Question

A bullet of mass 10g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 sec. Calculate the distance of penetration of the bullet into the

Answer 1

GIVEN ;-

⇒ Mass of the bullet = 10 g

⇒Velocity of the bullet = 150 m / s

⇒ Tie taken by the bullet to take rest when it strikes the wooden block = 0.03 s

TO FIND ;-

⇒ Distance of the penetration of the bullet = ?


SOL :-

⇒ Distance is also reffered to Displacement.


⇒Displacement is the shortest distance travelled by an object.


Therefore ,

          ⇒Displacement                              = Velocity × Time 
                      
                       where average velocity is = (u + v) / 2

⇒ So displacement    =  [(u + v) / 2] × time
                                
                                 = [(150 + 0) / 2] × 0.03
         
                                 = 2.25 m

Distance of the penetration of the bullet is 2.25 m

Answer 2

Explanation:

Mass of bullet => (m) = 10g

=> 10/1000 = 0.01 kg

U(final velocity) = 150m/s

Time taken = 0.03 s

Therefore, bullet comes to rest,

v( initial velocity) = 0/ms

Finding a (acceleration) using 1st equation of motion:

v= u+at

0= 150+a × 0.03

0.03 a= -150

3/100 × a= -150

=> a= -150× 100/3

Therefore, a (acceleration) = -5000m/s^2

Finding distance(s) using the 3rd equation of motion:

v^2 - u^3 = 2as

=> (0)^2 -(150)^2 = 2×(-5000) × s

=> -22500 = 10000 s

=> s = -22500/10000 = 2.25 m

Therefore, Distance of penetration of bullet= 2.25 m

Finding force :

=> F= m× a

=> 0.01 × -5000

= -50 N

Hence, the magnitude of force = 50 N.

(Without unit magnitude is just a value).