There is a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0430 kg and...

Question:

There is a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0430 kg and is moving along the x-axis with a velocity of +4.36 m/s. It makes a collision with puck B, which has a mass of 0.0860 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with puck A going left at an angle of 65 degrees and puck B going right at an angle of 37 degrees. Find the speed of (a) puck A and (b) puck B.

Collision

When two or more bodies exerts force on each other due to motion of each one of them is the collision. The collision is of two tyes,

1). Elastic collision: In this type the kinetic energy and momentum both are conserved of whole system.

2).Inelastic collision: In this type only the momentum is conserved of whole system.

Answer and Explanation:

Given:-

  • Mass of Puck A {eq}m_{a}{/eq}
  • Mass of Puck B {eq}m_{b}{/eq}
  • Velocity of Puck A before collision {eq}u_{a}=+4.36\ m/s{/eq}
  • Velocity of Puck B before collision {eq}u_{b}=0{/eq}
  • Let velocity of Puck A after collision {eq}v_{a}{/eq}
  • Let velocity of Puck B after collision {eq}v_{b}{/eq}
  • Angel of path of Puck A after collision {eq}\theta_{a}=65^\circ{/eq}
  • Angel of path of Puck B after collision {eq}\theta_{b}=37^\circ{/eq}

So, according to Law of Conservation of Momentum,

Along x-axis,

{eq}m_{a}\times u_{a}+m_{b}\times u_{b}= m_{a}\times v_{a} \times \cos \theta_{a}+m_{b}\times v_{b}\times \cos \theta_{b} \\ 0.043\times 4.36+0 = 0.043\times v_{a} \times \cos 65 +0.086 \times v_{b}\times \cos 37 \\ 0.187= 0.0181\times v_{a} +0.068 \times v_{b} \\ 0.0181\times v_{a}=0.187 -0.068 \times v_{b} \\ v_{a}=\dfrac{0.187-0.068 \times v_{b}}{0.0181} \\ v_{a}= 10.33-3.75 \times v_{b} \ \ \ \ \ \ eq.1{/eq}

Law of conservation of momentum,

Along y-axis,

{eq}0=m_{a}\times v_{a} \times \sin \theta_{a}-m_{b}\times v_{b}\times \sin \theta_{b} \\ 0= 0.043\times v_{a} \times \sin 65 -0.086 \times v_{b}\times \cos 37 \\ 0= 0.0181\times v_{a} -0.049 \times v_{b} \\ 0.0181\times v_{a} -0.049 \times v_{b}=0 {/eq}

Putting value of {eq}v_{a}{/eq} from equation 1 we get,

{eq}0.0181\times (10.33-3.75 \times v_{b}) -0.049 \times v_{b}=0 \\ 0.186-0.488\times v_{b}-0.067 \times v_{b}=0 \\ 0.186-(0.488-0.067)\times v_{b}=0 \\ 0.186-0.421\times v_{b}=0 \\ 0.421\times v_{b}= 0.186 \\ v_{b}= \dfrac {0.186}{0.421} \\ v_{b}=0.441 \ m/s {/eq}

Now putting the values {eq}v_{b}{/eq} in equation 1 we get,

{eq}v_{a}= 10.33-3.75 \times v_{b} \\ v_{a}=10.33-3.75\times 0.441 \\ v_{a}=10.33-1.65 \\ v_{a}=8.68\ m/s{/eq}


Learn more about this topic:

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Analyzing Elastic & Inelastic Collisions

from College Physics: Help & Review

Chapter 12 / Lesson 5
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