Question

36 In a game of billiards, the cue ball strikes another ball initially at rest. After the collision, the cue ball moves at 3.20 m/s along a line making an angle of 60.0º with its original direction of motion. The second ball acquires a speed of 7.0 m/s. Using momentum conservation, find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

37 A 40-kg uniform square sign, 1.0 m on a side, is hung from a 1.5-m rod of negligible mass. A cable is attached to the end of the road and to a point on the wall 2.0 m above the point where the road is fixed to the wall as shown below (a) What is the tension in the cable? (b) What are the horizontal and vertical components of the force exerted by the wall on the rod?

ANSWERS 35 A) = -23.3 B) = 8.03 m/s 26 A) = 327 B) Fy = 130.5N Fx = 196N

I need a clear and nice explanation with graphs. It's hard to understand when there are just only numbers. I will appreciate very much all your work, more details will help me to understand better. Thanks a lot.

Answer 1

36)

A)

Using law of conservation of momentum  

Along horizontal direction,

m u1 = mv cos(x1) + mv2 cos(x2) ..... (1)

Along vertical direction,

0 = - sin(x1) + m v2 sin(x2) ....... (2)

Sin(x2) = (v/v2) sin(x1) = (3.2/7) sin(60)

X2 = 23.32 degree

B)

In equation (1)

Velocity, v1 = (3.2) cos60 + 7 cos(23.32) = 8.03 m/s

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