COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 7, Problem 73P

(a)

To determine

The direction and speed of the acrobats after they grab onto each other.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The velocity of the acrobats after grabbing onto each other is 0.64 m/s and is at 73° above the x-axis.

Explanation of Solution

Write an expression using conservation of angular momentum in the x-direction

    m1v1cosθ1m2v2cosθ2=(m1+m2)vxvx=m1v1cosθ1m2v2cosθ2(m1+m2)        (I)

Here, m1 is the mass of the first acrobat, m2 is the mass of the second acrobat, v1 is the velocity of the first acrobat and v2 is the final velocity of the combined mass and θ1 is the angle of first acrobat with the horizontal and θ2 us the horizontal.

Write the expression for final velocity of the combined mass conservation of momentum in the y-direction

    m1v1sinθ1+m2v2sinθ2=(m1+m2)vyvy=m1v1sinθ1+m2v2sinθ2(m1+m2)        (II)

Here, vy is the y-component of the final velocity of the combined mass.

Write the expression for magnitude of the velocity of the combined mass

    v=(vx)2+(vy)2        (III)

Here, v is the final velocity of combined mass.

Write the expression for direction of the velocity of the combined mass

    θ=tan1(vyvx)        (IV)

Here, θ is the direction of the velocity of the combined mass.

Conclusion:

Substitute 60 kg for m1, 80 kg for m2, 3.0m/s for v1, 2.0m/s for v2, 10° for θ1 and 20° for θ2 in equation (I) to find vx

    vx=(60 kg)(3.0m/s)cos(10°)(80 kg)(2.0m/s)cos(20°)(60 kg)+(80 kg)=0.19 m/s

Substitute 60 kg for m1, 80 kg for m2, 3.0m/s for v1, 2.0m/s for v2, 10° for θ1 and 20° for θ2 in equation (I) to find vy

    vy=(60 kg)(3.0m/s)sin(10°)+(80 kg)(2.0m/s)sin(20°)(60 kg)+(80 kg)=0.61 m/s

Substitute 0.19 m/s for vx and 0.61 m/s for vy in equation (III) to find vB

    v=(2.34 m/s)2+(0.61 m/s)2=0.64 m/s

Substitute 0.19 m/s for vx and 0.61 m/s for vy in equation (IV) to find θ

    θ=tan1(0.61 m/s2.34 m/s)=73°

Therefore, the velocity of the acrobats after grabbing onto each other is 0.64 m/s and is at 73° above the x-axis.

(b)

To determine

The nature of the collision whether, elastic or inelastic.

(b)

Expert Solution
Check Mark

Answer to Problem 73P

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:

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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