Question

A bullet travelling at 360 m/s; strikes a block of soft wood. The mass of the bullet
is 2.0g. The buliet comes to rest after penetrating 10cm into the wood?
a) Find the average deceleration force exerted by the wood.
b) Find the time taken by the bullet to come to rest.​

Answer 1

Answer:

(a). The force is 1296 N.

(b). The time is 0.00055 sec.

Explanation:

Given that,

Mass = 2.0 g

Speed of bullet = 360 m/s

Distance = 10 cm

We need to calculate the average deceleration force exerted by the wood

Using equation of motion

$v^2=u^2-2as$

Put the value in the equation

$360^2=0-2\times a\times10\times10^{-2}$

$a=\dfrac{360^2}{2\times10\times10^{-2}}$

$a=-648000\ m/s^2$

The deceleration is 658000 m/s².

(a). We need to calculate the force

Using formula of force

$F=ma$

Put the value into the formula

$F=2.0\times10^{-3}\times648000$

$F=1296\ N$

The force is 1296 N.

(b). We need to calculate the time

Using equation of motion

$v = u+at$

Put the value in the equation

$360=0+648000\times t$

$t=\dfrac{360}{648000}$

$t=0.00055\ sec$

Hence, (a). The force is 1296 N.

(b). The time is 0.00055 sec.

Answer 2

Answer:

May I know from which book you got this question? PLEASE!

Explanation:

ANSWERS:

A) 1296 N

B) 0.00055 s