A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of ? = 52.0

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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2