I have no idea to solve from 4 to 8 I got 1. -12 2. 40 3. 49.76 Prelab 2DC 2D co

Question

I have no idea to solve from 4 to 8

I got

1. -12

2. 40

3. 49.76

Prelab 2DC 2D collisions Credit will only be given when answers are entered on the Blackboard "test". A boxindicates that an individualized numerical value will be given for that quantity in that "test". Consider the collision of two identical hockey pucks on a smooth ice rink. Various but not all components of the velocities of the two pucks before and after the collision have been determined and are recorded in the table below BEFORE BEFORE AFTER AFTER Vx (cm/s)Vy (cm/s)Vx (cm/s) Vy (cm/s) Puck 1 21 35 -13 10 Puck 2 -46 15 Assume that friction with ice is negligible. Complete the missing entries. (NOTE: The masses of the pucks are not provided because they are not needed... Do you see why?) 1. Puck 2, Vx (in cm/s) after collision: 2. Puck 2, Vy (in cm/s) after collision: 3. What percentage of kinetic energy is lost in the collision? Prelab 2DC- Page 1

Explanation / Answer

Puck 1, u1 = 1 cm*60 /s = 0.6 m/s along x axis

Puck 2, u2 = 0.6 cm/s along y axis

m1*u1 = m2*v2*cos(22)
m2*u2 = m1*v1 + m2*v1*sin(22)

(4) the speed of puck 2 after the collision is
v2 = 0.6*(cos(22) i +sin(22) j)
speed = 0.6 m/s

(5)  the angle between the x-axis and direction of motion of puck 2 after the collision is

22 degrees

the components of the velocity of puck 2 after the collision is

(6)

Vx = 55.63 cm/s

(7)

Vy = 22.47 cm/

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