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A small steel ball is rolled down an inclined plane from a table. The inclined plane is a
slotted ruler. One end of the ruler is 10cm off of the table. The ball leaves the table
horizontally and lands 46cm from the table on the floor. The table is 72 cm from the
ground.
1. Using projectile motion equations, determine the speed of the ball the instant it
leaves the table.
The initial vertical component of the ball as it leaves the table is zero and the
acceleration is g = 9.8 m/s2 in downward direction, hence the time taken by the ball to
reach the ground t is given by using the equation of motion {s = ut + (1/2)*at 2} for vertical
direction as
0.72 = 0*t + (1/2)*9.8*t2
Gives t 
2 * 0.72
 0.3833 s.
9.8
Now as there is no force is acting on the ball in horizontal direction the horizontal velocity
v0 remains constant and hence the horizontal distance covered in this time t during the
fall is given by the equation
x = v0*t
Gives v0 = x/t = 0.46/0.3833 = 1.2000 m/s
Alternative method
Using the equation of trajectory of a projectile also we can calculate the same velocity as
y  x tan  
gx 2
2v02 cos 2 
where y = 0.72 m is the vertical displacement, x = 0.46 m is the horizontal displacement,
 = 0 is the angle of projection. Substituting the values we get
 0.72  0.46 * tan 0 
or
9.8 * 0.46 2
2 * 0.72
v 0  1.2000m / s
v 02 
9.8 * 0.46 2
2v02 cos 2 0
2. Compare the potential energy of the ball prior to its release to the kinetic energy
of the ball just as it leaves the table.
The kinetic energy of the ball at the time it leaves the table is given by
KE = ½ mv02 = 0.5*m*1.20002 = 0.72*m J
Where m is the mass of the ball
The height of the ball when it is released with respect to the table surface is given by
PE = m*g*h = m*9.8*0.1 = 0.98*m J
So the ratio of the two is
PE/KE = 0.98*m/(0.72*m) = 1.36
Or PE is 1.36 times the kinetic energy
(this is because the rolling ball will have some kinetic energy in form of rotational
kinetic energy as well)
--------------------------------------A second ball is placed in the slot of the ruler at the edge of the table (the end of the
inclined plane). The first ball rolls down the incline and collides with the second ball.
The ball lands 44cm from the table on the floor.
3. Determine the speed at which the ball is projected off the table.
Again solving in the similar way the initial speed v1of second ball is given by
y  x tan  
gx 2
2v12 cos 2 
or
 0.72  0.44 * tan 0 
or
9.8 * 0.44 2
v 
2 * 0.72
v1  1.1478m / s
9.8 * 0.44 2
2v12 cos 2 0
2
1
4. Using the conservation of energy, determine the speed of the first ball at the
bottom of the ruler just before it hits the second ball.
As the first ball looses the height of 0.1 m on rolling down the incline, and as nonconservative forces are absent the potential energy lost will convert in the kinetic energy
of the first ball and hence using the law of conservation of mechanical energy the kinetic
energy at the end of the incline is given by
½ mv2 = m*g*h
or
v2 = 2.g*h = 2*9.8*0.1
or
v = 1.4 m/s
5. Since the balls are exactly the same, the conservation of momentum dictates that the
momentum of the first ball is transferred to the second ball. Using the above data, show
this relationship.
If the collision is perfectly elastic and the mass of the two balls are equal let the two balls
are moving with velocities v1 and v2, and as there is not force on the system in horizontal
direction, according to the law of conservation of linear momentum we have
Or
Or
Final momentum of the system = Initial momentum
mv1 + mv2 = m*1.4 + m*0
v1 + v2 = 1.4
---------- (1)
Now applying the law of conservation of energy we have
½ mv12 + ½ mv22 = ½ m*1.42
or
v12 + v22 = 1.42
------------ (2)
Substituting the value of v2 in equation 2 from equation 1 we have
v12 + (1.4 - v1)2 = 1.42
or
v12 + 1.42 – 2*1.4*v1 + v12 = 1.42
or
2*v12 – 2*1.4*v1 = 0
or
v1 = 0, and
v1 = 1.4 m/s (not possible)
hence v1 = 0 and hence v2 = 1.4 m/s
This shows that the whole momentum of the first ball is transferred to the second ball.
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