Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 7, Problem 63P

(a)

To determine

The x and y component of momentum change of the ball of mass m1.

(a)

Expert Solution
Check Mark

Answer to Problem 63P

The x and y components of momentum change of the ball of mass m1 are Δp1x=1.00m1vi_ and Δp1y=0.751m1vi_ respectively.

Explanation of Solution

The momentum change of the ball of mass m1 is equal but opposite to that of the ball of mass m2.

Δp1x=Δp2x (I)

Δp1y=Δp2y (II)

Write the expression for the momentum change of the ball of mass m2 in x and y directions.

Δp2x=m2v2fxm2v2ix (III)

Δp2y=m2v2fym2v2iy (IV)

Here, v2i is the initial speed of the ball of mass m2, and v2f is the final speed of the ball of mass m2.

Use equation (III) in (I).

Δp1x=m2v2ixm2v2fx (V)

Use equation (IV) in (II).

Δp1y=m2v2iym2v2fy (VI)

The initial speed of the ball of mass m2 is zero in both x and y directions. Thus, equations (V) and (VI) can be modified as,

Δp1x=m2(0v2fx)=m2v2fx (VII)

Δp1y=m2(0v2fy)=m2v2fy (VIII)

Use m2=5m1 in equations (VII) and (VIII).

Δp1x=5m1v2fx (IX)

Δp1y=5m1v2fy (X)

Write the expression for the x and y components of the final velocity of the ball of mass m2.

v2fx=14vicos(36.9°) (XI)

v2fy=14visin(36.9°) (XII)

Use equation (XI) in equation (IX).

Δp1x=5m1[14vicos(36.9°)]=1.00m1vi

Δp1y=5m1[14visin(36.9°)]=0.751m1vi

Conclusion:

Therefore, the x and y components of momentum change of the ball of mass m1 are Δp1x=1.00m1vi_ and Δp1y=0.751m1vi_ respectively.

(b)

To determine

The x and y component of momentum change of the ball of mass m2 and the relation between the momentum changes.

(b)

Expert Solution
Check Mark

Answer to Problem 63P

The x and y components of momentum change of the ball of mass m2 are Δp2x=+1.00m1vi_ and Δp2y=0.751m1vi_ respectively. The momentum changes for each mass are equal in magnitude and opposite.

Explanation of Solution

The momentum change of the ball of mass m2 is equal but opposite to that of the ball of mass m1.

Δp2x=Δp1x (XIII)

Δp2y=Δp1y (XIV)

In part (a) it is obtained that the change in momentum of the ball of mass m1 in x and y directions are as, Δp1x=1.00m1vi and Δp1y=0.751m1vi.

Thus, from equations (XIII) and (XIV) it can be deduced that,

Δp2x=+1.00m1vi

Δp2y=0.751m1vi

The momentum changes for each mass are equal in magnitude and opposite.

Conclusion:

Therefore, the x and y components of momentum change of the ball of mass m2 are Δp2x=+1.00m1vi_ and Δp2y=0.751m1vi_ respectively. The momentum changes for each mass are equal in magnitude and opposite.

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Chapter 7 Solutions

Physics

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