I don\'t understand the steps of this problem. Please Help! The white ball in th

Question

I don't understand the steps of this problem. Please Help!

The white ball in the figure has a speed of 1.86 m/s and the yellow ball is at rest prior to an elastic glancing collisions. After the collision the white ball has a speed of 1.37 m/s, to the nearest tenth of a degree, measured counterclockwise from east, what angle does it scatter at if the yellow ball is scattered at 280 degrees?

My teacher gave me these formulas...I really need help working these.

m1(vo1)2+m2(vo2)2=m1(v1)2+m2(v2)2 He said solve for v2 then plug v2 into the following.

0=m1(v1)sin(theta)+m2(v2)sin280

Explanation / Answer

In an elastic collision, total kinetic energy and momentum is conserved. In the initial conditions, the white ball has all of the kinetic energy, so plug m=2kg, vi=1.86 m/s into K=(1/2)mv2. This is the total kinetic energy of the system before the collision. Since energy is conserved, it is also the the amount of kinetic energy after the collision. Since both balls are moving after the collision, this energy is split between them. Calculate the white ball's final speed using the same equation, but with v=1.37. The find the energy of the yellow ball by Kinitial - Kwhite/final = KYellow/final. Then using the K = (1/2)mv2 equation, find the yellow ball's speed. Find the y component of the yellow ball's speed using vsin() where = 280. Since there was no motion in the y direction before the collision, the momentum in the y direction of the white ball will be equal to that of the yellow ball. Use m1v1=m2v2 to find the y component of the velocity of the white ball. Since you know total velocity and y-velocity of the white ball, use sin = opposite/adjacent to calculate the that the what ball is moving in.

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