Question

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg.She finds that the target ball travels a distance of 1.50 m after it is struck. Assume

g = 9.80 m/s².

What is the kinetic energy (in joules) of the target ball just after it is struck?

Answer 1

Here,

Height above ground = y = 0.860 m
mass of steel ball = 0.0120 Kg
D = 1.50m
g = 9.8 m/s^2

assuming ball travels only horizonatlly,
time taken by ball to reach the Ground,

y = 1/2 * a * t^2
0.860 = 1/2 * 9.8 *t^2

t = sqrt(0.860/4.9)
t = 0.41 s

Speed = Distance travelled / time
V = 1.5/0.41
V = 3.65 m/s

ThereFore

KE = 1/2 * M * V^2
KE = 1/2 * 0.0120 * 3.65 *3.65
KE = 0.079 J

the kinetic energy of the target ball just after it is struck is 0.079 J