# -Momentum-in-2-Dimensions-1-Solution ```Le Fevre High School
SACE Stage 2 Physics
Motion in 2-Dimensions 1 Solution
1. An empty freight car A of mA = 2.5 x 104kg coasts along a horizontal track at 2.0 m s-1 until it
couples to a stationary car B of mB = 5.0 x 104kg. There is negligible friction with the track,
and the brakes are off.
(a) What is the initial momentum of car A?
p = m v = 2.5 x 104 x 2 = 5 x 104 kg m s-1 in the direction of motion of the car.
A1
A
A1
(b) What is the speed of the two cars as they move along the track after the interaction?
p = 0 total momentum before the collision is
B1
p = p + p = 5 x 104 kg m s-1 in direction of car A.
1
A1
B
As there are no external forces acting the total momentum after the collision is equal to
the total momentum before the collision.
 p 2 = 5 x 104 kg m s-1 in direction of car A.
But p 2 = (mA + mB) v 2
 v2 =
as the cars couple.
4
5 x 10
2
=
= 0.67 m s-1 in the direction of the initial direction of car A.
4
mA  mB
7.5 x 10
(c) What is the total kinetic energy before and after the impact? Is this an example of an elastic
collision?
Kinetic Energy Before,
1
1
K1 = 2 mA vA2 = 2 x 2.5 x 104 x 22 = 5 x 104 J
Kinetic Energy After,,
K2 = 1/2(mA + MB) x 0.6672
= 1 x 7.5 x 104 x 0.6672 = 1.7 x 104 J
2
Kinetic energy is lost because heat and sound is generated as the cars link
together.
2. A high powered rifle whose mass is 5kg fires a 15g bullet with a muzzle velocity of
3 x 104 cm s-1.
(a) Calculate the recoil velocity of the rifle?
p = 0
1
p
2
= p
R2
+ p
B2
As there are no external forces the total momentum after the interaction equals the total
momentum before the interaction.
 p 1= p2
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0 = p
R2
+ p
B2
= mR v R2 + mB v B2
= 5 v + 15 x 10-3 x 3 x 102
= 5 v + 4.5
-4.5
 v = 5 = - 0.9 m s-1
The rifle will recoil at 0.9 m s-1 in the opposite direction to the bullet.
(b)Calculate the total kinetic energy of the system before and after firing.
Kinetic Energy before,
KT1 = 0
Kinetic Energy After,
KT2 = KR2 + KB2
1
1
= 2 mR VR22 + 2 mB VB22
1
1
= 2 x 5 (- 0.9)2 + 2 x 15 x 10-3 x (3 x 102)2
= 2.025 + 675
= 677.025
 680 J
(c) Where does the final kinetic energy come from?
The final K comes from chemical energy from the explosion when the bullet is fired.
3. A small cart of mass 300 g when empty can move freely on a horizontal surface. The cart
contains 100 g of sand and moves initially with a speed of 10 cm s-1.
(a) What would be the speed of the cart if a further 200 g of sand were dropped vertically
into it from a stationary container?
The sand initially has no horizontal component of velocity. The cart exerts a force on the
sand to accelerate it and consequently the sand exerts a force on the cart to slow it down.
These forces are internal to the cart sand system and so momentum will be conserved.
(neglect friction).
 momentum before = momentum after
m1 v1 = m2 v2
 v2 =
=
m1v1
m2
(300  100) x 10
400  200
400
= 600 x 10 = 6. 7 cm s-1
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(b) If, instead of adding sand to the cart, the original 100 g of sand were allowed to pour out
through a hole in the bottom of the cart, what would be the final speed of the cart?
As the sand pours out the bottom it retains its velocity of 10 cm s-1 as the cart does not
exert any force on it. Hence the sand exerts no force back on the cart. So the cart retains
its original velocity. Momentum is still conserved horizontally because the sand continues
to move through the air with a horizontal velocity of 10 cm s-1. The vertical momentum of
the falling sand is not conserved because there is an external vertical force acting on the
falling sand via. gravity. Hence vertical momentum increases.
4. A ball A with a mass of 1.0 kg and moving at 4.0 m s-1 strikes a glancing blow on a second ball
B which is initially at rest. Assume no external forces act. After the collision, ball A is
moving at right angles to its original direction at a speed of 3.0 m s-1, as illustrated below:
(a) Calculate the initial momentum of the system.
total momentum of system
= momentum of A before collision
= mA vA
=4 kg m s-1 to right.
(b) Determine the magnitude of the momentum of B after the collision?
Also p total = p B2 + p A2
p total = 4 kg m
37s-1o
pB2 = 5 kg m
s-1
 p
= p total - p
= 5 kg m s-1
B2
A2
- pA2 = 3 kg m
s-1
(vector subtraction)
(c) In what direction is B moving after the collision?
37o to right of A's initial direction
(d) If B has a mass of 5.0 kg, what is its speed after the collision?
v=
p
m
=
5
5
= 1 m s-1
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(e) If the impact lasted 0.020 s, calculate the magnitude of the average force exerted on B
during the collision.
Fave =
=
m v
p

t
t
5
= 250 N (2sf)
0.020
(f) Was this a collision in which the total initial kinetic energy is the same as the total final kinetic
2
1
Kbefore = KA =
mv
2
1
= 2 x 1 x (4)2 = 8 J
Kafter = KA + KB
1
1
= 2 x1x32+2 x5x12
= 4.5 +2.5
= 7J
No as kinetic energy is lost in collision.
