Preparation problems

11-40. A block of mass m1 undergoes a one-dimensional elastic collsion with an initially stationary block of mass m2. Find an expression of the fraction of the initial kinetic energy transferred to the second block, and plot your results as a function of mass ratios m1/m2 from 0 to 20. Show also that, for a given mass ratio, the energy transfer is the same no matter which mass is initially at rest.

Solution:

We are asked to find the fraction of the intial kinetic energy transferred to the second block. To do this, we need to solve for the velocity of the second block in terms of the initial velocity of the first block. Since we are dealing with an elastic collision in one-dimension, we know both energy and momentum will be conserved in the collision. With the first block moving, and the second block stationary, we can write down an equation demonstrating conservation of momentum and solve for the final velocity of the first block. Doing this again for conservation of kinetic energy will allow us to set them equal and get an expression for the second block's final velocity

The conservation of momentum equation gives us the expression:

pi
=
pf
m1 v1i
=
m1 v1f + m2 v2f
m1 v1f
=
m1 v1i - m2 v2f
v1f
=
v1i - m2
m1
 v2f
(1)

The conservation of energy equation gives us the expression

Ki
=
Kf
1/2 m1 v1i2
=
1/2 m1 v1f2 + 1/2 m2 v2f2
m1 v1f2
=
m1 v1i2 - m2 v2f2
v1f2
=
v1i2 - m2
m1
 v2f2
(2)

Now we can set the square of the right side of Eq. (1) to the right side of Eq. (2). This gives us

é
ê
ë 		v1i - m2
m1
 v2f ù
ú
û 		2

 
=
v1i2 - m2
m1
 v2f2
v1i2 - 2 m2
m1
 v1iv2f + ( m2
m1
 )2 v2f2
=
v1i2 - m2
m1
 v2f2
é
ê
ë 		( m2
m1
 )2 + m2
m1
 ù
ú
û 		v2f2 - é
ê
ë 		2 m2
m1
 v1i ù
ú
û 		v2f
=
0
é
ê
ë 		( m2
m1
 )2 + m2
m1
 ù
ú
û 		v2f
=
2 m2
m1
 v1i
v2f
=
2 m2
m1
é
ê
ë 		( m2
m1
 )2 + m2
m1
 ù
ú
û
·v1i
v2f
=
2
( m2
m1
 ) + 1
·v1i
v2f2
=
4
é
ê
ë 		( m2
m1
 ) + 1 ù
ú
û 		2

 
·v1i2
(3)

Now we can look at the ratio of the kinetic energy of the second block to that of the first block. This will be

K2f
Ki
=
1/2 m2 v2f2
1/2 m1 v1i2
=
m2
m1
 v2f2
v1i2
We can substitute our result from Eq. (3) into this equation to find the desired result.
K2f
Ki
=
m2
m1
 v2f2
v1i2
K2f
Ki
=
m2
m1
 4
é
ê
ë 		( m2
m1
 ) + 1 ù
ú
û 		2

 

We can plot this for function for values of [(m2)/( m1)] ranging from 0 to 20. This looks like this:



Follow-up problems

11-27. Two identical pendulums are suspended from strings of equal length, and one is released from a height h as shown in Fig. 11-16 (p. 269). When the first bob hits the second, the two stick together. Show that the maximum height to which the combination rises is 1/4 h.

Solution:


To solve this problem, we need to break down what happens into as several individual steps, shown in the picture above. If we consider what happens between (a) and (b), we see that we are converting potential energy of the white bob into kinetic energy. Thus, we can apply the principle of conservation of energy to solve for the velocity of the white bob when just before it hits the black one. Labeling the kinetic and potential energy by the letter of the picture to which they correspond, we see
0
=
DK + DU
0
=
(Kb - Ka) + (Ub - Ua)
0
=
(1/2 m vb2 - 0) + (0 - m g h)
vb
=
  ___
Ö2gh
 
(1)

Now we must look at the collision between the two bobs. Since they stick together, it must be a totally inelastic collision. This means we can apply conservation of momemtum to the instances just before the collision, (b), to just after, (c). Doing this, and recalling that we have solved for the incoming velocity of the white bob in Eq. (1), we get

pb
=
pc
m vb + 0
=
(m + m)vc
vb
=
2 vc
vc
=
  ____
Ögh/2
 
(2)

