Two wooden pucks approach each other on an ice rink as shown in the figure. Puck
ID: 777712 • Letter: T
Question
Two wooden pucks approach each other on an ice rink as shown in the figure. Puck #2 has an initial speed of 4.96 m/s and a mass that is some fraction f = (2/3) that of puck #1. Puck #1 is made of a hard wood and puck #2 is made of a very soft wood. As a result, when they collide, puck #1 makes a dent in puck #2 and 10.0% of the initial kinetic energy of the two pucks is lost. Before the collision, the two pucks approach each other in such a manner their momentums are of equal magnitude and opposite directions. Determine the speed of the two pucks after the collision
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Explanation / Answer
mass of Puck 1 = m
mass of puck2 = 2m/3
vel. of puck = 4.96 m/s
both have same momentum
vel. of puck 1 = (2m/3 *4.96 )/m = 3.31 m/s
Totel KE befors collision = 0.5m *3.312 + 0.5* 2m/3 * 4.962
= 13.68m J
10% KE is lost in collision
Total KE after collision = 12.31m J
let v1 and v2 be the velocities after collision
0.5m v12 + 0.5 2m/3 *v22 = 12.31m
3v12 + 2v22 = 73.86 ---(1)
momentum is conserved
Total momentum along x-axis before collision =0
after collision
0.5 mv1Cos(50) - 0.5 2m/3 *v2 Cos(50) =0
3v1 = 2v2 --- (2)
solving 1 ,2
v1 = 3.14 m/s , v2 = 4.71 m/s
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