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Chapter 9: Q35P (page 250)

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脳102N.

Step by step solution

01

Understanding the given information

  1. The mass of the superball,mis58脳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鈫�=鈭�p鈫�=m鈭�v鈫�=Favg鈭�t

03

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

The impulse 鈥� linear momentum theorem gives

J=m鈭�v=Favg鈭�t

The impulse can be calculated as

J=m鈭�v=58脳10-3脳(34-(-34))=58脳10-3脳68=3.94N.s

From the graph we can see that

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

Using the two equations of the 鈭�p, we get

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

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Most popular questions from this chapter

In Figure, two particles are launched from the origin of the coordinate system at timet=0. Particle 1 of massm1=5.00gis shot directly along the xaxis on a frictionless floor, with constant speed10.0m/s . Particle 2 of massm2=3.00gis shot with a velocity of magnitude20.0m/s, at an upward angle such that it always stays directly above particle 1. (a) What is the maximum height Hmax reached by the com of the two-particle system? In unit-vector notation, (b) what are the velocity and (c) what are the acceleration of the com when the com reaches Hmax

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