PHY201
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
• For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
V0 = 20cm/s
Displacement = 120cm
A=9.8m/s^2
• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
V^2 = V0^2+2a(x-x0)=20cm^2+2*9.8m/s^2*120cm = 2752cm/s
Your units are not correct so your result is very likely incorrect. You need to recalculate vf.
m/s^2 * cm is not equal to cm/s.
Also note that the initial velocity is not 20 cm, and that the entire velocity is squared and should therefore be in parentheses.
Vf=52.5cm/s
Displacement = 120cm
Change in velocity = 52.5cm/s – 20cm/s = 32.5cm/s
Average velocity = (52.5+20)/2 = 36.25cm/s
• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
A=0
V0 = 80cm/s
T=sqrt(2y/-g) = sqrt((2*120)/-9.8)= 4.9s
The quantities in sqrt((2*120)/-9.8) which have units need to be expressed in units. Also there is no quantity in the horizontal direction that has magnitude 9.8. 9.8 m/s^2 is an acceleration in the vertical direction, and does not apply to horizontal motion.
• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
displacement=vx0t = 80cm/s * 4.9s = 396cm
final velocity = vx = vx0 = 80cm/s
average velocity = (80cm/s+80cm/s)/2 = 80cm/s
change in velocity = 80 – 80=0cm/s
You haven't calculated the horizontal displacement.
• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No because the force exerted by the ball on the floor is met by the force of the floor on the ball.
• Why does this analysis stop at the instant of impact with the floor?
Because acceleration and velocity are all at 0 at the instant of impact with the floor.
The analysis assumes uniform acceleration. Acceleration ceases to be uniform when the ball contacts the floor.
** **
1hr
** **
Overall your approach is good, but there are some errors in details.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#