Chapter 9; Linear Momentum and Collisions
 ToC, Chapter 9  Course Calendar 
D9.1
What is the momentum of a 1200 kg sedan traveling at 90 km/hr? At what speed
must a 3600 kg truck travel to have the same momentum?
First, change the speed to units of m/s,
D9.2 A 150 gram baseball initially traveling at 30 m/s is struck by a bat and leaves in the opposite direction at 35 m/s.
a) What is its change in momentum?
b) Is the change in momentum directed along the initial velocity, the final velocity, or some other direction?
c) What is the impulse delivered by the bat?
d) In hitting a baseball, why is it important to "follow through," that is, keep the bat moving and make a full swing, rather than to stop right after the ball is hit?Initially, . . .
Finally, . . .
a) p = p_{f}  p_{i} = (0.150 kg)( 35 m/s)  (0.150 kg)(30 m/s)
= (0.150 kg)( 65 m/s) =  9.75 kg m/s
b) The change in momentum is opposite to the initial momentum and/or along the same direction as the final momentum.
c) "Impulse" is but another word for "change in momentum. Therefore, the impulse is is  9.75 kg m/s
d) Impulse = p = F t
By "following through" there is contact for a longer time so that a given force can cause a greater change in momentum. Or a smaller force can cause the same change in momentum.
D9.3 A 600 kg cannon fires a 5 kg cannonball with a horizontal muzzle velocity of 120 m/s. What is the recoil velocity of the cannon?
Initially, . . .
Finally, . . .
Initially, with cannon and cannon ball both at rest, the total momentum is zero.
P_{Tot, initial} = 0
By conservation of momentum, we later expect the total momentum to still be zero. The cannon ball carries momentum to the right so it is positive. The cannon carries momentum to the left so it is negative.
P_{Tot, final} = p_{ball} + p_{cannon} = (5 kg)(120 ) + (600 kg)(v_{cannon}) = 0
(600 kg)(v_{cannon}) =  (5 kg)(120 )
v_{cannon} =  1
D9.4 An astronaut of mass 105 kg carrying an empty oxygen tank of mass 40 kg is stationary relative to a nearby space shuttle. She throws the tank away from herself with a speed of 2 m/s (measured relative to the shuttle). With what velocity relative to the shuttle does the astronaut start to move through space?
Initially, with astronaut and oxygen tank both at rest, the total momentum is zero.P_{Tot, initial} = 0
After the astronaut throws the tank, the total momentum is still zero.
P_{Tot, final} = p_{astro} + p_{tank} = (105 kg)(v_{astro}) + (40 kg)( 2.0 ) = 0
(105 kg)(v_{astro}) =  (40 kg)( 2.0 )
v_{astro} =  0.76
D9.5 A 10 000 kg railroad grain car and its load of 3 000 kg of grain coast along a level track at 3.0 m/s. A door is open slightly and lets the grain pour out at a rate of 100 kg/s. What is the speed of the grain car after the grain has all emptied from the car? What has happened to the initial kinetic energy of the railroad car?
As the grain falls out of the car, it does not change the speed of the railroad car and/or the remaining grain in the car. Much of the initial kinetic energy of the grain and car is "lost". As the grain falls from the car it carries kinetic energy with it that is lost to heat as the grain finally comes to rest on the ground.
D9.6 An inflated rubber raft of mass 30 kg carries two swimmers of mass 50 kg and 70 kg. The raft and swimmers are initially floating at rest when the swimmers simultaneously dive off from the midpoints of opposite ends of the raft, each with a horizontal velocity of 3 m/s. The 50 kg swimmer dives to the left; the 70 kg swimmer dives to the right. With what speed and in what direction does the raft start to move?
Initially, with both divers and the raft at rest, the total momentum is zero.P_{Tot, initial} = 0
After the dive, the total momentum is still zero:
P_{Tot, final}
= p_{LeftDiver} + p_{raft}
+ p_{RightDiver}
P_{Tot, final} = (50
kg)( 3 m/s) + (30 kg)(v) + (70 kg)(3 m/s) = 0
 150 kg m/s + (30 kg)(v) + 210 kg m/s = 0
(30 kg)(v) =  60 kg m/s
v =  2 m/s
That is, the raft moves to the left at 2 m/s.
Initially,
After the collision,
P_{Tot, final}
= (850 g) v = 6250 gm cm/s = P_{Tot, initial}
v = 7.35 cm/s
As the block with embedded bullet starts to move, it has KE given by
Relative to the bottom of its swing, its PE is now
zero
So its total energy is E = KE + PE = 0.62 J
At the top of its swing, the block (with embedded bullet) momentarily comes
to rest so its KE there is zero and all of its energy is now PE = m g h
By energy conservation, this final energy must still be 0.62 J
After the radioactive decay, the alpha carries momentum to the right and the thorium nucleus recoils and carries momentum to the left. The total momentum is still zero.
where m = 1.67 x 10^{27} kg
or, if we drop the units, this becomes
This is one equation with two unknows, so we must find additional information. Since this is an elastic collision, we also know that the Kinetic Energy is conserved.
= (^{1}/_{2})(600 g)(10 cm/s)^{2} + (^{1}/_{2})(400 g)( 20 cm/s)^{2} = KE_{Tot,in}_{ }
Now we have two equations in two unknowns. We can solve for v_{2f} from the earlier one and substitute it into this equation,
There are two possible solutions, due to the ± in the equation;
Either  
or 

v_{1f}(1)
=  14 
v_{1f}(2)
= 10 

Each of these provides a possible solution for v_{2f}:  
v_{2f}(1)
=  1.5 v_{1f} 
 5 v_{2f}(2)
=  1.5v_{1f }
 5 

v_{2f}(1)
=  1.5 ( 14
) 
 5 v_{2f}(2)
=  1.5( 10
)  5 

v_{2f}(1)
= 21 
 5 v_{2f}(2)
= (  15  5 ) 

v_{2f}(1)
= 16 
v_{2f}(2)
=  20 

v_{2f}(1)
= 16

v_{2f}(2)
=  20 
That is, either
v_{1f}(1) =  14 and v_{1f}(2) = 10
or
v_{2f}(1) = 16
and v_{2f}(2) =  20
The second set is somewhat interesting for
v_{1f}(2) = 10
and v_{2f}(2) =  20
are the initial conditions repeated. This solution is as if the collision
had not occured!
The solution we seek  the real, physical solution  is the first set,
glider #1 
glider #1 
glider #2 
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The text gives the velocity v_{2f} for just such a case, . Therefore,
 ToC, Chapter 9  Course Calendar 
(c) 2005, Doug Davis; all rights reserved