Someone please help me with Physics!!?

Question

I have no idea how to do this problem, please help!

Two objects of masses m and 4m are moving toward each other along the x-axis with the same initial speeds v0 = 40.0 m/s. The object with mass m is traveling to the left, and the object with mass 4m is traveling to the right. They undergo an elastic glancing collision such that mass m is moving downward after the collision at right angles from its initial direction.

(a) Find the final speeds of the two masses.

(b) What is the angle from its initial direction of motion at which the mass 4m is scattered?

Answer 1

An elastic collision conserves both momentum and kinetic energy. The initial kinetic energy of the system is 0.5*m(40^2) + 0.5*4m(40^2) = 4000m, so this must also be the final kinetic energy of the system. The initial momentum of the system in the x-direction is 4m*40 - m*40 = 120m, and the initial momentum in the y-direction is zero.

After the collision, the mass m has zero momentum in the x-direction, momentum in the y-direction of -mv, and kinetic energy of 0.5*mv^2, where v is the magnitude of its downwards velocity. Since the total momentum of the system in the x-direction must be 120m, but the mass m has zero momentum in the x-direction, the mass 4m must have this entire momentum, requiring a velocity of 30 m/s in the positive x direction. In addition, since the total momentum in the y-direction must be zero, and the mass m has momentum in the y-direction of -mv, the mass 4m must have a momentum in the y-direction of mv, and thus a velocity of v/4 in the upwards or positive y-direction. The mass 4m thus has a magnitude of velocity equal to sqrt(30^2 + (v/4)^2) by the Pythagorean theorem.

We can find v by using our knowledge that the final kinetic energy of the system must be 4000m. The final kinetic energy of the system can be written as 0.5*mv^2 + 0.5*4m(30^2 + (v/4)^2), which we set equal to 4000m and solve for v, an exercize that I leave to you. The final speeds, then, are v for the mass m and sqrt(30^2 + (v/4)^2) for the mass 4m.

Finally, use trigonometry to find the angle at which the mass 4m is traveling. It will be arctan((v/4) / 30) above the original horizontal direction of motion.

Answer 2

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Answer 3

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