Question

3. A ball of mass 5 kg moving with a speed of 20 m/s rebounds after striking normally against a perfectly
elastic wall. If contact time with the wall is 10 s, the force exerted by the wall will be______:

Answer 1

Given:

• mass of the ball (m) = 5kg
• initial velocity (u) = 20m/s
• final velocity (v) = -20m/s (as it rebounds elastically)

To find:-

• Force exerted by the wall (f)

Solution:-

• According to the formula, ∆P/t where t= 10s and ∆P = change in momentum.
• ∆P = m*v-m*u

             = 5*(-20)-5*(20)

             = -200 kg.m/s

• f = ∆P/t

          = -200/10

          = -20 N

Answer:- Thus the force exerted by the wall will be 20 N in the opposite direction of the motion.

Answer 2

Answer:

Force exerted by the wall= -20N

Explanation:

As we know that the force(F)= Rate of change of Momentum=ΔP/t

Change in momentum= final momentum-Initial Momentum

                                     = Mv-Mu

(M= mass of the ball, v= final velocity and u=Initial velocity)

Here we have taken the final velocity v as negative because the collision was elastic so initial velocity changes to final velocity with a opposite direction.

so here v= -u

               = -20 m/s

Putting the values we get that,

F= [5×(-20)-5×(20)]/10

 =[-100-100]/10

 =-200/10

= -20N