Question

two small spheres hold electric charge of opposite signs, which means they are attracted to each other. these spheres collide with each other, but the collision is too hard for the spheres to remain bound to each other by their mutual attraction, and they bounce off. a little while later, however, the attraction once again draws them together, and this time when they collide, they do bond. their first collision occurs on what we will call the x - axis, and when this collision occurs, their velocity vectors make the same angle with the x - axis: θ=21.1°. the sphere below the x - axis has 4.2 times the mass of the sphere above the x - axis, and both spheres have the same speed of 7.8m/s at the time of the collision. find the speed of the two - sphere combination after they are bound to each other. provide at least two decimal places

Explanation:

Step1: Calculate the x - components of momentum

Let the mass of the sphere above the \(x\) - axis be \(m\), then the mass of the sphere below the \(x\) - axis is \(4.2m\).
The speed of each sphere is \(v = 7.8\frac{m}{s}\).
The x - component of momentum of the sphere above the \(x\) - axis: \(p_{1x}=mv\cos\theta\)
The x - component of momentum of the sphere below the \(x\) - axis: \(p_{2x}=4.2mv\cos\theta\)
The total initial x - component of momentum \(p_{ix}=mv\cos\theta + 4.2mv\cos\theta=(1 + 4.2)mv\cos\theta=5.2mv\cos\theta\)

Step2: Calculate the y - components of momentum

The y - component of momentum of the sphere above the \(x\) - axis: \(p_{1y}=mv\sin\theta\)
The y - component of momentum of the sphere below the \(x\) - axis: \(p_{2y}=- 4.2mv\sin\theta\)
The total initial y - component of momentum \(p_{iy}=mv\sin\theta-4.2mv\sin\theta=(1 - 4.2)mv\sin\theta=- 3.2mv\sin\theta\)

Step3: Calculate the total mass after collision

The total mass after collision \(M=m + 4.2m=5.2m\)

Step4: Calculate the x - and y - components of velocity after collision

By conservation of momentum:
\(p_{ix}=Mv_{fx}\), so \(v_{fx}=\frac{p_{ix}}{M}=\frac{5.2mv\cos\theta}{5.2m}=v\cos\theta\)
\(p_{iy}=Mv_{fy}\), so \(v_{fy}=\frac{p_{iy}}{M}=\frac{-3.2mv\sin\theta}{5.2m}=-\frac{3.2}{5.2}v\sin\theta\)

Step5: Calculate the speed after collision

The speed \(v_f=\sqrt{v_{fx}^2 + v_{fy}^2}\)
Substitute \(v = 7.8\frac{m}{s}\) and \(\theta = 21.1^{\circ}\)
\(v_{fx}=7.8\cos(21.1^{\circ})\)
\(\cos(21.1^{\circ})\approx0.933\), so \(v_{fx}\approx7.8\times0.933 = 7.28\)
\(v_{fy}=-\frac{3.2}{5.2}\times7.8\sin(21.1^{\circ})\)
\(\sin(21.1^{\circ})\approx0.360\), \(\frac{3.2}{5.2}\times7.8 = 4.8\), so \(v_{fy}\approx-4.8\times0.360=-1.73\)
\(v_f=\sqrt{(7.28)^2+(-1.73)^2}=\sqrt{52.9984 + 2.9929}=\sqrt{55.9913}\approx7.48\)

Answer:

\(7.48\frac{m}{s}\)