Lesson 6 – 8 questions – Conservation of Momentum
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1. State the principle of conservation of momentum and explain what is meant by
elastic collisions and inelastic collisions.
……The momentum before a collision or explosion = the momentum after that
collision or explosion………………………………………………………………..
…………………………………………………………………………………………..
……In elastic collisions momentum and KE are conserved.………..
…………………………………………………………………………………………..
……In inelastic collisions momentum is conserved but the KE before the collision is
greater than afterwards.…………..
……………………………………………………………………………………… (3)
2 In the fission of a uranium 235 nucleus a neutron collides with the nucleus causing
it to break up. The particles formed are two smaller nuclei and three neutrons. Is the
linear momentum of the system conserved during the collision? Explain your answer.
…………………………………………………………………………………………..
Yes – momentum is conserved in all explosions and collisions as long as all particles
involved are taken into account
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
MOST
3A ball of mass 210 g moving at a speed of 23ms–1 hits a wall at right-angles and
rebounds at the same speed. The ball is in contact with the wall for 0.31 s.
a) Calculate the change in momentum of the ball.
Δp=mΔv = 0.210 × (–23 – 23) (original direction taken as ‘positive’) [1]
Δp= –9.66kgms–1 ≈ –9.7kgms–1 [1]
(The minus implies that the force exerted by the wall on the ball is in the opposite
direction to its initial direction of travel.)
change in momentum = ……–9.7kgms–1 …………..(2)
b) Is the momentum of the ball conserved? Explain your answer.
The momentum of the ball itself is not conserved. [1]
The total momentum of the wall and the ball is conserved. The wall gains momentum
equal to 9.7kgms–1 but because it is massive its velocity is negligible. [1] ……… (2)
c) Calculate the magnitude of the average force acting on the ball.
F = Δp / Δt [1]
Δp= –9.66kgms–1, Δt = 0.31 s
F = –9.66 /0.31 (magnitude only) [1]; F ≈ 31N [1]
Average Force = ……31N …………….(3)
4 A ball mass 2 kg moving left to right at 1.5 ms -1 collides with a ball of mass 3 kg
moving from right to left at 0.2 ms-1. After the collision the 2 kg ball moves from left
to right at 1 ms-1.
Draw a diagram.
Appropriate diagram drawn
Calculate:
(a) the change of momentum of the 2 kg ball
= m[v – u]
= 2[1 - 1.5]
= -1 Ns
Change in momentum = ………-1 Ns…………… (2)
(b) the momentum of the 3 kg ball after the collision
Momentum change
Momentum of 3 kg ball after collision
= -3x0.2 + 1
= 0.4 Ns
momentum = …………0.4 Ns ………… (2)
(c) the velocity of the 3 kg ball after the collision
Velocity of 3 kg ball = 0.4/3
= 0.13 ms-1
velocity = ………0.13 ms-1…………… (2)
5 A 1 kg ball of plasticene moving at 4 ms-1 collides with another ball of mass 2 kg
initially at rest. They stick together and move off.
Draw a diagram
Appropriate diagram drawn
Calculate:
(a) the total momentum of the system before the collision
Momentum
= 1x4 + 2x0
= 4 kgms-1
total momentum =………4 Ns ……………… (2)
(b) the velocity of the combined balls after impact
Velocity
= Momentum/mass
= 4/3
= 1.33 ms-1 in the direction that the 1 kg was moving before the
collision
velocity = ……1.33 ms-1 ………… (2)
SOME
(c) the kinetic energy converted to other forms
Kinetic energy converted
= 0.5x1x16 – 0.5x3x1.77
= 8 – 2.65
kinetic energy = …5.35 J …………… (2)
6 A radium 226 nucleus decays by the emission of an alpha particle into a radon 222
nucleus. If the velocity of the alpha particle is 2x106 ms-1 what is the recoil velocity of
the nucleus?
222xv = 4x2x106
Therefore: v = 8x106/222
Recoil velocity = …3.6x104 ms-1……………… (2)
7The diagram shows two toy trains T and R held in place on a level track against the
force exerted by the compressed spring.
When the trains are released, R moves to the right at a speed of 3.8ms–1. The spring
takes 0.25 s to uncoil to its natural length. Calculate:
a) the velocity of train T;
Initial momentum = final momentum [1]
Moving towards the right is taken as the ‘positive’ direction.
0=(0.500×3.8)+(0.310×v) (v is the velocity of T) [1]
v = – 0.500×3.8 /0.310 (the minus sign means that T moves to the left) [1]
v = –6.13ms–1 ≈ –6.1ms–1 [1]
velocity = ……–6.1ms–1……… (4)
b) the average force exerted by the spring on each train.
F = Δp / Δt [1]
Δp= 0.500×3.8=1.9kgms–1, Δt =0.25 s [1]
F =1.9/0.25 [1];
average force = …7.6N …[1]………. (4)
8A 850kg cannon fires a 20kg shell at a velocity of 180ms–1.
a) Calculate the final momentum of the shell.
p=mv = 20×180 [1]
p= 3.6×103 kgms–1 [1]
final momentum = ……3.6×103 kgms–1 …………….(2)
b) What is the magnitude of the momentum of the cannon immediately after
the shell is fired? (You may assume that the cannon is initially at rest.)
The momentum is conserved in this explosion. The momentum of the cannon is equal
in magnitude but opposite in direction to that of the shell. [1]
Momentum of the cannon= 3.6×103 kgms–1 [1]
magnitude of the momentum = ……3.6×103 kgms–1 ………………(2)
c) Calculate the recoil velocity V of the cannon.
Using the answer from b, we have:
850×V= 3.6×103 [1]
V = 3.6×103 / 850[1]; V≈4.2ms–1 [1]
recoil velocity = ……4.2ms–1 ………………[3]
9The diagram shows flour falling onto a horizontally moving conveyor.
The flour falls vertically onto the conveyor belt at a constant rate of 3.2kg s–1. The
conveyor belt is moving at a constant speed of 1.5ms–1. Calculate the horizontal force
required to keep the belt moving.
In a time interval of 1 s we have:
change in the horizontal momentum of the flour Δp= 3.2 × 1.5= 4.8kgms–1 [2]
Δt =1.0 s
Using Newton’s second law F Δp / Δt = [1]; so we have F =4.8/1.0 = 4.8 N [1]
Horizontal force = ……4.8 N …………(4)
10 stationary radioactive nucleus of mass M ejects an alpha particle of mass m at a
speed of 2.0x107ms–1. Given M = 55m, calculate the kinetic energy of the alpha
particle as a percentage of the final total kinetic energy.
kinetic energy of the alpha particle = ……98………..% [6]