Question

A bullet of mass 10g traveling with a velocity of 100 m/s penetrates in a wooden plank and is brought to rest in 0.01s Find :
(a) the distance through which the bullet penetrates in the wooden plank
(b) the force exerted on the bullet.

Answer 1

Explanation:

Given,

mass m= 10g = 0.01 kg

initial velocity u= 150 m/sec

Bullet finally come in rest, so final velocity, v= 0 m/sec

Time t= 0.03 sec

Now, from equation

v= u+at

0= 150+a(0.03)

acceleration a= -150/0.03 = -5000 m/sec2

From equation v2 -u2 = 2as

02 - (150)2 = 2×(-5000)×s

So, s= 150×150/(2×5000)

s= 2.25 m

The penetration distance of the bullet in the wooden block =2.25 m

and magnitude of force, F = ma

=0.01×5000 = 50 N

Answer 2

Answer:

a) 0.5m

b) -100N

Explanation:

Mass of the bullet (m) = 10g = 0.01kg

Time taken to be brought to rest (t) = 0.01s

Velocity of the bullet before penetrating (u) = 100 m/s

Velocity of the bullet after penetrating (v) = 0 m/s

a) Now, we need to find the distance it penetrated in the wooden plank. So we can use the formula:

v² - u² = 2as

But first, we need to find acceleration (a).

Using v = u + at to find acceleration:

0 = 100 + (0.01)(a)

-100 = (0.01)(a)

Hence acceleration is - 10,000 m/s²

Mow substitute the values in v² - u² = 2as

0 - 10000 = 2(-10000)s

-10000 = -20000(s)

=> s = 0.5 meters

b) Now, we know F = ma

Hence:

F = (0.01)(-10000)

=> F = -100 N

Hope that you will find this useful :)

Mark me as the brainliest!!