Introduction
When a ball rolls off a table, gravity will act on it and pull it down towards the ground. While the ball is rolling over the table, it has an initial velocity that is horizontal in direction before it leaves the table. As it leaves the table and gravity starts acting on the ball, it will follow a parabolic trajectory as shown in the following figure.
Figure 1 : Ball Rolling Off a Table Path of Motion
The first problem is how to find where the ball will bounce off the ground the first time. The second problem is where the ball bounces off a second time with the following assumptions: same angle it hits the floor, 20% less energy.
Analysis
As shown in figure 1, the velocity of the ball as it rolls off from the table has two components: horizontal component, and vertical component. The horizontal component is the initial velocity of the ball when it is still travelling over the table. The vertical component is the effect of gravity on the ball. As illustrated in figure 1, the initial vertical velocity before the ball rolls off from the table is zero. As the ball leaves the table, the vertical velocity is slowly increasing until it hits the ground.
Horizontal:
x=vxt
vfx=vix=vx
ax=0
Vertical:
y=viyy+12gt2
vfy=viy+gt, vfy2=viy2+2gt
ay=g=-9.81ms2
In these equations, the variable t is the time, the variables x and y are the horizontal and vertical positions respectively, the variables vx and vy are the horizontal and vertical velocities respectively, the variables ax and ay are the horizontal and vertical accelerations respectively, and the subscripts i and f stand for initial and final respectively.
At this point, the formula for kinetic energy must be noted:
Kinetic Energy KE=12mv2
Where m is the mass of the ball, and v is the velocity of the ball.
The assumptions of this problem are the following: initial vertical velocity is zero viy=0, the mass of the ball is constant, the surfaces touched by the ball are frictionless and the effect of wind is negligible, thus the horizontal velocity is constant vfx=vix=vx. In this problem, the values given are the initial horizontal velocity of the ball vx and the height of the table. Let the height of the table be equal to h.
The time it takes for the ball to hit the ground can be computed by simplifying the equation:
h=12gt2
After finding the time t, the distance from the table can be computed by:
x=vxt
All the necessary details are now available to solve for the problem questions.
Solutions
Assume that the initial horizontal velocity of the ball is vx=3.2ms , and the table height is h=1.5 m. The time it takes for the ball to hit the ground is:
h=12gt2
-1.5 m=12-9.81ms2t2
t2=-1.5m-4.905ms2
t2=0.3058 s2→t=0.5530 s
The horizontal distance travelled by the ball is:
x=vxt
x=3.2ms0.5503s
x=1.7696 m
If the energy is 20% less, it will be translated to a decrease in the velocity. The velocity of the ball as it hits the ground is:
vy=gt=-9.81ms20.5503s=5.398ms
v=vx2+vy2=3.22+5.3982=6.275ms
The KE at this point is:
KE=12mv2=12m6.2752=19.69m
With a 20% decrease, the velocity becomes
Interpretation
Discussion
References
Naigle, D 2009, Module 3: Horizontally Launched Projectiles, viewed 6 June 2016, <http://homepage.usask.ca/~dln136/projectile/pages/module3.html>.
Nave, R 2014, Kinetic Energy, viewed 6 June 2016, <http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html>.
