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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A cue ball traveling at 4.00 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The target ball deflects the cue ball so that its subsequent motion makes an angle of 30.0° with respect to its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

(a)

To determine
The angles between the velocity vectors of two balls after collision.

Explanation

Given info:

The speed of ball is 4.00m/s .

The incident angle is 30.0° .

The Formula to calculate the conservation of momentum along x- axis is,

mvTfcosθ'+mvcfcosθ=0+mvci

  • vci is the initial speed of the cue ball
  • vcf is the final speed of the cue ball
  • vTf is the final speed of the target
  • θ' is the angle target’s final velocity makes with x- axis
  • θ is the incident angle

Rearranging the formula gives,

vTfcosθ'=vcivcfcosθ (I)

The Formula to calculate the conservation of momentum along y- axis is,

mvTfsinθ'+mvcfsinθ=0+0

Rearranging the formula gives,

vTfsinθ'=vcfsinθ (II)

Squaring and adding equation (I) and (II) gives,

(vTfcosθ')2+(vTfsinθ')2=(vcivcfcosθ)2+(vcfsinθ)2vTf2(cos2θ'+sin2θ')=vci22vcivcfcosθ+vcf2(cos2θ+sin2θ)vTf2=vci22vcivcfcosθ+vcf2 (III)

The Formula to calculate the conservation of kinetic energy is,

12mvTf2+12mvcf2=12mvci2

Rearranging the formula gives,

(b)

To determine
The speed of each ball after collision.

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