Now we must consider what happens between (c) and (d). We see that we are converting kinetic energy of the combined bob into potential energy. Thus, we can apply the principle of conservation of energy to solve for the height of the combined bob when it reaches the top of its swing, which we call y. Labeling the kinetic and potential energy by the letter of the picture to which they correspond, we see

0
=
DK + DU
0
=
(Kd - Kc) + (Ud - Uc)
0
=
(0 - 1/2 m vc2) + (m g y - 0)
m g y
=
1/2 m (g h/2)
y
=
h/4
This was the result we desired.

11-38. A 59.1-g tennis ball is moving at 14.5 m/s when it collides elastically and head-on with a basketball moving in the opposite direction at 9.63 m/s. If the tennis ball rebounds at twice its initial speed, find the mass and final velocity of the basketball.

Solution:

This is an example of a 1-dimensional elastic collision. We need to apply the principles of conservation of momentum and conservation of kinetic energy in order to solve the problem. First, let's set up our knowns and unknowns:

Tennis Ball Basketball
Mass mt = 0.0591 kg mb = unknown
Init. Velocities vt,i = 14.5 m/s vb,i = -9.63 m/s
Final Velocities vt,f = -2vt,i = -29 m/s vb,f = unknown

Now we can write down our two equations and employ a strategy to solve them both simultaneously. I will solve the conservation of momentum equation for the mass of the basketball in terms of its final velocity, and then plug that result into the conservation of energy equation. Applying conservation of momentum to our collision, we get

pi
=
pf
mt vt,i + mb vb,i
=
mt vt,f + mb vb,f
mt vt,i + mb vb,i
=
mt (-2vt,i) + mb vb,f
mb vb,f - mb vb,i
=
mt vt,i + 2 mt vt,i
mb [vb,f - vb,i]
=
3mt vt,i
mb
=
3mt vt,i
vb,f - vb,i
(1)

Now we can look at the conservation of kinetic energy in the collision. This equation looks like

Kt,i + Kb,i
=
Kt,f + Kb,f
1/2 mt vt,i2 + 1/2 mb vb,i2
=
1/2 mt vt,f2 + 1/2 mb vb,f2
mt vt,i2 + mb vb,i2
=
mt (-2vt,i)2 + mb vb,f2
mt vt,i2 - 4 mt vt,i2
=
mb vb,f2 - mb vb,i2
mb [vb,f2 - vb,i2]
=
- 3 mt vt,i2
mb
=
- 3 mt vt,i2
vb,f2 - vb,i2
Setting this equation equal to Eq. (1), we get the equation
3mt vt,i
vb,f - vb,i
=
- 3 mt vt,i2
vb,f2 - vb,i2
3mt vt,i [ vb,f2 - vb,i2 ]
=
- 3 mt vt,i2 [ vb,f - vb,i ]
Now divide both sides by 3 mt [ vb,f - vb,i], recalling that
A2 - B2 = (A - B) (A + B)
and solve for vb,f, giving us
vt,i [ vb,f + vb,i ]
=
- vt,i2
vt,i vb,f
=
- vt,i2 - vb,ivt,i
vb,f
=
- vt,i2 - vb,ivt,i
vt,i
=
- vt,i - vb,i
=
- (14.5  m/s) - (-9.63  m/s)
vb,f
=
-4.87  m/s

Finally, we can plug this result into Eq. (1) to find the mass of the basketball, giving

mb
=
3mt vt,i
vb,f - vb,i
=
3(0.059  kg) (14.5  m/s)
(-4.87  m/s) - (-9.63  m/s)
mb
=
0.540  kg

11-44. On ice, a 3.2-kg rock moving at 1.0 m/s collides elastically with a 0.35-kg rock initially at rest. The smaller rock goes off at an angle of 45° to the larger rock's initial motion. What are the final speeds of the two rocks? What is the direction of the larger rock?

