COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 75P
To determine

The velocity of the birds after collision.

Expert Solution & Answer
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Answer to Problem 75P

The ratio of the speed of the balls is 0.781.

Explanation of Solution

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

    (m1+M1)u1(m2+M2)u2=(m1+m2)vxM1v1sinθ1+M2v2sinθ2

Here, m1, m2 are the masses of the swallows, M1, M2 are the masses of the coconut, u1 is the mass of the first swallow and the first coconut, u2 is the mass of the second swallow and the second coconut, vx is the horizontal velocity of the combined mass of the bird, θ1 is the angle subtended by the first coconut and θ2 is the angle subtended by the second coconut.

Rearrange the above equation

    vx=(m1+M1)u1(m2+M2)u2+M1v1sinθ1M2v2sinθ2(m1+m2)        (I)

Write an expression using conservation of momentum in the y-direction

    0=(m1+m2)vyM1v1cosθ1+M2v2cosθ2

Here, vy is the vertical velocity of the combined mass of the birds.

Rearrange the above equation

    vy=M1v1cosθ1M2v2cosθ2(m1+m2)        (II)

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

    v=vx2+vy2        (III)

Here, v is the magnitude of the velocity of the combined mass of the birds

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

    θ=tan1(vyvx)        (IV)

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

Conclusion:

Substitute 20 m/s for u1, 15 m/s for u2, 0.270 kg for m1, 0.220 kg for m2, 0.80 kg for M1, 0.70 kg for M2, 10° for θ1, 30° for θ2, 13m/s for v1 and 14m/s for v2 in equation (I) to find vx

    vx=[((0.270 kg)+(0.80 kg))(20 m/s)((0.220 kg)+(0.70 kg))(15 m/s)+(0.80 kg)(13m/s)sin10°(0.70 kg)(14m/s)sin30°](0.270 kg+0.220 kg)=1.27 m/s

Substitute 0.270 kg for m1, 0.220 kg for m2, 0.80 kg for M1, 0.70 kg for M2, 10° for θ1, 30° for θ2, 13m/s for v1 and 14m/s for v2 in equation (I) to find vx

    vy=(0.80 kg)(13m/s)cos30°(0.80 kg)(14m/s)cos30°(0.270 kg+0.220 kg)=2.39 m/s

Substitute 1.542 m/s for vx and 1.828 m/s for vy in equation (III) to find vB

    vB=(1.542 m/s)2+(1.828 m/s)2=2.39 m/s

Substitute 1.542 m/s for vx and 1.828 m/s for vy in equation (IV) to find θ

    θ=tan1(1.828 m/s1.542 m/s)=49.8°

Therefore, the ratio of speed of the balls is 0.781.

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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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