sorry for the bad quality. incline is a 3,4,5 which makes an angle of 36.83 degr

Question

sorry for the bad quality. incline is a 3,4,5 which makes an angle of 36.83 degrees. Ball is dropped from a height of 4 ft and it rebounds, travels a distance d down the incline and lands at B.

I am posting this because I had a hard time understanding the textbook solution on this site so please dont copy and paste that solution. I want to see it done from someone else.

The 1 lb ball is dropped from rest and falls a distance of 4 ft before striking the smooth plane at A. If it rebounds and in t = 0.5 s strikes the plane at B, determine the coefficient of restitution e between the ball and the plane. Also, what is the distance d?

Thanks

Explanation / Answer

Velocity at time of hitting v = (2gh) = (2*32.2*4) = 16.05 ft/s

KE when it hits the plane for first time is 1/2*m*v^2 = 1/2*1*16.05^2 = 128.8 J

After bouncing the ball will get redirected. Angle between normal to plane and "velocity vector after hitting" will be 36.83 deg. (Imagine reflection of light rays). Angle between horizontal and normal to plane is 90- = 53.17 deg. Therefore, Angle between horizontal and "velocity vector after hitting" will be 53.17 - 36.83 = 16.34 deg above horizontal.

If velocity after hitting is u. Horizontal component of this velocity = u Cos 16.34 and vertical component is u Sin 16.34.

In 0.5 s, horizontal distance travelled = 0.5*u Cos 16.34.

Hence, d*Cos 36.83 = 0.5*u Cos 16.34

or d = 0.6u...........(1)

Time taken to reach max. height after hitting is t1 = (u Sin 16.34)/g

Max. height after hitting = (u Sin 16.34)^2/(2g)

Time taken to drop a height of (u Sin 16.34)^2/(2g) + d Sin 36.83 is t2 = [2[(u Sin 16.34)^2/(2g) + d Sin 36.83]/g]

We have t1 + t2 = 0.5

So, (u Sin 16.34)/g + [2[(u Sin 16.34)^2/(2g) + d Sin 36.83]/g] = 0.5

Putting g = 32.2 ft/s^2 and solving:

u*0.008737 + (0.093472*u^2 + 0.03723*d) = 0.5

(0.093472*u^2 + 0.03723*d) = 0.5 - 0.008737*u

Squaring both the sides, 0.093472*u^2 + 0.03723*d = 0.25 + 0.00007633*u^2 - 0.008737*d

Simplifying: 0.00866*u^2 = 0.25 - 0.045967*d............(2)

Putting d from (1), 0.00866*u^2 = 0.25 - 0.0275802*u

0.00866*u^2 + 0.0275802*u - 0.25 = 0

Solving this quadratic eqn, u = 4.011 ft/s

a) So, KE after hitting = 1/2*mu^2 = 1/2*1*4.011^2 = 8.0462 J

Coefficient of restitution e = KE after hitting/KE before hitting = 8.0462/128.8 = 0.062

b) From (1), d = 0.6*4.011 = 2.4 ft

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