Question

a bullet of mass 10 gram traveeling horizently with the velocity of 150m.strikes a stationary wooden block and comes to rest in 0.03 .calculate the distance penitration of the bullet into the block also calculate the magnitude of force excerted by the wooden block on the bullet​

Answer 1

Hi, here it is -

Mass of bullet=10g

=0.01kg

Initial velocity =150m/s

Time=0.03s

Final velocity =0m/s

Fron newton's first equation of motion

v=u+at

0=150+a(0.03)

-150/0.03=a

a=-15000m/s^2

now, from second equation of motion

s=ut + 1/2at^2

s=150(0.03)+1/2(-5000)(0.03)

s=4.50+(-2500)(0.0009)

s=4.5+(-2.2500)

s=4.5-2.25

s=2.25m

Magnitude of force applied by the bullet on the block -

F=m(a)

F=0.01(-5000)

F=-50.00N

F=-50N

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Hope it helped!