The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.
Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.
The given parameters are;
The mathematical representation of the above question will be in two components.
Horizontal component
M1U1 - M2U2 = M1V1cosФ - M2V2cosФ
Substitute all the parameters
0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42
0.1443 - 0.1936 = 0.13V1cosФ - 0.1379
0.13V1cosФ = 0.0886
V1cosФ = 0.0886/0.13
V1cosФ = 0.6815 ........ (1)
Vertical component
0 = M1V1sinФ - M2V2sinФ
M1V1sinФ = M2V2sinФ
Substitute all the parameters
0.13 x V1 sinФ = 0.16 x 1.16sin42
V1 sinФ = 0.1242/0.13
V1 sinФ = 0.9553 ......... (2)
Divide equation 2 by 1
V1 sinФ / V1 cosФ = 0.9553/ 0.6815
Tan Ф = 1.40
Ф = [tex]Tan^{-1}[/tex](1.4)
Ф = 54.5°
Substitute Ф into equation 2
V1 sin54.5 = 0.9553
V1 = 0.9553 / 0.8141
V1 = 1.17 m/s
Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
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