COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 71P

(a)

To determine

The velocity of puck B after collision.

(a)

Expert Solution
Check Mark

Answer to Problem 71P

The speed of puck B is 1.7m/s and is at 30° below the x-axis.

Explanation of Solution

Write the expression for conservation of momentum in the x-direction

    mvAx+mvBx=mvAcosθ+mvBxvBx=vAx+vBxvAcosθ        (I)

Here, m is the mass of puck A which is same as puck B, vAx is the x-component of the initial velocity of puck A, vBx is the x-component of the initial velocity of puck B, vA is the final velocity of puck A, vBx is the x-component of the final velocity of puck B and θ is the angle at which puck A moves after collision.

Write the expression for conservation of momentum in the y-direction

    mvAy+mvBy=mvAsinθ+mvByvBy=mvAy+mvBymvAsinθ        (II)

Here, vAy is the y-component of the initial velocity of puck A, vBy is the y-component of the initial velocity of puck B, vA is the final velocity of puck A and vBy is the y-component of the final velocity of puck B.

Write the expression for magnitude of the velocity of puck B

    vB=(vBx)2+(vBy)2        (III)

Here, vB is the final velocity of puck B

Write the expression for direction of the velocity of puck B

    θ=tan1(vByvBx)        (IV)

Conclusion:

Substitute 0 for vBx, 2.0m/s for vAx, 1.0m/s for vA and 60° for θ in equation (I) to find vBx

    vBx=(2.0m/s)+(0 m/s)(1.0m/s)cos(60°)=1.5 m/s

Substitute 0 for vBy, 0 for vAy, 1.0m/s for vA and 60° for θ in equation (II) to find vBy

    vBy=(0)+(0)(1.0m/s)sin(60°)=0.866 m/s

Substitute 1.5 m/s for vBx and 0.866 m/s for vBy in equation (III) to find vB

    vB=(1.5 m/s)2+(0.866 m/s)2=1.729 m/s1.7 m/s

Substitute 1.5 m/s for vBx and 0.866 m/s for vBy in equation (IV) to find θ

    θ=tan1(0.866 m/s1.5 m/s)=30.0°

Therefore, the speed of puck B is 1.7m/s and is at 30° below the x-axis.

(b)

To determine

The nature of the collision whether, elastic or inelastic.

(b)

Expert Solution
Check Mark

Answer to Problem 71P

The collision is elastic.

Explanation of Solution

Write the expression for change kinetic energy

    ΔK=12m((vA)2+(vB)2(vA)2)

Here, ΔK is the change in kinetic energy.

Conclusion:

for vA

Substitute 1.0 m/s for vA, 1.7 m/s for vB and 2.0 m/s for vA in the above equation to find ΔK

    ΔK=12m((1.0 m/s)2+(1.73 m/s)2(2.0 m/s)2)0

Since the change in kinetic energy is zero, the collision is elastic.

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Chapter 7 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 7.7 - Prob. 7.7ACPCh. 7.7 - Prob. 7.7BCPCh. 7.7 - Instead of colliding elastically, suppose the two...Ch. 7.8 - Prob. 7.11PPCh. 7 - Prob. 1CQCh. 7 - Prob. 2CQCh. 7 - Prob. 3CQCh. 7 - Prob. 4CQCh. 7 - Prob. 5CQCh. 7 - Which would be more effective: a hammer that...Ch. 7 - Prob. 8CQCh. 7 - Prob. 10CQCh. 7 - Prob. 11CQCh. 7 - Prob. 1MCQCh. 7 - Prob. 2MCQCh. 7 - Prob. 3MCQCh. 7 - Prob. 4MCQCh. 7 - Prob. 5MCQCh. 7 - Prob. 6MCQCh. 7 - Prob. 7MCQCh. 7 - Prob. 8MCQCh. 7 - Prob. 9MCQCh. 7 - Prob. 10MCQCh. 7 - Prob. 11MCQCh. 7 - Prob. 12MCQCh. 7 - Prob. 1PCh. 7 - Prob. 2PCh. 7 - Prob. 3PCh. 7 - Prob. 4PCh. 7 - Prob. 5PCh. 7 - Prob. 6PCh. 7 - Prob. 7PCh. 7 - Prob. 8PCh. 7 - A ball of mass 5.0 kg moving with a speed of 2.0...Ch. 7 - Prob. 10PCh. 7 - Prob. 11PCh. 7 - Prob. 12PCh. 7 - Prob. 13PCh. 7 - Prob. 14PCh. 7 - Prob. 15PCh. 7 - Prob. 16PCh. 7 - Prob. 17PCh. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - Prob. 20PCh. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - Prob. 23PCh. 7 - Prob. 24PCh. 7 - Prob. 25PCh. 7 - Prob. 26PCh. 7 - Prob. 27PCh. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - Prob. 31PCh. 7 - Prob. 32PCh. 7 - Prob. 34PCh. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Prob. 47PCh. 7 - Prob. 48PCh. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 56PCh. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - Prob. 73PCh. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - Prob. 82PCh. 7 - Prob. 83PCh. 7 - Prob. 84PCh. 7 - Prob. 85PCh. 7 - Prob. 86PCh. 7 - Prob. 87PCh. 7 - Prob. 88PCh. 7 - Prob. 89PCh. 7 - Prob. 90PCh. 7 - Prob. 91PCh. 7 - Prob. 92PCh. 7 - Prob. 93PCh. 7 - Prob. 94PCh. 7 - Prob. 95PCh. 7 - Prob. 96PCh. 7 - Prob. 97PCh. 7 - Prob. 98PCh. 7 - Prob. 99PCh. 7 - Prob. 100PCh. 7 - Prob. 101PCh. 7 - Prob. 102PCh. 7 - Prob. 103PCh. 7 - Prob. 104PCh. 7 - Prob. 105PCh. 7 - Prob. 106PCh. 7 - Prob. 107PCh. 7 - Prob. 109PCh. 7 - Prob. 110PCh. 7 - Prob. 111P
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