Question

A 15.0 g bullet is moving to the right with speed 270 m/s when it hits a target and travels an additional 25.0 cm into the target. What are the magnitude (in N) and direction of the stopping force acting on the bullet

Answer 1

Answer:

$F=2187\ N$

Explanation:

Given:

• mass of bullet,$m=15\ g=0.015\ kg$

• initial velocity of bullet, $u=270\ m.s^{-1}$

• displacement of the bullet in the target, $s=25\ cm=0.25\ m$

Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.

• so, final velocity of the bullet, $v=0\ m.s^{-1}$

Now using the equation of motion:

$v^2=u^2+2a.s$

where:

$a=$ acceleration of the bullet

$0^2=270^2+2a\times 0.25$

$a=145800\ m.s^{-2}$

Now the force of resistance offered by the target in stopping it:

$F=m.a$

$F=0.015\times145800$

$F=2187\ N$