Question

A bullet of mass 10g travelling horizontally with a velocity of 150 metre per second strikes a stationary wooden block and comes to rest in 0.03 seconds. Calculate the distance of penetration of the bullet into the block .Also calculate the magnitude of the force exerted by the wooden block on the bullet ​

Answer 1

Answer:

According To 1st equation of Motion

V = u - at

V = 0

u = 150

t = .03sec

a = 5 × 10^3m/s^2

F = ma

F = 50N

3rd Equation

S = u^2/2a

S = 2.25m

or

Work done by all force = ∆KE

Thanks for Reading

Answer 2

Answer:

given, m=10g, u=150m/s,v=0m/s, t=0.03s

let be the acceleration,

then,

a=v-u/t

= 0-150/0.03 = -5000m/s²= 5000m/s²

(-) sign indicates retardation, so a is taken in positive.

according to equation of motion,

s=ut+1/2at²

 =150*0.03+1/2*5000*0.0009

  =4.5+2.25

  =6.75metres

magnitude of force:

f=m*a

 = 10/1000*5000

  =50Newton.