Question

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A steel ball of mass 0.5 kg is dropped from a height of 4 m on to a horizontal heavy steel slab. The
ball strike the slab and rebounds to its original height (take g = 10 m/sec2)
(a) Calculate the impulse delivered to the ball during impact.
(b) If the ball is in contact with the slab for 0.002s, find the average reaction force on the sam
during impact​

Answer 1

Answer:

sorry I am not understanding this question

Answer 2

Impulse, $J = 8.85\ N/s$

F = 4425 N

Explanation:

It is given that,

Mass of the steel ball, m = 0.5 kg

Height, h = 4 m

(a) Let v is the speed with which it will move. It can be calculated using conservation of energy as :

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 4}$

v = 8.85 m/s

As it rebounds, final speed of the ball will be (-8.85 m/s). The impulse delivered to the ball during impact is given by the change in momentum as :

$J=m(v-u)$

$J=0.5\times (8.85-(-8.85))$

$J = 8.85\ N/s$

(b) time for contact, t = 0.002 s

Let F is the average reaction force on the Sam  during impact​. Impulse is given by the product of force and time. It is given by :

$J=F\times t$

$F=\dfrac{J}{t}$

$F=\dfrac{8.85}{0.002}$

F = 4425 N

So, the average reaction force on the Sam  during impact is 4425 N. Hence, this is the required solution.

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