Name
Date
Pd
Impulsive Force Model Worksheet 3:
Conservation of Momentum I
(All units of momentum in the momentum conservation diagrams are kgsm .)
1. In a railroad yard, a train is being assembled. An empty boxcar, coasting at 3 m/s, strikes a loaded
car that is stationary, and the cars couple together. Each of the boxcars has a mass of 9000 kg when
empty, and the loaded car contains 55,000 kg of lumber.
a. Complete the momentum conservation diagram.
event: A totally inelastic collision.
initial
object/mass/velocity
final
object/mass/velocity
9,000 kg at 3.0 m/s
73,000 kg at ? m/s
64,000 kg at 0 m/s
Momentum conservation equation:
m9000iv9000i  0  (m9000  m64,000 )vf
+
0
–
27,000
b.
27,000
0
–
+
c. Find the speed of the coupled boxcars.
9,000kg(3.0 ms )
m9000i v9000i
vf 
 vf 
 0.37 ms
(m9000  m64,000 )
(9,000kg  64,000kg )
2. An astronaut of mass 80 kg carries an empty oxygen tank of mass 10 kg. By pushing the tank away
with a speed of 2.0 m/s, the astronaut recoils in the opposite direction.
a. Complete the momentum conservation diagram.
event: Two objects in contact at rest, one object throws the other.
initial
object/mass/velocity
final
object/mass/velocity
80 kg at 0 m/s
80 kg at ? m/s
10 kg at 0 m/s
10 kg at -2.0 m/s
b. Momentum conservation equation:
+
–
0
20
0
−20
–
+
m80iv80i  m10iv10i  m80f v80f  m10f v10f
c. Find the speed with which the astronaut moves off into space.
0  0  80kg(v80f )  10kg( 2.0 ms )  v80f 
©Modeling Instruction – AMTA 2013
1
20 kgsm
 0.25 ms
80kg
U9 Momentum – ws 3 v3.1
3. A tennis player returns a 30 m/s serve straight back at 25 m/s, after making contact with the ball for
0.50 s. The ball has a mass of 0.20 kg.
a. Use a momentum conservation diagram to show the change in momentum of the ball.
event: A tennis racquet provides an impulse to a ball
initial
object/mass/velocity
final
object/mass/velocity
ball, 0.20 kg, 30 m/s
ball, 0.20 kg, -25 m/s
0
–
b. Impulse-Momentum equation:
6
+
– -6
0
+
F t  mv
c. How much force did the racket exert on the ball?
F t  mv  F 
0.20kg( 25 ms  30 ms ) 0.20kg( 55 ms )

 22N
0.50s
0.50s
4. A 50 kg cart is moving across a frictionless floor at 2.0 m/s. A 70 kg boy, riding in the cart, jumps
off so that he hits the floor with zero velocity.
a. Complete the momentum conservation diagram.
event: Two objects moving together, one pushes off other.
initial
object/mass/velocity
final
object/mass/velocity
50 kg at 2.0 m/s
50 kg at ? m/s
70 kg at 2.0 m/s
70 kg at 0 m/s
+
–
0
140
0
140
–
+
b. Momentum conservation equation:
(m50i  m70i )v i  m50f v 50f  0
c. How large an impulse did the boy give to the cart?
The cart gave an impulse on the boy of 70kg(0-2.0 ms ) or −140 Ns. Therefore the boy gave an equal
and opposite impulse on the cart of 140 Ns. (Using answer from d.: 50kg(4.8 ms -2.0 ms )=140 Ns
d. What was the velocity of the cart after the boy jumped?
(50kg  70kg )2.0 ms  (50kg )v 50f  v 50 f  4.8 ms
©Modeling Instruction – AMTA 2013
2
U9 Momentum – ws 3 v3.1
5. Two girls with masses of 50 kg and 70 kg are at rest on frictionless in-line skates. The taller girl
pushes the shorter girl so that the shorter girl rolls away at a speed of 10 m/s.
a. Show the effect of the push on both girls with a momentum conservation diagram.
event: Two objects in contact at rest, one object pushes the other.
initial
object/mass/velocity
final
object/mass/velocity
50 kg at 0 m/s
50 kg at 10 m/s
70 kg at 0 m/s
70 kg at ? m/s
0
+
–
500
–
0
+
b. Momentum conservation equation:
0  0  m50 f v50 f  m70 f  v 70 f
c. Calculate the impulse that each girl imparts to the other.
The tall girl gave an impulse on the short girl of 50 kg(10 ms -0) or 500 Ns. Therefore the short
girl gave an equal and impulse of −500 Ns.
6. A 2 kg melon is balanced on a circus performer's head. An archer shoots a 50 g arrow at the melon
with a speed of 30 m/s. The arrow passes through the melon and emerges with a speed of 18 m/s.
a. Draw a momentum conservation diagram for the stunt.
event: A collision between melon and arrow where arrow is not stopped.
initial
object/mass/velocity
final
object/mass/velocity
2 kg at 0 m/s
2 kg at ? m/s
0.050 kg at 18 m/s
0.050 kg at 30 m/s
+
–
0
1.5
0
1.5
–
+
b. Momentum conservation equation:
m0.050i v 0.050 i  0  m0.050 f v 0.050 f  m2i v 2i
c. Find the speed of the melon as it flies off the performer's head.
0.050kg (30 ms )  0  0.050kg (18 ms )  (2.0kg )v 2f
v 2f 
1.5 kgsm  0.90 kgsm
 0.30 ms
2.0kg
©Modeling Instruction – AMTA 2013
3
U9 Momentum – ws 3 v3.1