Question

A tennis ball of mass m = 0.060 kg is traveling with a speed of 28 m/s and θ = 35 degrees. It strikes a wall and bounces off with a speed of 28 m/s. What is the magnitude of the force, in Newtons, exerted on the ball if the collision time is 0.1 s? what direction was the force exerted by the wall?

Answer 1

Answer:

19.15 Newton

Explanation:

Force on a body is rate of change of momentum.

Thus force on wall is =$\frac{dP}{dt}$

 ΔP = P(final) - P(initial) [in vector]

Note : take component of velocity normal to wall (assuming wall smooth)

 P(final) = m*v*sin(Ф) [in negative x direction] = -0.06*28*sin(35)

 P(initial) = m*v*sin(Ф) [in positive x direction] = 0.06*28*sin(35)

Thus ΔP=2(0.06*28*sin(35))

Now ΔT is given =0.1 s

Force on wall = $\frac{ΔP}{Δt}$ =$\frac{2(0.06*28*sin(35))}{0.1}$ =19.15 Newton