Youssouf www.yahoo.com The Expert TA Human-like A 5.75-kg Bowling Ball Mo C Secu

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Youssouf www.yahoo.com The Expert TA Human-like A 5.75-kg Bowling Ball Mo C Secure https://usu34ny.theexpertta.com/Commonw TakeTutorialAssignment.aspx Home I Student: kantB16401redonia.edu My Account Log out Class Management Help Ch 9 assignment Begin Date: 47/2017 12:00:00 AM Due Date: 4/14/2017 59:00 PM End Date 4/2017 59:00 PM (9%) Problem 10: A 5.75. kg bowling ball moving at 8.5 m/s collides with a 0.825. kg bowling pin, which is scattered at an angle of 870 rom the initial direction of the bowling ball, with a speed of 15.5 m/s. Assignment Status Click here for A 50% Part (a) Calculate the direction, in degrees, ofthe final velocity of the bowling ball. This angle should be measured in the same way detailed view that is, Grade Summary Deduct Problem Status Potential Completed sino cos0 tan0 7 8 9 Submissions Partial Attempts remaining: 3 cotan() asino acos0 4 5 6 3 Completed per attempt atano acotano sinh0 detailed view 4 Completed 1 2 3 tanho cotanh0 Completed ODegrees O Radians Completed 7 Completed Completed 9 Completed Hints: 3 duction per hint. Hints remaining: 2 Feedback: 0 deduction per feedback. 10 Submission History Feedback Totals Totals 0% 0% 0% 50% Part Calculate the magnitude of the final velocity, in meters per second, of the bowling ba

Explanation / Answer

Let us take the direction of the bowling ball before collision to be along x-axis.

Let us suppose i and j be unit vectors along x and y axis respectively.

Let mass of the bowling ball be M = 5.75kg

Let mass of the bowling pin be m = 0.825kg

Momentum of ball+pin system before collision = Mvx = (5.75kg)(8.5m/s) = 48.875kg-m/si

After collision let us consider the bowling pin. It moves at a speed of 15.5m/s at an angle = 870 with the initial direction of the ball which is also the x-axis direction. So, after collision the pin moves at an angle = 870 with the x-axis.

So momentum of the pin after collision = mvxi + mvyj = (0.825kg)[(15.5m/s)cos870i + (15.5m/s)sin870j]

or, momentum of the pin after collision = 12.7875(cos870i + sin870j)

Let, the ball moves with a speed v at an angle 0 with the x-axis after collision, then,

momentum of the ball after collision = M[vcos0i + vsin0j]

Now we apply conservation of linear momentum:

Net momentum before collision = Net momentum after collision

or 48.875kg-m/si = 12.7875(cos870i + sin870j) +  M[vcos0i + vsin0j], ------------ (1) which gives:

48.875 = 12.7875cos870 + Mvcos0

Mvcos0 = 48.2057kg-m/s --------- (2)

equation (1) also gives:

0 = 12.7875sin870 + Mvsin0

or Mvsin0 = -12.7699 ----------- (3)

from equation (2) and (3) we get

tan0 = -12.7699/48.2057

0 = -14.8370

So, the final velocity of the ball is at an angle 14.8370 with the initial direction of the ball measured in opposite direction to the angle measured for the pin after collision. So if the angle is measures in the same way that is, which is in counterclockwise direction with the initial direction of the ball, then the value of 0 can be taken as either -14.8370 or +345.1630.

b)from equation (2) and (3) above, we get:

(Mv)2 = (-12.7699)2 + (48.2057)2

or Mv = 49.868

or v = 8.672m/s

So, the magnitude of the final velocity of the ball is  v = 8.672m/s

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

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