ - 2} }}}$

We Know that,

$\huge{ \sf{f = ma}}$

Putting the values,we get:

$\sf{f = (0.03)( - 1041666.6)} \\ \\ \implies \: \huge{\sf{f = - 3124999.8 N}}$

The force on the bullet due to the wooden block is -3124999.8 N