Answer:
u2 = 0.266 m/s
Explanation:
Let the left Puck mass at rest = m1 =0.307 Kg
mass of the right puck m2 = 0.439 kg
velocity of m1 before collision v1= 2.19 m/s
velocity of m2 before collision v2 = 0m/s
velocity of m1 after collision u1 =1.19 m/s
velocity of m2 after collision u2 = ? m/s
θ = 37°
<u>Solution:</u>
Before collision:
Momentum (y-axis ) before collision= 0 Kgm/s
Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.307 Kg x 2.19 m/s + 0
= 0.672 Kgm/s
After collision:
Momentum (y-axis ) after collision= m1u1 sinθ + m2u2 sinθ
= 0.307 x 1.19 m/s sin 37 ° + 0.439 x u2 sin 37°
= 0.22 + 0.26 u2
Momentum (x-axis ) after collision= m1u1 cosθ + m2u2 cos θ
= 0.307 x 1.19 m/s cos 37 ° + 0.439 x u2 cos 37°
= 0.29 + 0.35 u2
According to law of conservation momentum
momentum before collision = momentum after collision.
0 + 0.672 Kgm/s = 0.22 Kgm/s + 0.26 kg u2 + 0.29 Kgm/s + 0.35 kg u2
0.672 Kgm/s = 0.51 Kgm/s + 0.61 u2
u2 = 0.266 m/s
