I Help Chapter 9: Linear Momentum and Center of Mass Begin Date: 4/18/2016 12:00
ID: 1509095 • Letter: I
Question
I Help Chapter 9: Linear Momentum and Center of Mass Begin Date: 4/18/2016 12:00:00 AM-Due Date: 4262016 3:00.00 PM End Date:4/27/2016 1:00:00 PM (10%) Problem 11: The eight ball, which has a mass of m 0.5 kg, is initially moving with a velocity v = 4.9i ms. It the strikes the six ball, which has an identical mass and is initially at rest. After the collision the eight ball is deflected by an angle of 17° and the six ball is deflected by an angle of = 33.5°, as shown in the figure. -170 33% Part (a) Write an expression for the magnitude of six ball's velocity, in terms of the angles given in the problem and the magnitude of the eight balls initial velocity, v. Grade Ssmmary cos() cos( D) c0s() sin(sin() sin() cotan@) sin(0) 4 5 6 tan(8) detailed view 1% Tm tan( ) Hines: 2 for a doduction. Hints remaining: Feedback:--dedactice per foodback - - Momentum will be conserved in this collision, and is conserved separately in cach direction -Weite the conservation of momentum in both the x and y directions, wsing the components of the velocity. Then find the final velocity by using Pythagorean theorem 33% Part (b) What is the magnitude of the velocity, in meters per second, ofthe six ball? 33% Part (c) What is the magnitude of the velocity of the eight ball, in meters per second, after the collision?Explanation / Answer
a)
since in a collision there is conservation of linear momentum
so initial momentum of system = final momentum
let final velocity of 8 ball = v1 and 6 ball = v2
taking conservation of linear momentum along vertical direction
initial momentum of system = final momentum
0 = 0.5 x v1 sin 17 - 0.5 x v2 sin 33.5
=> v1 = v2 sin 33.5/ sin 17
taking conservation of linear momentum along horizontal direction
initial momentum of system = final momentum
=> 0.5 x 4.9 = 0.5 x v1 cos 17 + 0.5 x v2 x cos 33.5
substituting v1 in terms of v2
=> 0.5 x 4.9= 0.5 x v2 sec 17 sin33.5+ 0.5 x v2 x cos 33.5
=> v2 = 4.9 (sec 17 sin33.5+ cos 33.5)
b)
calculating v2 we get
v2 = 3.472 m/s
c)we know
v1 = v2 sin 33.5/ sin 17
=>v1 = 3.472 x sin 33.5/ sin 17
= 6.555 m/s
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