Model Answers
1a
CIE AS Physics
Topic Questions
(a) Determine the velocity v of ball X before the collision:
the known quantities:
ASList Physics
CIE Topic Questions
3.2 Linear
Momentum
Mass
of X = 140 g& Conservation
3. Dynamics
Mass of Y = 620 g
Velocity of X before the collision = v
Velocity of X after the collision = 0
Velocity of Y before the collision = 0
Velocity of Y after the collision = 0.8 m s−1
3.2 Linear Momentum &
Conservation
Use conservation of momentum to write
an expression for v:
Medium
Easy
Momentum before collision: m X v X + m Y v Y = 140v + 0
Momentum after collision: m X v X
Hard
+ Answers
m Y v Y = 0 + (620 × 0 . 8)
Conservation of momentum: Momentum before collision = Momentum after
collision
2
3
1
Conservation of momentum: 140v
1a Rearrange: v
=
= 620 × 0 . 8 [1 mark]
620 × 0 . 8
[1 mark]
140
v = 3.543 = 3.5 m s−1 [1 mark]
3 marks
A ball X of mass 140 g travelling at constant velocity v collides head-on with a
[Total: 3 marks]
stationary ball Y of mass 620 g, as shown in Fig. 1.1.
1b
(b)
(i) Show that the collision in (a) is inelastic:
List the known quantities:
Mass of X = 140 g = 0.14 kg
Mass of Y = 620 g = 0.62 kg
Velocity of X before the collision = 3.5 m s−1
Velocity of X after the collision = 0
After the collision, ball X comes to a stop. Ball Y moves off in the same initial
direction
as X with a velocity of 0.8 m s−1.
Model Answers
Determine the velocity v of ball X before the collision.
1a
(a) Determine the velocity v of ball X before the collision:
List the
quantities:
How
didknown
you do?
Mass of X = 140 g
Mass of Y = 620 g
Velocity of X before the collision = v
Velocity
of X after
the collision = 0
Stuck?
View related
notes
Velocity of Y before the collision = 0
Velocity of Y after the collision =View
0.8 mAnswer
s−1
Use conservation of momentum to write an expression for v:
Momentum before collision: m X v X
1b Momentum after collision: m v
X X
+ m Y v Y = 140v + 0
+ m Y v Y = 0 + (620 × 0 . 8)
4 marks
Conservation of momentum: Momentum before collision = Momentum after
(i) Show that the collision in (a) is inelastic.
collision
140vin=
× 0 . 8 [1 mark]
Conservation
momentum:
(ii) Explainof
why
the collision
(a)620
is inelastic.
620 × 0 . 8
Rearrange: v =
[1 mark]
140
[3]
[1]
How vdid
you do?
= 3.543
= 3.5 m s−1 [1 mark]
[Total: 3 marks]
1b
Stuck? View related notes
(b)
View Answer
(i) Show that the collision in (a) is inelastic:
List the known quantities:
Mass of X = 140 g = 0.14 kg
1c Mass of Y = 620 g = 0.62 kg
Velocity of X before the collision = 3.5 m s−1
Velocity of X after the collision = 0
3 marks
The balls are replaced by an aluminium sphere of mass 2.7 kg and a steel
sphere
7.9 kg as shown in Fig. 1.2.
ModelofAnswers
The aluminium sphere has an initial velocity of 10.0 m s–1 when it collides with
1a stationary steel sphere. Immediately after the collision, the velocity of
the
the steel sphere is 5.1 m s–1.
(a) Determine the velocity v of ball X before the collision:
List the known quantities:
Mass of X = 140 g
Mass of Y = 620 g
Velocity of X before the collision = v
Velocity of X after the collision = 0
Velocity of Y before the collision = 0
Velocity of Y after the collision = 0.8 m s−1
Use conservation of momentum to write an expression for v:
Momentum before collision: m X v X
+ m Y v Y = 140v + 0
Momentum
after collision:
v + m Y v sphere
= 0 + (620 × 0 . 8)
X X
Y
Calculate
the velocity
v of them aluminium
immediately after the
collision.
Conservation of momentum: Momentum before collision = Momentum after
collision
HowConservation
did you do? of momentum: 140v
Rearrange: v
=
= 620 × 0 . 8 [1 mark]
620 × 0 . 8
[1 mark]
140
v =View
3.543related
= 3.5 mnotes
s−1 [1 mark]
Stuck?
[Total: 3 marks]
View Answer
1b
(b)
1d
(i) Show that the collision in (a) is inelastic:
List the known quantities:
Verify that the collision in (c) is elastic.
