Chapter 6, Problem 62AP

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# Consider a frictionless track as shown in Figure P6.62. A block of mass m1 = 5.00 kg is released from . It makes a head-on elastic collision at  with a block of mass m2= 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.Figure P6.62

To determine
The height to which m1 rises after collision.

Explanation

Given Info:

Mass of the block at position A is 5.00â€‰kg , the mass of the block at B is 10.0â€‰kg , initial speed of mass at position B is zero.

According to conservation of energy , kinetic energy will be equal to the potential energy of the mass.

Formula to calculate the kinetic energy of the mass 1 is at position A is ,

KE=12m1v12 (I)

• m1 is the block released from top.
• v1 is the speed of the mass 1 at position A before collision

Formula to calculate the potential energy of the mass 1 at position A is,

PE=m1gh (II)

• g is the acceleration due to gravity
• h is the height of the mass m1

According to the principle of conservation of energy, Potential energy of mass 1 at position A is converted into its equivalent the kinetic energy at position B.

Equate (I) and (II) to calculate v1 .

12m1v12=m1ghv12=2ghv1=2gh

Substitute 9.80â€‰mâ‹…sâˆ’2 for g and 5.00â€‰m for h in the above equation to calculate v1 .

v1=2(9.80â€‰mâ‹…sâˆ’2)(5.00â€‰m)=98.0m2â‹…sâˆ’2=9.899â€‰mâ‹…sâˆ’1â‰ˆ9.90â€‰mâ‹…sâˆ’1

Collisions conserve momentum.

Apply conservation of momentum before and after collision for the masses m1 and m2

m1v1+m2v2=m1v1â€²+m2v2â€² (III)

• v2 is the speed of the mass m2 at position B
• v1â€² is the speed of mass m1 after collision
• v2â€² is the speed of mass m2 after collision
• m2 is the mass of the mass of the block at rest at bottom

Initial speed of mass2 is zero.

Substitute 0â€‰m/s for v2 in the above equation and rewrite in terms of vâ€²2 .

m1v1+m2(0â€‰m/s)=m1v1â€²+m2v2â€²m2v2â€²=m1v1âˆ’m1v1â€²=m1(v1âˆ’vâ€²1)vâ€²2=m1m2(v1âˆ’vâ€²1)

But for elastic collision

v1âˆ’v2=âˆ’(v1â€²âˆ’v2â€²)

Since initial speed of m2 is zero,

Substitute 0â€‰m/s for v2 in the above equation and rewrite in terms of vâ€²2 .

v2â€²=v1+v1â€² (IV)

Substitute the above equation (IV) in (III) to calculate v1â€²

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