Question

a)   A bullet of mass 0.25 kg travels at a speed of 180 ms-1 and deeply penetrates into a fixed target object. It brings to rest in 0.14 s. Find
i)   the distance of penetration of the target object,
ii)   the magnitude of the average retarding force exerted on the bullet.

b)   If a kid on an amusement rides moves at a constant speed in a horizontal circle of diameter 10.0 m,

i) making a complete circle (one complete revolution) in 3.0 s, what is the magnitude of the centripetal acceleration?

Answer 1

(a)

m = 0.25 kg

initial velocity, u = 180 m/s

final velocity, v = 0

t = 0.14 s

Deceleration, a = (v-u)/t = (0-180)/0.14 = -1285.71 m/s²

(i) The distance it penetrates is given by Newton's Equatiions

2 a s = v² - u²

s = - u² / 2* a = -(180)² / 2 * (-1285.71) = 12.6 m

(ii) Retarding Force = m a = 0.25 * -1285.71 = -321.43 N (The negative sign represents retardation)

(b)

Radius, R = 10/2 = 5 m

Time taken for 1 revolution, T = 3 s

Velocity of the ride = total distance covered in one round / time taken for 1 revolution = 2* pi* R / T = 2 * pi * 5 / 3 = 10.47 m/s

Centripetal acceleration, a = v² / R = (10.47)² / 5 = 21.93 m/s²