Mass of X = 140 g = 0.14 kg
Mass of Y = 620 g = 0.62 kg
of X before the collision = 3.5 m s−1
HowVelocity
did you do?
Velocity of X after the collision = 0
3 marks
Model Answers
Stuck? View related notes
1a
View Answer
(a) Determine the velocity v of ball X before the collision:
List the known quantities:
Mass of X = 140 g Did this page help you?
Mass of Y = 620 g
Velocity of X before the collision = v
Velocity of X after the collision = 0
Velocity of Y before the collision = 0
Velocity of Y after the collision = 0.8 m s−1
Yes
No
Next question
Use conservation of momentum to write an expression for v:
YOUR PROGRESS
before collision: m X v X
CIE Momentum
AS Physics
+ m Y v Y = 140v + 0
14%Momentum after collision: m X v X + m Y v Y = 0 + (620 × 0 . 8)
Conservation of momentum: Momentum before collision = Momentum after
collision
1. Physical Quantities & Units
16%
Conservation of momentum: 140v = 620 × 0 . 8 [1 mark]
2. Kinematics
Rearrange: v
=
620 × 0 . 8
[1 mark]
140
37%
v = 3.543 = 3.5 m s−1 [1 mark]
3. Dynamics
[Total: 3 marks]
37%
1b
3.1 Newton's Laws of Motion
(b)
36%
(i) Show that the collision in (a) is inelastic:
3.2 Linear Momentum & Conservation
List the known quantities:
39%
Multiple
Choice
Mass of
X = 140 g = 0.14 kg
Mass of Y = 620 g = 0.62 kg
Velocity of X before the collision = 3.5 m s−1
Velocity of X after the collision = 0
87%
Structured Questions
0%
Model Answers
4.1a
Forces, Density & Pressure
45%
(a) Determine the velocity v of ball X before the collision:
List
the known
quantities:
5.
Work,
Energy
& Power
0%
Mass of X = 140 g
Mass of Y = 620 g
6. Deformation
Solidsthe collision = v
Velocity of Xofbefore
Velocity of X after the collision = 0
Velocity of Y before the collision = 0
Velocity of Y after the collision = 0.8 m s−1
7. Waves
0%
0%
Use conservation of momentum to write an expression for v:
Momentum before collision: m X v X
8. Superposition
Momentum after collision: m X v X
+ m Y v Y = 140v + 0
0%
+ m Y v Y = 0 + (620 × 0 . 8)
9. Electricity
0%
Conservation of momentum: Momentum before collision = Momentum
after
collision
Conservation of momentum: 140v
10. D.C. Circuits
Rearrange: v
=
= 620 × 0 . 8 [1 mark]
0%
620 × 0 . 8
[1 mark]
140
11. Particle Physics −1
v = 3.543 = 3.5 m s [1 mark]
[Total: 3 marks]
1b
(b)
Resources
Members
(i) Show that the collision in (a) is inelastic:
Join
Launchpad
List the known quantities:
FAQ
Account
Mass of X = 140 g = 0.14 kg
Support
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Mass of Y = 620 g = 0.62 kg
Velocity of X before the collision = 3.5 m s−1
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Velocity of X after the collision = 0
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(a) Determine the velocity v of ball X before the collision:
Physics
List the known quantities:
Maths
2022 Advance Information
Mass of X = 140 g
Mass of Y = 620 g
Velocity of X before the collision = v
Velocity of X after the collision = 0
Velocity of Y before the collision = 0
Velocity of Y after the collision = 0.8 m s−1
Missing something? Tell us
Use conservation of momentum to write an expression for v:
© Copyright 2015-2023 Save My Exams Ltd. All Rights Reserved.
Momentum before collision: m v + m Y v Y = 140v + 0
IBO was not involved in the production of,XandX does not
endorse, the resources created by Save My
Exams.
Momentum after collision: m X v X + m Y v Y = 0 + (620 × 0 . 8)
Conservation of momentum: Momentum before collision = Momentum after
collision
Conservation of momentum: 140v
Rearrange: v
=
= 620 × 0 . 8 [1 mark]
620 × 0 . 8
[1 mark]
140
v = 3.543 = 3.5 m s−1 [1 mark]
[Total: 3 marks]
1b
(b)
(i) Show that the collision in (a) is inelastic:
List the known quantities:
Mass of X = 140 g = 0.14 kg
Mass of Y = 620 g = 0.62 kg
Velocity of X before the collision = 3.5 m s−1
Velocity of X after the collision = 0