Question

A 70 g ball B dropped from a height `h_o = 9m` and reaches a height `h_2 = 0.25m` after bouncing twice from identical 210 g plates. Plate A rests directly on the hard ground, while plate C, rests on a foam - rubber mat. Determine?

1) coefficient of restitution between the ball and the plate.
2) the height `h_1` of the ball's first bounce

Answer 1

Representative figure is given above.
Velocity `v` of ball as it approaches plate `A` can be found from the kinematic expression

`v^2-u^2=2gh` .....(1)
`=>v^2-0^2=2xx10xx9`
`=>v=sqrt180\ ms^-1`

The horizontal component of velocity of ball helps in horizontal displacement and remains unaffected due to collisions with plates.

If `e` is the coefficient of restitution between the ball and plates,

`e="relative velocity of separation"/"relative velocity of approach"=-[(v_p-v_b)/(u_p-u_b)]` .......(2)

Plate A rests directly on the hard ground, as such velocity of plate after collision is zero. The velocity of separation of ball can be calculated using (1) as the ball after collision rises to height `h_1`

`0^2-(v_(bA))^2=2xx(-10)h_1`
`=>20h_1=v_(bA)^2`
`=>v_(bA)=sqrt(20h_1)` .......(3)

Taking up as positive direction, using (1) we get

`e=-[(0-sqrt(20h_1))/(-(-sqrt180))]`
`=>e=sqrt(h_1)/3`
`=>h_1=9e^2` ........(4)

Velocity of ball as it approaches plate `C` is same as velocity of separation of ball at plate `A` equation (3), except for the change of direction.
The velocity of separation at plate `C` is found by use of expression (1), as it rises to height `h_2=0.25\ m`

`0^2-u^2=2xx(-10)xx0.25`
`=>v_(bC)=sqrt5\ ms^-1`

The plate rests on rubber-foam mat. As such it will have velocity `v_p` after the collision. From Law of conservation of momentum of plate ball system in SI units we have

`70/1000(-sqrt(20h_1))+0=70/1000(sqrt5)+210/1000(-v_p)`

Using (4) and simplifying it becomes

`(-3esqrt20)+0=(sqrt5)+3(-v_p)`
`=>v_p=1/3(3esqrt20+sqrt5)`

Using (3) for the second bounce

`e=-[(-1/3(3esqrt20+sqrt5)-sqrt5)/(0-(-3esqrt20))]`
`=>9e^2sqrt20=3esqrt20+4sqrt5`
`=>9sqrt20e^2-3sqrt20e-4sqrt5=0`

Using in-built Graphics tool to find solution of the quadratic, ignoring the negative root as `e` can not be negative, we get

`e=0.67`, rounded to two decimal places

Value of height `h_1` is found from (4) as

`=>h_1=9xx(0.6667)^2`
`=>h_1=4.0\ m`