Question

hockey pucks on a smooth ice rink. Assume that by analyzing the videotape of the collision, various but not all components of the velocities of the two pucks beforeand after the collision have been determined and are recorded in the table below. .
BEFORE
vx (cm/s) BEFORE
vy (cm/s) AFTER
vx (cm/s) AFTER
vy (cm/s)
Puck 1 21 35 –13 10
Puck 2 –46 15
1). Assume that friction with ice is negligible. Complete the missing entries. 1. Puck 2, after collision vx (in cm/s):(NOTE: The masses of the pucks are not providedbecause they are not needed... Do you see why?)
2. Puck 2, after collision vy (in cm/s):
Answer
3. What percentage of energy is lost in the collision? 3 (Note that the mass of the pucks is NOT needed for this question. If you are unsure why this is so, try usinga mass of 1.0 kg for each puck.
4. In the lab, you will be analyzing the collisions of two pucks on an air table. The pucks are connected to an electric pulsing system that provides a brief electricpulse 60 times per second. Each pulse results in a spark being produced between the puck and the table. By placing a sheet of paper between the puck and the table, wecan use the sparks to record the position of the pucks at regular intervals. The trails of spark marks look very much like the image in the Prelab page that you willfind at the beginning of the writeup for the collisions lab in the manual (or you can print it from here).Let us analyze this provided sample collision. Use the rulerand protractor to answer the following questions. What is the speed of puck 2 after the collision, in cm/s? (A 5% margin of error is allowed for this question)
5. Let us take the x axis along the initial direction of puck 1, and y axis perpendicular to that and pointing up. Use the protractor to determine the direction ofmotion of puck 2 after the collision, in degrees.
6. Now determine the components of the velocity of puck 2 after the collision. vx, in cm/s:
7 vy, in cm/s: 3
Answer
8)
As discussed in previous labs, uncertainties are propagated through calculations.


Let us assume that these 8.5 cm, with the corresponding uncertainty, correspond to a trail of 13 spark marks, or 12 time intervals of 1/60 s. The speed is thus:



Let us assume for simplicity that the rate of spark production is very precise (i.e., the uncertainty associated with the time intervals being 1/60 s introduces a verysmall contribution to the total error).


But the uncertainty associated with the use of a cheap plastic ruler should still be accounted for...

What is the uncertainty in the result above, in cm/s?
Answer
Selected Answer: [None Given]

Answer 1

The identical hockey pucks are colli de on smooth ice rink. Initi al velocity components of two pucks are 21m/s; v 35m/s Final velocity component of first puck is, 13m/s, -10m/s a) The x-component of velocity of second puck after collision is, Apply law of conservation of momentum along x-axis b) The y-component of velocity of second puck after colli si on is Apply law of conservation of momentum along x-axis %2,- ..ly+42,-w-35m/s +1 5 m/s_ (10m/s)-40m/s (c) The initial speed of the first puck is v-(21m/s)+(35m/s)? 40.82m/s The initial speed of the second puck is v,2-(-46m/s) +(15m/s)48.38m/s The final speed of the first puck is, vi-(-13m/s) +(10m/s)16.40m/s The initial speed of the second puck is %2-M-12m/s), (40m/s)' = 41.76 m/s Initi al kinetic energy of the system is, +7%y' = (nz/ 2)| (40.82m/s)' +(48.38m/s)' | = (nz/ 2)4006. 896J Final kinetic energy of the system is tmvm2) (16.40m/s)" +(41.76m/s)(m 2) (2012.857)J The percentage of energy lost K,-K, 2012.857J-4006.896. -49.7% (or) 50% K, 4006 896J

Answer 2

Using conservation of momentum, in x- direction,

$m_{1}u_{1}^{x} + m_{2}u_{2}^{x} = m_{1}v_{1}^{x}+m_{2}v_{2}^{x}$

$21m_{1} + (-46)m_{2} = (-13)m_{1}+m_{2}v_{2}^{x}$

$m_{1} =m_{2} = m$

$v_{2}^{x} =- 25+13 = -12cm/s$

similarly, conservation of momentum for, y directtion,,

$m_{1}u_{1}^{y} + m_{2}u_{2}^{y} = m_{1}v_{1}^{y}+m_{2}v_{2}^{y}$

$35m + 15m = 10m+mv_{2}^{y}$

$v_{2}^{y}= 30 cm/s$

Speed of puck 1 before collision = $\sqrt{21^2+35^2}}$ = 40.82 cm/s

speed of puck 2 before collision =$\sqrt{-46^2+15^2}}$ = 48.38 cm/s

speed of puck 1 after collision = $\sqrt{-13^2+10^2}}$ = 16.40 cm/s

speed of puck 2 after collision = $\sqrt{-12^2+40^2}}$ = 41.23 cm/s

now, kinetic energy before collison,

$KE_{initial} = \frac{1}{2}m(40.82)^2+\frac{1}{2}m(48.38)^2 = (2003.5)$ (for unit mass m = 1kg)

and

$KE_{final} = \frac{1}{2}m(16.40)^2+\frac{1}{2}m(41.23)^2 = 1006.5$ (for m =1kg)

$percentage of KE_{lost}= \frac{2003.5 - 1006.5}{2003.5}$

so percentage lost in kE is 49.76

Answer 3