Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 11, Problem 49PQ

Two skateboarders, with masses m1 = 75.0 kg and m2 = 65.0 kg, simultaneously leave the opposite sides of a frictionless half-pipe at height h = 4.00 m as shown in Figure P11.49. Assume the skateboarders undergo a completely elastic head-on collision on the horizontal segment of the half-pipe. Treating the skateboarders as particles and assuming they don’t fall off their skateboards, what is the height reached by each skateboarder after the collision?

Chapter 11, Problem 49PQ, Two skateboarders, with masses m1 = 75.0 kg and m2 = 65.0 kg, simultaneously leave the opposite

FIGURE P11.49

Expert Solution & Answer
Check Mark
To determine

The height reached by each skateboarder after the collision.

Answer to Problem 49PQ

The height reached by skateboarder of mass 75.0kg after the collision is 2.94m and the height reached by skateboarder of mass 65.0kg after the collision is 5.22m .

Explanation of Solution

Apply conservation of mechanical energy for each skateboarder as they slide down the surface and collide.

  Ki+Ugi=Kf+Ugf                                                                                                  (I)

Here, Ki is the kinetic energy of the skateboarder at the maximum height, Ugi is the potential energy of the skateboarder at the maximum height, Kf is the kinetic energy of the skateboarder just before collision and Ugf is the potential energy of the skateboarder at the maximum height.

The collision occur at h=0, therefore the final potential energy is zero. Let v be the speed of skateboarder just before collision. Since they started from rest, initial height kinetic energy is zero.

Write the expression for Ki .

  Ki=0J                                                                                                                 (II)

Write the expression for Ugi .

  Ugi=mgh                                                                                                              (III)

Here, m is the mass of skateboarder,g is the acceleration due to gravity and h is the height of the half pipe.

Write the expression for Kf .

  Kf=12mv2                                                                                                           (IV)

Here, and v is the speed of skateboarder just before collision.

Write the expression for Ugf .

  Ugf=0J                                                                                                               (V)

Substitute equations (II),(III),(IV) and (V) in equation (I) to get expression of v .

  0J+mgh=12mv2+0J12v2=ghv=2gh                                                                                           (VI)

Rearrange above equation to get h .

  h=v22g                                                                                                                  (VII)

Equation (VI) indicates that each skateboarder has the same speed when he meets the other skateboarder since they started from the same height.

Let v1i be the speed of first skateboarder with mass m1=75.0kg and v2i is the speed of second skater with mass m1=75.0kg just before collision.

Since both are at same speed but in opposite direction, v1i=v2i=v .

Substitute 9.81m/s2 for g and 4.00m for h in equation (VI) to get v .

  v=2(9.81m/s2)(4.00m)=8.86m/s

Thus, v1i=v2i=8.86m/s .

Write the expression for the velocity of skater with mass m1  after collision.

  v1f=(m1m2m1+m2)v1i+(2m2m1+m2)v2i                                                                      (VIII)

Here,v1f is the velocity of the first skater after collision, m1 is the mass of first skater and m2 is the mass of second skater.

Write the expression for the velocity of skater with mass m2  after collision.

  v2f=(m2m1m1+m2)v2i+(2m1m1+m2)v1i                                                                       (IX)

Here,v2f is the velocity of the second skater after collision.

Equation (VII) can be used to calculate height maximum height attained by skaters if initial velocity is given.

Write the expression for the maximum height attained by first skater.

  h1=v1f22g                                                                                                                (X)

Here, h1 is the maximum height attained by skateboarder with mass m1 .

Write the expression for the maximum height attained by second skater.

  h2=v2f22g                                                                                                                 (XI)

Here, h2 is the maximum height attained by skateboarder with mass m2 .

Conclusion:

Substitute 75kg for m1 , 65kg for m2 , 8.86m/s for v1i and 8.86m/s for v2i in equation (VIII) to get v1f .

  v1f=(75kg65kg75kg+65kg)(8.86m/s)+(2×65kg75kg+65kg)(8.86m/s)=7.59m/s

Substitute 75kg for m1 , 65kg for m2 , 8.86m/s for v1i and 8.86m/s for v2i in equation (IX) to get v2f .

  v2f=(65kg75kg75kg+65kg)(8.86m/s)+(2×75kg75kg+65kg)(8.86m/s)=10.1m/s

Substitute 7.59m/s for v1f and 9.81m/s2 for g in equation (X) to get h1 .

  h1=(7.59m/s)22(9.81m/s2)=2.94m

Substitute 10.1m/s for v2f and 9.81m/s2 for g in equation (XI) to get h2 .

  h2=(10.1m/s)22(9.81m/s2)=5.22m

Therefore, the height reached by skateboarder of mass 75.0kg after the collision is 2.94m and the height reached by skateboarder of mass 65.0kg after the collision is 5.22m .

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Chapter 11 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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