Question

plz solve the below question ■■■■

A bullet of mass 4 g fired with a velocity of 50 m/s enters a wall up to depth of 10 cm. Calculate the average resistance offered by the wall. ​

Answer 1

The resistance offered will be expressed in terms of retardation or negative acceleration (m/s²)

To find the resistance we use the 3rd equation of motion

v2=u2+2asv2=u2+2as

vv = final velocity

uu = initial velocity

aa = acceleration (the unknown variable)

ss = displacement

After plugging in the values we get this linear equation

0=502+2(0.1)a0=502+2(0.1)a

vv is 0 as the bullet has stopped and displacement is 0.1 because 10 cm is converted to metres

Solving this equation

a=−12500m/s²a=−12500m/s²

Therefore the wall retards the bullet by 12500m/s²12500m/s² or the resistance offered by the wall is 12500m/s²12500m/s²

Note: the mass of the bullet wont affect the answer and is just given to throw you off track I suppose

EDIT

As the unit of resistance isn't specified in the question the mass can be multiplied with the retardation to find the resistance in terms of Newtons

The answer comes out to be -50 Newtons therefore the resistance offered is 50 Newtons

Answer 2

Given: Mass of Bullet = 0.004 kg

Initial Velocity (u) = 50 m/s

Final velocity = 0m/s

Penetration (s)= 0.1 m

v² - u² = 2as

⇒a = [0 - 2500]/(2)(0.1) = -1250m/s²

F = ma

⇒F = - 0.004 x 1250 = - 100N

The Force Exterted By Bullet on The wall is -100 N SO Acc to Newtons Third's Law The Force Exerted by wall on Bullet is +100N