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Figure 9-53 shows an approximate plot of force magnitude F versus time t during the collision of a 58 gSuperball with a wall. The initial velocity of the ball is 34 m/sperpendicular to the wall; the ball rebounds directly back with approximately the same speed, also perpendicular to the wall. What isFmaxthe maximum magnitude of the force on the ball from the wall during the collision?

Fig. 9-53

Short Answer

Expert verified

The maximum magnitude of the force on the ball from the wall is Fmaxis9.9×102N.

Step by step solution

01

Understanding the given information

  1. The mass of the superball,mis58×10-3kg.
  2. The initial speed of the ball= final speed of the ball= 34 m/s.
02

Concept and formula used in the given question

The graphical presentation of force and time indicates the way an impulse act on an object. Using the graph, we can determine the momentum change or force in a given situation.

J→=∆p→=m∆v→=Favg∆t

03

Calculation for the maximum magnitude of the force on the ball from the wall

The impulse – linear momentum theorem gives

J=m∆v=Favg∆t

The impulse can be calculated as

J=m∆v=58×10-3×(34-(-34))=58×10-3×68=3.94N.s

From the graph we can see that

role="math" localid="1661240669180" ∆p=areaunderthecurvefromt=0tot=6.0s=areaoftrapezoid=12×(a+b)×h=12×(2+6)×10-3×Fmax

Using the two equations of the ∆p, we get

3.94=12×(2+6)×10-3×FmaxFmax=3.944×10-3=9.9×102N

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