Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 9, Problem 9.77AP

Two blocks of masses m1 = 2.00 kg and m2 = 4.00 kg are released from rest at a height of h = 5.00 m on a friction less track as shown in Figure P9.77. When they meet on the level portion of the track, they undergo a head-on, elastic collision. Determine the maximum heights to which m, and rise on the curved portion of the track after the collision.

\Chapter 9, Problem 9.77AP, Two blocks of masses m1 = 2.00 kg and m2 = 4.00 kg are released from rest at a height of h = 5.00 m

Expert Solution & Answer
Check Mark
To determine

The maximum height attained by block m1 and m2 on the curved portion of the track after collision.

Answer to Problem 9.77AP

The maximum height attained by block m1 is 13.9m and block m2 is 0.556m

Explanation of Solution

Given info: The mass of block m1 is 2.00kg , mass of block m2 is 4.00kg and height between the block and surface level is 5.00m .

Write the expression for potential energy of block is,

U=mgh

Here,

h is the height between the block and surface level.

m is the mass of block.

g is the acceleration due to gravity.

Write the expression for kinetic energy of block is,

K=12mv2

Here,

m mass of block.

v is the speed of block.

Write the expression for velocity of blocks using conservation of energy,

U=K

Here,

U is the potential energy of block.

K is the kinetic energy of block.

Substitute mgh for U and 12mv2 for K in above expression.

mgh=12mv2v=2gh

Thus, the expression for velocity of block is 2gh .

Write the expression for initial velocity of block 1 and block 2 (as it is clear from the expression for velocity of block that the velocity does not depend upon mass of block. Hence, the initial velocity of block 1 will be equal to block 2 ),

v=2gh

Here,

v is the initial speed of block m1 and block m2 .

g is the acceleration due to gravity.

h is the height between the block and surface level.

The value of acceleration due to gravity is 9.8m/s2

Substitute 5.00m for h and 9.8m/s2 for g in above expression.

v=2×9.8m/s2×5.00m=9.90m/s

Thus, the initial speed of block m1 and block m2 is 9.90m/s .

Write the expression for conservation of linear momentum of block.

m1v+m2(v)=m1v1+m2v2

Here,

m1 is the mass of block m1 .

m2 is the mass of block m2 .

v is the initial speed of block m1 and block m2 .

v1 is the final speed of block m1 .

v2 is the final speed of block m2 .

Substitute 2.00kg for m1 , 4.00kg for m2 and 9.90m/s for v in above expression.

(2.00kg×9.90m/s)+(4.00kg×(9.90m/s))=(2.00kg)v1+(4.00kg)v22v1+4v2=19.8v1+2v2=9.9v1=9.92v2 (1)

Thus, the expression for final velocity of block m1 is 29.72v2 .

Write the expression for conservation of kinetic energy of the block.

12m1v2+12m2(v)2=12m1v12+12m2v22(m1+m2)v2=m1v12+m2v22

Here,

m1 is the mass of block m1 .

m2 is the mass of block m2 .

v is the initial speed of block m1 and block m2 .

v1 is the final speed of block m1 .

v2 is the final speed of block m2 .

Substitute 2.00kg for m1 , 4.00kg for m2 , 9.90m/s for v and 29.72v2 for v1 in above expression.

(2.00kg+4.00kg)×(9.90m/s2)2=(2.00kg×(9.92v2)2)+(4.00kg×v22)3×9.902=(9.9v2)2+2v22v22+6.6v2196=0v2=3.30m/s

Thus, the final velocity of block m2 is 3.30m/s .

Write the expression for final velocity of block m1 .

v1=9.92v2

Here,

v1 is the final speed of block m1 .

v2 is the final speed of block m2 .

Substitute 3.30m/s for v2 in above expression.

v1=9.9(2×3.30)=16.5m/s

Thus, the final velocity of block m1 is 16.5m/s .

Write the expression to calculate the maximum height attained by block m1 after collision.

h1=v122g

Here,

h1 is the maximum height attained by block m1 after collision.

v1 is the final speed of block m1 .

g is the acceleration due to gravity.

The value of acceleration due to gravity is 9.8m/s2 .

Substitute 16.5m/s for v1 and 9.8m/s2 for g in above expression.

h1=(16.5m/s)22×9.8m/s2=13.9m

Thus, the maximum height attained by block m1 after collision is 13.9m .

Write the expression to calculate the maximum height attained by block m2 after collision.

h2=v222g

Here,

h2 is the maximum height attained by block m2 after collision.

v2 is the final speed of block m2 .

g is the acceleration due to gravity.

The value of acceleration due to gravity is 9.8m/s2 .

Substitute 3.30m/s for v2 and 9.8m/s2 for g in above expression.

h2=(3.30m/s)22×9.8m/s2=0.556m

Thus, the maximum height attained by block m2 after collision is 0.556m .

Conclusion:

Therefore, the maximum height attained by block m1 is 13.9m and block m2 is 0.556m .

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Chapter 9 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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