PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 8, Problem 89P
To determine

The Speed of 2.0 kg ball and the direction of the 3.0 kg ball after collision.

Expert Solution & Answer
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Answer to Problem 89P

The speed of 2.0 kg ball after collision is 5.3434 m/s and the direction of 3.0 kg ball after collision from incident direction is 26.43° .

Explanation of Solution

Given:

The mass of ball 1 is m1=2.0kg .

The mass of ball 2 is m2=3.0kg .

The speed of ball 1 before collision is  u1=10 m/s .

The speed of ball 2 before collision is  u2=0 m/s .

The speed of ball 2 after collision is v1=4.0 m/s .

The direction of ball 1 after collision from incident direction is θ1=30° .

Formula Used:

The expression for conservation of momentum is X-direction is given by,

  m1u1cos(0°)+m2u2cos(0°)=m1v1cos(θ1)+m2v2cos(θ2)

The expression for conservation of momentum is Y-direction is given by,

  m1u1sin(0°)+m2u2sin(0°)=m1v1sin(θ1)+m2v2sin(θ2)

The expression for kinetic energy after collision is,

  K2=12m1v12+12m2v22

The expression for kinetic energy before collision is,

  K1=12m1u12+12m2u22

Calculation:

The expression for conservation of momentum is X-direction is calculated as,

  m1u1cos(0°)+m2u2cos(0°)=m1v1cos(θ1)+m2v2cos(θ2)210+30=2v1cos( 30)+34cos(θ2) 20=2v132+12cos(θ2)cos(θ2)=203v112 ....... (1)

The expression for conservation of momentum is Y-direction is calculated as,

  m1u1sin(0°)+m2u2sin(0°)=m1v1sin(θ1)+m2v2sin(θ2)20.+300=2v1sin( 30)34sin(θ2) 2v112=12sin(θ2)sin(θ2)=v112 ....... (2)

Square Equation (1) and (2) and add them

  (cos( θ 2 )= 20 3 v 1 12)2(sin( θ 2 )= v 1 12)2cos2(θ2)+sin2(θ2)=( 20 3 v 1 12)2+( v 1 12)2( 20 3 v 1 12)2+( v 1 12)2=1

Further simplify the above,

  ( 20 3 v 1 12)2+( v 1 12)2=1400+3v12403v1 ( 12 )2+v12 ( 12 )2=1400+3v12403v1+v12=144v12103v1+10036=0

Further simplify the above,

  v12103v1+64=0v1=( 10 3 )± ( ( 10 3 ) 2 4641 )21v1=103± ( 4( 253 )464 )2v1=103±2 ( 7564 )2

This implies,

  v1=53±11

So,

  v1=11.9766 m/s and v1=5.3434 m/s

The value of total kinetic energy ffter collision when v1=11.9766 m/s is calculated as,

  K2=12m1v12+12m2v22=122(11.9766)2+123(4)2=143.43+24=167.43 J

The value of total kinetic energy after collision when v1=5.3434 m/s is calculated as,

  K2=12m1v12+12m2v22=122(5.3434)2+123(4)2=28.55+24.00=52.55 J

The value of total kinetic energy before collision is calculated as,

  K1=12m1u12+12m2u22=122(10)2+1230=100 J

When v1=11.9766 m/s kinetic energy after collision increases, this is not possible. So v1=5.3434 m/s is correct answer.

From Equation (2),

  sin(θ2)=v112=5.343412 θ2=sin1( 5.3434 12)=26.43°

Conclusion:

Therefore, the speed of 2.0 kg ball after collision is 5.3434 m/s and the direction of 3.0 kg ball after collision from incident direction is 26.43° .

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Chapter 8 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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