Question

A body of mass 3 kg moving with velocity 10 m/s
hits a wall at an angle of 60`and returns at the
same angle. The impact time was 0.2 s. Calculate
the force exerted on the wall.​

Answer 1

Explanation:

velocity perpendicular to the wall is 10*sin60

5√3m/s

impulse =3*5√3

=15√3kgm/s.

I=fxt

15√3=fx0.2

f=75√3N

Answer 2

Answer:

Force, F = 150√3 N

Explanation:

[Refer to the attached image to visualize the case]

Given;-  

        Mass, m = 3 kg

    Velocity, v = 10 m/s

      Angle, ∅ = 60°

Time of contact, t = 0.2 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 60i + mv cos 60j ____(1)

Also, for final momentum, we have;-

        Pf = - mv sin 60i + mv cos 60j ____(2)

Now, we find the change in the momentum;-

      ΔP = Pf - Pi

      ΔP = (- mv sin 60i + mv cos 60j) - (mv sin 60i + mv cos 60j)  [From (1) and (2)

      ΔP = - 2 mv sin 60i

    |ΔP| = 2 mv sin 60°

Now, we know that;-

    Force, F = ΔP/ Δt

So,           F =  2 mv sin 60°/ 0.2

                F = 2 × 3 × 10 × √3/2  / 0.2

                F = 30√3 / 0.2

                F = 300√3 /  2

                F = 150√3 N

Hence, the force exerted on the wall is 150√3 N.

Hope it helps! ;-))