Answer:
u2 = 0.897 m/s
Explanation:
Note: Question is incomplete ( upon the missing of velocity of the 0.300 kg puck after collision -- i have tried to find as below)
Let the left Puck mass at rest = m1 =0.300 Kg
mass of the right puck m2 = 0.200 kg
velocity of m1 before collision v1= 2.00 m/s
velocity of m2 before collision v2 = 0m/s
velocity of m1 after collision u1 =1.00 m/s
velocity of m2 after collision u2 = ? m/s
θ = 52°
Solution:
Before collision:
Momentum (y-axis ) before collision= 0 Kgm/s
Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.300 Kg x 2.00 m/s + 0
= 0.600 Kgm/s
After collision:
Momentum (y-axis ) after collision= m1u1 sinθ + m2u2 sinθ
= 0.300 x 1.00 m/s sin 52 ° + 0.2 x u2 sin 52°
= 0.236 + 0.158 u2
Momentum (x-axis ) after collision= m1u1 cosθ + m2u2 cos θ
= 0.300 x 1.0 m/s cos 52 ° + 0.200 x u2 cos 52°
= 0.184 + 0.123 u2
According to law of conservation momentum
momentum before collision = momentum after collision.
0 + 0.60 Kgm/s = 0.236 Kgm/s + 0.158 kg u2 + 0.184 Kgm/s + 0.123 kg u2
0.672 Kgm/s = 0.420 Kgm/s + 0.281 u2
u2 = 0.897 m/s
