Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 93P
To determine

The speed of the target after collision if the collision is perfectly elastic.

Expert Solution & Answer
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Answer to Problem 93P

The speed of the target after collision if the collision is perfectly elastic is 2.8m/s_.

Explanation of Solution

Given that the mass of the object is 2.0kg, the object is approaching the target with speed 8.0m/s, the angle of deflection is 90.0°, and speed of the object after collision is 6.0m/s.

Since the collision is perfectly elastic, the kinetic energy and momentum of the system is conserved.

Write the expression indicating the momentum conservation along x direction, to the given system.

m1v1ix+m2v2ix=m1v1fx+m2v2fx (I)

Here, m1 is the mass of the object, m2 is the mass of the target, v1ix is the initial speed of the object in x direction, v2ix is the initial speed of the target in x direction, v1fx is the final speed of the object in x direction, and v2fx is the final speed of the target in x direction.

Write the expression indicating the momentum conservation along y direction, to the given system.

m1v1iy+m2v2iy=m1v1fy+m2v2fy (II)

Here, v1iy is the initial speed of the object in y direction, v2iy is the initial speed of the target in y direction, v1fy is the final speed of the object in y direction, and v2fy is the final speed of the target in y direction.

Since the collision is elastic, v1ix=v1i, and v1fy=v1f.

The initial speed of the target in both x and y directions are zero. The final speed of the object in x direction is zero. Moreover, the initial speed of the object in y direction is zero. Thus, equation (I) and (II) can be modified as,

m1v1ix=m2v2fx (III)

m1v1fy=m2v2fy (IV)

Square and add equation (III) and (IV).

(m1v1ix)2+(m1v1fy)2=(m2v2fx)2+(m2v2fy)2m22v2f2=m12v1i2+m12v1f2m2v2f2=m12m2(v1i2+v1f2) (V)

Write the expression indicating the conservation of kinetic energy of the system.

12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f212m1v1i2+0=12m1v1f2+12m2v2f2m2v2f2=m1v1i2m1v1f2v2f=m1m2(v1i2v1f2) (VI)

Use equation (VI) in (V) and solve for m2.

m2(m1m2(v1i2v1f2))=m12m2(v1i2+v1f2)m2=m1(v1i2+v1f2v1i2v1f2) (VII)

Use equation (VII) in (VI).

v2f=m1(v1i2v1f2)m1(v1i2+v1f2v1i2v1f2)=(v1i2v1f2)2v1i2+v1f2=v1i2v1f2v1i2+v1f2 (VIII)

Conclusion:

Substitute 8.0m/s for v1i, and 6.0m/s for v1f in equation (VIII) to find v2f.

v2f=(8.0m/s)2(6.0m/s)2(8.0m/s)2+(6.0m/s)2=2.8m/s

Therefore, the speed of the target after collision if the collision is perfectly elastic is 2.8m/s_.

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Chapter 7 Solutions

Physics

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