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A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:

What are its final velocity in the vertical direction and its average velocity in the horizontal direction? </p>
 

We analyze the vertical motion first:

The vertical velocity of the ball is initially zero, its vertical displacement is 122 cm downward and its acceleration in the vertical direction is 980 cm/s^2 in the downward direction.  Choosing the downward direction as positive we therefore have v0 = 0, a = 980 cm/s^2 and `ds = 122 cm.

We use the fourth equation of unif accel motion:

vf^2=v0^2+2a`ds so that

vf=+-sqrt(2(980cm/s^2)*122cm) = +- 489cm/s. The final velocity is clearly downward, which is in the positive vertical direction, so we discard the negative solution.

Using the first equation of motion:

vf=v0+a`dt so that

`dt = (vf - v0) / a = (489cm/s-0cm/s) / (980cm/s^2) =.5s.

Note that `dt could have also been reasoned out from the definition of acceleration.

We now analyze the horizontal motion:

Acceleration in the horizontal direction is zero, so horizontal velocity doesn't change. 

Thus, for the horizontal motion,

• vAve = `ds/`dt = 40cm/.5s = 80cm/s

Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?

What are its speed and direction of motion at this instant?

The x and y components of the velocity relative to a right-hand coordinate system in which y is vertically upward are respectively 80 cm/s and -490 cm/s.  Thus the final velocity has magnitude 

final resultant velocity magnitude= sqrt( (80 cm/s)^2 + (-490 cm/s)^2 ) = 495 cm/s, approx.

and is directed at angle

theta = arcTan(-490 cm/s^2 / (80 cm/s^2) ) = -84 deg,

i.e., at angle 84 deg below horizontal. 

COMMON STUDENT ERROR: 

Magnitude = 1.2839 m
Direction = tan^-1 (1.22m/.4m)
= 71.85 degrees

INSTRUCTOR RESPONSE

You have calculated the angle of the displacement vector from the end of the ramp to the landing position. This isn't the angle of the velocity vector, but for the moment let's consider a little more closely the angle you have calculated.

Using a standard x-y coordinate system, the y displacement would be negative. The x might or might not also be negative, depending on whether your initial velocity was to the right or left (could be drawn either way).

The angle would then come out either -72 deg (same as 288 deg), or 252 deg (72 deg + 180 deg).

 

Now let's consider the question of the velocity vector, which is the vector that indicates speed and direction of motion.

• I believe your figures for x and y velocities came out vf_y = 4.8 m/s and vf_x = .8 m/s.

• As mentioned previously, the y component will be negative, the x component could be either positive or negative.

• It is these components, not the components of the displacement vector, that should be used to calculate the direction of motion.

What is its kinetic energy at this instant? </p>
</p>
answer/question/discussion:</p>
</p>
</p>
 

The .07 kg mass is moving at 495 cm/s, which is 4.95 m/s, so its KE is

• KE = 1/2 m v^2 = 1/2 * .07 kg * (4.95 m/s)^2 = .86 Joules, approx.

</p>
</p>
</p>
What was its kinetic energy as it left the tabletop? </p>
</p>
answer/question/discussion:</p>
</p>
 

As it left the tabletop is was moving at 80 cm/s = .8 m/s so its KE was 

KE at tabletop = 1/2 * .07 kg * (.8 m/s)^2 = .02 Joules, approx.

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</h3>
</p>
</p>
What is the change in its gravitational potential energy from the tabletop to the floor? </p>

From the tabletop to the floor the force exerted on the ball by gravity is in the same direction as the displacement. The work done on the ball by gravity is therefore positive. The work done by the ball against gravity is therefore negative.

The change in PE is equal to the work done by the object against conservative forces, so that in this case the change in gravitational potential energy is negative.

This -.84 J change in gravitational potential energy is equal to the change in the KE (from about .02 J to about .6 J).

Note how energy conversion works here.  From tabletop to floor, KE increases from .02 J to .86 J while PE decreases by .84 J.  We have assumed that the only force is the gravitational force, so that no force is exerted by nonconservative forces, and thus `dW_NC_ON = 0. 

Since `dW_NC_ON = `dKE + `dPE, we expect that `dKE + `dPE = 0. 

This is easily verified: 

`dKE = .86 J - .02 J = .84 J, and

`dPE = -.84 J, so

`dKE + `dPE = +.84 J - .84 J = 0.