Question

A ball of mass 1 kg is dropped from a height 5m on a large horizontal surface. If time of contact between ball and surface is 0.02s and collision is perfectly elastic, then the average force exerted by the surface on the ball is?

Answer 1

height of ball from teh horizontal surface , h = 5m

so, speed of ball before collision, v = √{2gh}

= √{2 × 10 × 5} = 10m/s

for perfectly elastic collision,

if v is the velocity of ball before striking the large horizontal surface , then after collision , velocity of ball will be -v .

so, change in momentum of ball = m(-v) - mv = -2mv

so, change in momentum of wall = -(-2mv) = 2mv

so, impulse = 2mv

now, force exerted by wall = impulse/time

= 2mv/t

where , m = 1kg, v = 10m/s and t = 0.02 sec

so, force exerted by wall = 2 × 1 × 10/0.02

= 1000 N

Answer 2

Answer:

1000 N

Explanation:

Given S = 5 m, u = 0 and g = 10 then the appropriate kinematics equation to use is :

V² = u² + 2gs

Doing the substitution we have:

V² = 0 + 2 × 10 × 5

V = (100)^½

V = 10 m/s

The ball strikes the floor at 10 m/s.

For a perfectly elastic collision after collision the balls will bounce back at a velocity (-V) m/s.

In this case we have : - V = - 10 m/s

Momentum = mass × velocity.

Initial momentum = 10 × 1 = 10

Final momentum = - 10 × 1 = - 10

Change in momentum = final momentum - initial momentum.

= - 10 - 10 = 20 kgm/s

Change in momentum = impulse

Impulse = 20 kgm/s

Force = Impulse / time

Force = 20/0.02 = 1000 N

The force exerted by the surface = 1000 Newtons.