Question

a bullet hits a 4cm thick wall at a velocity of 20 m/s and decelerates at the rate of 4000m/s^2. will the bullet cross to the other side of the wall or remain embedded in it?

Answer 1

Answer:

Explanation:

Let the mass of the bullet be m. The initial kinetic energy of the bullet= ½mv².

KE=½×m×20×20=200m.

This much energy is spent in penetrating the wall upto 5cm. Let the force be F.

Work done =Force×displacement.

W=F×5×10-².

So, 200m= ma×0.05 (F=ma)

Therefore answer= a= 4000 ms-².

Answer 2

Answer:

Explanation:

We basically need to find at what distance(s), the final velocity becomes zero, in the brick medium.

(When it hits the brick):

u = 20m/s

a = -4000m/s²

v = 0m/s (comes to rest)

We have the equation:

v²-u²=2as

So, s = (v²-u²)/2a

s = (0-400)/2(-4000)

= 400/8000

= 0.05m

= 5 cm

Since the thickness of the brick is 4cm, and the bullet comes to rest at 5cm (in the brick medium), the bullet crosses to the other side