A ball of 1 kg drops vetically on to the floor with a speed of `25 m s^(-1)` . It rebounds with an initial velocity of `10 m s^(-1)` . (a) What impulse acts on the ball during contact ? (b) If the ball is in contact for `0.02` s , what is the average force exerted on the floor ?

Options

A. `0.1, 55 kg m/s, 1550 N`
B. `0.2, 45 kg m/s, 1650 N`
C. `0.3, 30 kg m/s, 1550 N`
D. `0.4, 35 kg m/s, 1750 N`

Answer

`(4)`

Explanation

` m = 1.0 kg, u = 25 m/s, v = - 10 m/s,`
`t = 0.02 s`
Coefficient of restitution
`e = ("relative velocity of separation")/("relative velocity of approach")`
`10/25 = 0.4`
Impulse = Change in momentum = m(v - u)
`= 1.0 [10 -(-25)]`
`= 1.0 xx (+35) = 35 kg m/s`
Average force = Rate of change of momentum = `35/0.02 = 1750 N`