A particle with an initial linear momentum of 3.44 kg · m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of 6.88 kg · m/s, directed along the positive y-axis. The final momentum of the first particle is 5.16 kg · m/s, directed 45.0° above the positive x-axis. Find the final momentum of the second particle.
The law of conservation of linear momentum
vector p₁ +vector p₂ =vectorp₁`+vector p₂`
Projections on the axis x and y
p₁⒳ +p₂⒳= p₁`⒳+p₂`⒳
p₁⒴ +p₂⒴= p₁`⒴+p₂`⒴
3.44=5.16•cos45 + p₂`⒳
6.88=5.16•sin45 + p₂`⒴
p₂`⒳ =3.44-5.16•0.707= - 0.17
p₂`⒴=6.88-5.16•0.707= 3.28
p₂=sqrt{ p₂`⒳²+p₂`⒴²}=…
tanβ = p₂`⒴/ p₂`⒳
To find the final momentum of the second particle, we need to use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
Let's break down the given information:
Particle 1:
- Initial momentum (before collision): 3.44 kg·m/s along the positive x-axis.
- Final momentum (after collision): 5.16 kg·m/s at an angle of 45.0 degrees above the positive x-axis.
Particle 2:
- Initial momentum (before collision): 6.88 kg·m/s along the positive y-axis.
Now, let's solve the problem step by step:
Step 1: Resolve the initial momentum of particle 1 into its x and y components:
Given that the initial momentum of particle 1 is along the positive x-axis, we can break it down into x and y components as follows:
Initial momentum of particle 1 in the x-direction (Px1) = 3.44 kg·m/s
Initial momentum of particle 1 in the y-direction (Py1) = 0 kg·m/s
Step 2: Resolve the final momentum of particle 1 into its x and y components:
Given that the final momentum of particle 1 is at an angle of 45.0 degrees above the positive x-axis, we can calculate its x and y components as follows:
Final momentum of particle 1 in the x-direction (Px1') = Final momentum of particle 1 * cos(45°) = 5.16 kg·m/s * cos(45°) = 5.16 kg·m/s * √(2)/2 = 5.16 * √(2)/2 ≈ 3.65 kg·m/s
Final momentum of particle 1 in the y-direction (Py1') = Final momentum of particle 1 * sin(45°) = 5.16 kg·m/s * sin(45°) = 5.16 kg·m/s * √(2)/2 = 5.16 * √(2)/2 ≈ 3.65 kg·m/s
Step 3: Apply the law of conservation of momentum:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
Initial momentum of particle 1 + Initial momentum of particle 2 = Final momentum of particle 1 + Final momentum of particle 2
(Px1) + (Py1) + (Px2) + (Py2) = (Px1') + (Py1') + (Px2') + (Py2')
Substituting the given values:
3.44 kg·m/s + 0 kg·m/s + 6.88 kg·m/s + 0 kg·m/s = 3.65 kg·m/s + 3.65 kg·m/s + (Px2') + (Py2')
Simplifying the equation:
10.32 kg·m/s = 7.3 kg·m/s + (Px2') + (Py2')
Now, we can isolate the variables and solve for the final momentum of particle 2:
(Px2') + (Py2') = 10.32 kg·m/s - 7.3 kg·m/s
(Px2') + (Py2') = 3.02 kg·m/s
Finally, the final momentum of particle 2 is equal to the vector sum of its x and y components:
Final momentum of particle 2 = √((Px2')^2 + (Py2')^2) = √(3.02^2) = √9.12 ≈ 3.02 kg·m/s
Therefore, the final momentum of the second particle is approximately 3.02 kg·m/s.