Solution:


Picture Omitted
This is an example of a 2-dimensional elastic collision. We need to apply the principles of conservation of momentum in both the x- and the y-direction, as well as conservation of kinetic energy in order to solve the problem. First, let's set up our knowns and unknowns:

Big rock (1) Little rock (2)
Mass m1 = 3.2 kg m2 = 0.35 kg
Init. Velocities (x-dir) v1x,i = 1.0 m/s v2x,i = 0 m/s
Init. Velocities (y-dir) v1y,i = 0 m/s v2y,i = 0
Final Velocities (x-dir) v1y,f = v1,f cosq v2x,f = v2,f cosf
Final Velocities (y-dir) vt,f = -v1,f sinq v2y,f = v2,f sinf

Now we can write down our three equations and employ a strategy to solve them all simultaneously. I will solve the conservation of momentum equations for the components of the final velocity of the first rock. Then I can use this result in the conservation of energy equation. Applying conservation of momentum to our collision in the x-direction, we get

px,i
=
px,f
m1 v1x,i + m2 v2x,i
=
m1 v1x,f + m2 v2x,f
m1 v1x,i
=
m1 v1x,f + m2 v2x,f
m1 v1x,f
=
m1 v1x,i - m2 v2x,f
v1x,f
=
v1x,i - m2
m1
 v2x,f
v1x,f
=
v1x,i - m2
m1
 v2,f cosf
(1)

Applying conservation of momentum to our collision in the y-direction, we get

py,i
=
py,f
m1 v1y,i + m2 v2y,i
=
m1 v1y,f + m2 v2y,f
0
=
-m1 v1y,f + m2 v2y,f
m1 v1y,f
=
m2 v2y,f
v1y,f
=
m2
m1
 v2y,f
v1y,f
=
m2
m1
 v2,f sinf
(2)

Now we know we must use the conservation of kinetic energy to finish solving the problem. This will be easier if we combine Equations (1) and (2) remembering the Pythagorean Theorem and some trigonometry:

v1x,f2 + v1y,f2
=
v1,f2
sin2 f+ cos2 f
=
1

Substituting from Equations (1) and (2) and using these relations allow us to determine

v1,f2
=
v1x,f2 + v1y,f2
=
[v1x,i - m2
m1
 v2,f cosf]2 + [ m2
m1
 v2,f sinf]2
=
v1x,i2 - 2 m2
m1
 v2,fcosfv1x,i + ( m2
m1
 )2 v2,f2 cos2f + ( m2
m1
 )2 v2,f2 sin2f
v1,f2
=
v1x,i2 - 2 m2
m1
 v2,fcosfv1x,i + ( m2
m1
 )2 v2,f2
(3)

Now we can look at the conservation of kinetic energy in the collision. This equation looks like

K1,i + K2,i
=
K1,f + K2,f
1/2 m1 v1,i2 + 1/2 m2 v2,i2
=
1/2 m1 v1,f2 + 1/2 m2 v2,f2
m1 v1,i2
=
m1 v1,f2 + m2 v2,f2
m1 v1,f2
=
m2 v2,f2 - m1 v1,i2
v1,f2
=
v1,i2 - m2
m1
 v2,f2
v1,f2
=
v1x,i2 - m2
m1
 v2,f2
(4)

We can set Equation (3) equal to Equation (4) to get

v1x,i2 - m2
m1
 v2,f2
=
v1x,i2 - 2 m2
m1
 v2,fcosfv1x,i + ( m2
m1
 )2 v2,f2
- m2
m1
 v2,f2
=
- 2 m2
m1
 v2,fcosfv1x,i + ( m2
m1
 )2 v2,f2
v2,f2 [( m2
m1
 )2 + m2
m1
 ]
=
2 m2
m1
 v1x,i cosfv2,f
v2,f [( m2
m1
 )2 + m2
m1
 ]
=
2 m2
m1
 v1x,i cosf
v2,f
=
2 m2
m1
 v1x,i cosf
[( m2
m1
 )2 + m2
m1
 ]

Plugging in our numbers, we get

v2,f = 1.166  m/s
Putting this into Equation (4), we get
v1,f = 0.923  m/s
Finally, we can put this into Equation (2) to get
q = sin-1 é
ê
ë 		m2 v2f
m1 v1f
 sinf ù
ú
û 		= 5.17°

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Solutions translated from TEX by TTH, version 1.57.