5 A car of mass 1000 kg which has a velocity of 10 m s-1 east collides at a crossroads with a car of
mass 800 kg which has a velocity of 5.0 m s-1 north. If the cars lock together after the
collision, calculate their velocity immediately after the collision. How much kinetic energy is
converted to other forms in the collision? Neglect external forces. (Calculate Total Kinetic
energy before and after)
p
p
1car2
1car1
= m v 1 = 1000 x 10 = 10 000 kg m s-1 East
= m v 1 = 800 x 5 = 4000 kg m s-1 North
 total momentum before collision given by vector addition
 4000 kg m s1
10000 kg m s-1
tan  
10000
4000
 = 68.0 o
total momentum after=
4000 2 + 100002
1000 4 2 + 10 2
=
= 1000 116
= 10770 kg m s-1 68.2 o T
 total momentum after collision = 10770 kg m s-1 68.2o T
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 velocity =
10770
1800
= 5.98
i.e. v = 6.0 m s-1 68o T
2
2
1
1
Kbefore = 2 x 1000 x 10 + 2 800 x 5
= 60 000 J
2
1
Kafter = 2 x 1800 x 5.98 = 32 400 J
i.e. 27 600 Joules converted to other forms.
6. (a) State the law of conservation of linear momentum.
When no external forces act on a system of objects the total momentum of that system is
conserved independent of any interaction of its parts.
(b) In a certain road accident a car of mass 2.0 x 103 kg, travelling south, collided in the
middle of an intersection with a truck of mass 6.0 x 103 kg, travelling west. The vehicles
locked and slid freely (that is with negligible friction) off the road along a line pointing
southwest. The truck entered the intersection at 32 kilometre per hour.
The truck entered the intersection at 32 kilometre per hour.
As no external forces act (negligible friction) momentum is conserved.
Let vc be the initial speed of the car.
Then total momentum before collision. =
= p car + p truck
= mc vc1 + mt vt1
3
= p car + 6.0 x 10 x 32 West
=p
5
car
+ 1.92 x 10 kg km hr-1 West
total momentum after collision
= (mc + mt ) v Southwest where v is the velocity of the car and truck.
3
= 8 x 10 x v
p truck
total momentum before = total
momentum after = ??
pcar
But we do know it is South-west and
= 8 x 103 x v
45 o
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(i) Determine the speed at which the car entered the intersection.
Since triangle is right- angled with acute angle of 45o
p
= p car
 1.92 x 10 5 = 2 x 10 3 x vc1
 vc1 =
truck
1.92 x 105
2 x 103
= 96 km hr-1(26.7 ms-1)
(ii) Calculate the total initial and final kinetic energies of the vehicles. Account for any
difference.
From diagram
cos 45o =
truck
8 x 103 x v
 speed of car and truck after collision
v=
=
truck
3
8 x 10 x cos 45o
1.92 x 105
8 x 103 cos 45o
= 33.9 km hr-1 (9.42 m s-1)
 Total initial K = Kcar + Ktruck
3
2
3
2
1
1
2 x 10 x 96 + 6 x 10 x 32
2
2
6
6
= 9.22 x 10 + 3.07 x 10
6
2
-2
= 12.3 x 10 kg km hr
=
Total final K = Kcar and truck combined
3
2
1
= 2 8 x 10 x 33.9
6
2
-2
= 4.6 x 10 kg km hr
(note no need to convert hours to seconds)
i.e. Kinetic energy has been converted to other forms during the collision. Some would have been
converted to heat when work was done against the resisting forces when the metal of the cars was
distorted. A small percentage would have been converted to sound.
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7. A ball of mass 50.0 g rolling on a smooth level floor at a speed of 12.5 m s-1 collides with a
wall at an angle of 45o and rebounds without changing speed. Calculate the change in
momentum of the ball.
45
45
pinitial = mv = 0.050x12.5 = 0.625 kg m s-1
pfinal = 0.625 kg m s-1
p = pfinal - pinitial
45o
45o
90o
p = 62.5 2  62.5 2
= 0.884 kg m s-1 at 90o and away from the wall
8. A radio-active nucleus decays by emitting an electron and a neutrino at right angles to
each other. The momentum of the electron is 1.2 x 10-22 kg m s-1 and that of the
neutrino is 6.0 x 10-23 kg m s-1.
electron
neutrino
nucleus
BEFORE
(nucleus at rest)
AFTER
(&quot;explosion&quot; occurs, nucleus recoils
as particles emitted)
Le Fevre High School
(a) Assuming that the nucleus was initially at rest, find the momentum of recoil of the
nucleus.
the total momentum is conserved
 p T(after) = p T(before) = 0 as nucleus initially at rest.
p T(after) = p e + p neutrino + p recoiled nucleus
But
=0
 Three momenta after must add up as vectors to give zero. This will occur if the three
vectors complete the three sides of a triangle so that the resultant is of zero length
p e = 1.2 x 10-22 kg
m s-1
p neutrino
= 0.6 x 10-22 kg
m s-1
90o
p nucleus
-22
So p nucleus = (1.2 2 + 0.6 2 )
x 10
= 1.34 x 10-22 kg m s-1 at angle 
.6 x 10-22
1.2 x 10-22
tan  =
1
=2
  = 26.6o
pe
p nucleus
153

o
p neutrino
 p
nucleus
= 1.3 x 10
-22
o
kg m s-1 at 150 (2sf) to direction of recoil electron
(b) With what velocity does the nucleus begin to recoil?
(mass of nucleus = 3.35 x 10-25 kg).
22
v = 1.34 x 10
 25
3.35 x10
= 400 m s-1
o
i.e. v = 400 m s-1 (2 s.f.) at an angle 153 to direction of recoil electron
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