Explanation:
Given that,
Mass of the, m = 0.256 kg
Initial peed of the ball, v = 11.8 m/s
The ball is in contact with the wall for t = 0.066 s
(1) The magnitude of the initial momentum of the racquet ball is given by :
p = mv
[tex]p=0.256\ kg\times 11.8\ m/s\\\\p=3.02\ kg-m/s[/tex]
(2) It is projected at angle of 29 degrees. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal.
The change in momentum of the racquet ball is given by :
[tex]\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = 2mv\cos \theta\\\\\Delta p = 2\times 0.256 \times 11.8\times \cos (29)\\\\\Delta p=5.28\ kg-m/s[/tex]
(3) Impulse is equal to the change in momentum.
[tex]\Delta p=Ft\\\\F=\dfrac{\Delta p}{t}\\\\F=\dfrac{5.28}{0.066 }\\\\F=80\ N[/tex]
Hence, this is the required solution.
