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## Homework Statement

The diagram shows a section of a curtain track in a vertical plane. The curved section, CDE, forms a circular arc of radius of curvature 0.75 m and the point D is 0.25 m higher than B. A ball-bearing of mass 0.060 kg is released from A, which is 0.50 m higher than B. Assume that rotational and frictional effects can be ignored and that the ball-bearing remains in contact with the track throughout the motion.

(a) Calculate the speed of the ball-bearing (i) at B, (ii) at D.

(b) Draw a diagram showing the forces acting on the ball-bearing when it is at D and calculate the reaction between the track and the ball-bearing at this point.

Answers: (a) (i) 3.2 m s

^{-1}, (ii) 2.2 m s

^{-1}, (b) 0.2 N

**2. The attempt at a solution**

For (a) I used the following formula: v

^{2}= u

^{2}+ 2 a s

(i) u / v

_{0}= 0, a = 10 m s

^{-2}(positive, since the ball is moving downwards), s = h

_{A}= 0.5 m

Plug in and get: v

^{2}= 0 + 2 * 10 * 0.5 = 10 → v = 3.2 m s

^{-1}

(ii) u = 3.2 m s

^{-1}, a = -10 m s

^{-2}(since the ball is moving upwards), s = h

_{D}= 0.25 m

v

^{2}= 3.2

^{2}+ 2 * (-10) * 0.25 = 5 → v = 2.2 m s

^{-1}

I think the (a) part should be correct.

In terms of (b) I am not sure. I used the F = m v

^{2}/ r formula. m = 0.060 kg, v = 2.2 m s

^{-1}, r = 0.75 m.

Plug in and get: F = (0.060 * 2.2

^{2}) / 0.75 = 0.4 N, however, the answer is 0.2 N.

And this is how I see the forces acting on the ball, however, again, not sure.

I have the weight (mg) and the normal reaction as the vertical forces. But regarding the horizontal ones I've got two question marks.

Any help please?