Answer to Question #102693 in Mechanics | Relativity for sakshi

Question #102693
A stationary ball, with a mass of 0.5 kg, is struck by an identical ball moving at 60m/s . After the collision, the second ball moves 30° to the left of its original direction. The stationary ball moves 60° to the right of the moving ball’s original direction. What is the velocity of each ball after the collision?
1
Expert's answer
2020-02-21T10:09:14-0500

We can find the velocity of each ball after the collision from the law of conservation of momentum. Let's apply the law of conservation of momentum along the "x"- and "y"-axis:


"mv_{1,i} + mv_{2,i} = mv_{1,f}cos\\alpha + mv_{2,f}cos\\theta, (1)""0 = mv_{1,f}sin\\alpha - mv_{2,f}sin\\theta, (2)"

here, "m = 0.5kg" is the mass of the first (stationary) and the second ball, respectively, "v_{1,i} = 0 \\dfrac{m}{s}" is the initial velocity of the stationary ball, "v_{2,i} = 60 \\dfrac{m}{s}" is the initial velocity of the second ball, "v_{1,f}" is the final velocity of the stationary ball, "v_{2,f}" is the final velocity of the second ball, "\\alpha = 30^{\\circ}" is the recoil angle of the first ball (which is moving to the left of the "x"-axis) and "\\theta = 60^{\\circ}" is the scattering angle of the second ball (which is moving to the right of the "x"-axis).

Then, we get:


"v_{2i} = v_{1,f}cos30^{\\circ} + v_{2,f}cos60^{\\circ},""60 = 0.87v_{1,f} + 0.5v_{2,f}, (3)""0 = v_{1,f}sin30^{\\circ} - v_{2,f}sin60^{\\circ},""0 = 0.5v_{1,f} - 0.87v_{2,f} (4)"

Let's express "v_{1,f}" from the equation (4) and substitute it into the equation (3):


"v_{1,f} = \\dfrac{0.87}{0.5}v_{2,f} = 1.74v_{2,f},""60 = 0.87 \\cdot (1.74v_{2,f}) + 0.5v_{2,f},""v_{2,f} = 29.8 \\dfrac{m}{s}."

Then, we can find the final velocity of the first ball:


"v_{1,f} = 1.74v_{2,f} = 1.74 \\cdot 29.8 \\dfrac{m}{s} = 52 \\dfrac{m}{s}."

Answer:

"v_{1,f} = 52 \\dfrac{m}{s}."

"v_{2,f} = 29.8 \\dfrac{m}{s}."


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