Question

A coconut of mass 1.2kg is shot at with a 20g bullet. The bullet approaches the coconut at a velocity
v0=(400,0) m/s and exits at a velocity v1=(220,-80) m/s. (a) Calculate the velocity of the coconut
immediately after the collision. (b) Calculate the amount of kinetic energy lost in the collision.

Answer 1

here,

mass of coconut , m1 = 1.2 kg

mass of bullet , m2 = 20 g = 0.02 kg

the intial speed of bullet , u2 = 400 m/s i

the final speed of bullet , v2 = 220 i m/s - 80 j m/s

|v2| = sqrt(220^2 + 80^2) = 234.1 m/s

a)

let the veocity of coconut after the collison be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

0.02 * 400 i = 1.2 * v1 + 0.02 * ( 220 i - 80 j)

solving for v1

v1 = 3 i m/s + 1.33 j m/s

the velocity of coconut , v1 = ( 3 , 1.33) m/s

the magnitude of velocity , |v1| = sqrt(3^2 + 1.33^2) = 3.28 m/s

b)

the amount of kinetic energy lost in the collision, KE = (0.5 * m2 * u2^2 - 0.5 * m2 * v2^2 - 0.5 * m1 * v1^2)

KE = 0.5 * ( 0.02 * 400^2 - 0.02 * 234.1^2 - 1.2 * 3.28^2)

KE = 1045.5 J