Question

A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms.
The ball rebounds elastically and returns to its original height. The time interval for the round trip is 3.00 s. What is the magnitude of the
average force that the plate exerted on the ball?
2490 N
1500 N
2000 N
3500 N
3000 N

Answer 1

Answer:

F = 3000 N

Explanation:

Let's use the relationship between momentum and momentum

         I =  Δp

         F t = m v_f - m v₀

As the body when bouncing reaches the same height from which it begins to fall, the modulus of speed is the same, but in the opposite direction,

         v_f = - v₀ = v

         F t = m 2v

         F = 2 m v / t          (1)

This is the contact time t = 0.5 ms = 0.5 10⁻³ s

Let's use the kinematics to find the velocity at the point of touching the plate, as the total travel time is 3.0 s, the descent time must be half the total time

          t = 1.5 s

as the body solved its initial velocity is zero

         v = v₀ + g t

         v = g t

         v = 9.8  1.5

         v = 14.7 m / s

we substitute in equation 1

         F = 2 0.051 14.7 / 0.5 10-3

         F = 2,999 103 N

        F = 3000 N

Answer 2

The average force that the plate exerted on the ball is 3,000 N.

The given parameters;

• mass of the rubber ball, m = 51 g = 0.051 kg
• time in contact, t = 0.5 ms
• time interval for rebound, Δt = 3 s

The velocity of the ball is calculated as follows;

$v = g\Delta t\\\\v = 9.8 \times 3\\\\v = 29.4 \ m/s$

The average force that the plate exerted on the ball is calculated as follows;

$F = \frac{mv}{t} \\\\F = \frac{0.051 \times 29.4}{0.5 \times 10^{-3} } \\\\F = 3,000 \ N$

Thus, the average force that the plate exerted on the ball is 3,000 N.

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