Question

A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Answer:

50N

Explanation:

Given, mass = 10 × 10^-3

velocity = 150m/a

time = 3 × 10^-2 seconds

Now, acceleration = velocity / time

= 150/3×10^-3

Force = mass × acceleration

= 10 × 10^-3 × 150/3×10^-3

= 150/3

= 50 N

HOPE THIS HELPS YOU.......

Answer 2

$\huge\bigstar\sf\underline\red{GIVEN:-}$

$\large\sf{v=0}$

$\large\sf{u=150m/s}$

$\large\sf{t=0.03s}$

$\small\sf\orange{According\:of\:the\:first\:equation\:of\:motion,}$

$\large\sf\orange{v=u+at}$

$\small\sf\blue{Acceleration\:of\:the\:bullet,a}$

$\longrightarrow$$\large\sf{0=150+(a×0.03s)}$

$\longrightarrow$$\large\sf{a=\frac{-150}{0.3}}$

$\longrightarrow$$\large\sf{-5000m/s}$

$\small\sf\green{According\:of\:the\:first\:equation\:of\:motion,}$

$\large\sf\green{{v}^{2} = {u}^{2} + 2as}$

$\longrightarrow$$\large\sf{0 = {(150)}^{2} + 2( - 5000)s}$

$\longrightarrow$$\large\sf{s = \frac{ - {(150)}^{2} }{ - 2(5000)} }$

$\longrightarrow$$\large\sf{\frac{22500}{10000}}$

$\longrightarrow$$\large\sf{2.25m}$

$\small\sf\orange{From\:Newton's\:second\:law\:of\:motion,}$

$\large\sf{F=ma}$

$\large\sf{m=10g=0.01kg}$

$\large\sf{a=5000m/s}$

$\large\sf{F=ma}$

$\large\sf{0.01×5000}$

$\large\sf{50N}$