Question

A steel ball of mass 0.5 kg is dropped from a height of 4 m on to a horizontal heavy steel slap. the strile the slap and reounds to its original height.
(a) calculate the impulse delivered to the ball during impact.
(b) if the ball is in contact with the slab for 0.002 , find the average reaction force on the ball during impact.

Answer 1

Applying equation of motion $V^2=u^2+2as$

v = ?

u = 0

a = g = 10

s = 4 put in above

$V_1=4\sqrt{5}$

as ball rebounds to same height changed velocity ($V_2$ ) again will be $4\sqrt{5}$

Now impulse = $m(V_2-V_1)$= 2m$V_1$ = 2 ( 0.5 ) $4\sqrt{5}$

= $4\sqrt{5}$ ( answer )

Now

Force = $\frac{impulse}{times}$

= $\frac{4\sqrt{5}}{0.002}$

= $2000\sqrt{5}$