Question

A 0.439 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.307 kg puck moving initially along the x axis with a speed of 2.19 m/s. After the collision, the 0.307 kg puck has a speed of 1.19 m/s at an angle of 37◦ to the positive x axis. Determine the magnitude of the velocity of the 0.439 kg puck after the collision

Answer 1

u2 = 0.266 m/s

Step-by-step 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°

Solution:

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