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Alchen [17]
3 years ago
10

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 wit

h 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?

Physics
1 answer:
Sergio039 [100]3 years ago
6 0

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

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We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

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                                                    k=\frac{2mgh}{d^{2} }

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                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

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