Question

(3)
(4)
A 0.5 kg ball moving with a speed of 12 m/s strikes
a hard wall at an angle of 30° with the wall. It is
reflected with the same speed and at the same
angle. If the ball is in contact with the wall for
0.25 s, the average force acting on the wall is
IT
rolimel 2006​

Answer 1

Answer:

24N

Explanation:

Refer to the material.

Answer 2

Answer:

Average Force, Favg = 24 N

Explanation:

[Refer to the attached image to visualize the case]

Given;-  

     Mass, m = 0.5 kg

  Velocity, v = 12 m/s

     Angle, ∅ = 30°

Time of contact, t = 0.25 seconds

Now;-

It is given that equal speeds are acting on the projection and rebounding of the ball. So, we draw the components of velocity ( as in the image attached );-

       Vx = mv cos∅ & Vy = mv sin∅  

We attach mass m with the components since the mass m is also moving with the velocity of projection and rebound. Now;-

For, Initial momentum, we have;-

         Pi = mv sin 30i + mv cos 30j ____(1)

Also, for final momentum, we have;-

        Pf = - mv sin 30i + mv cos 30j ____(2)

Now, we find the change in the momentum;-

      ΔP = Pf - Pi

      ΔP = (- mv sin 30i + mv cos 30j) - (mv sin 30i + mv cos 30j)  [From (1) and (2)

      ΔP = - 2 mv sin 30i

    |ΔP| = 2 mv sin 30°

Now, we know that;-

Average Force, Favg = ΔP/ Δt

So,        Favg =  2 mv sin 30°/ 0.25

            Favg = 2 × 1/2 × 12 × 1/2/ 1/4

            Favg = 6 × 4

           Favg = 24 N

Hence, the average force acting on the wall is 24 N.

Hope it helps! ;-))