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A particle A of mass 2m is moving with speed 2u on a smooth horizontal table. The particle collides directly with a particle B of mass 4m moving with speed u in the same direction as A. The coefficient of restitution between A and B is 0.5.

(a) Show that the speed of B after the collision is 2/3u

(b) Find the speed of A after the collision. which im sure is just u

Subsequently B collides directly with a particle C of mass m which is at rest on the table. The coefficient of restitution between B and C is e. Given that there are no further collisions,

(c) find the range of possible values for e. (im not quite sure how to do this one, any help please?)

(a) Show that the speed of B after the collision is 2/3u

(b) Find the speed of A after the collision. which im sure is just u

Subsequently B collides directly with a particle C of mass m which is at rest on the table. The coefficient of restitution between B and C is e. Given that there are no further collisions,

(c) find the range of possible values for e. (im not quite sure how to do this one, any help please?)

Only maths exams, C2, C3 and M2 left; so i'll be in the maths forums often if you need my help.

m_{A} = 2m

m_{B} = 4m

u_{A} = 2u

u_{B} = u

v_{A} = vA

v_{B} = vB

Conservation of momentum

2m x 2u + 4m x u = 2m x vA + 4m x vB

4mu + 4mu = 2mvA + 4mvB

8mu = 2mvA + 4mvB

8u = 2vA + 4vB (1)

Coefficient of restitution

e = (vB - vA)/(2u - u) = (vB - vA)/u

vB - vA = eu

vA = vB - eu (2)

(2) into (1)

8u = 2(vB - eu) + 4vB

8u = 2vB - 2eu + 4vB

6vB = 8u + 2eu

6vB = 8u + 2(½)u

6vB = 9u

vB = 9u/6 = 3u/2

I think this is right.

vA = vB - eu

vA = vB - ½u

vA = 3u/2 - ½u

vA = u

mB = 4m

uB = 3u/2

vB = vB

mC = m

uC = 0

vC = vC

e = e

If there are no further collisions after this collision vB > u because otherwise A would catch up with B and collide again.

Conservation of momentum

4m x 3u/2 + m x 0 = 4m x vB + m x vC

6mu = 4mvB + mvC

6u = 4vB + vC

vC = 6u - 4vB (3)

Coefficient of restitution

e = (vC - vB)/3u/2

e = 2(vC - vB)/3u

3eu/2 = vC - vB

3eu/2 = 6u - 4vB - vB

5vB = 6u - 3eu/2

vB = 6u/5 - 3eu/10

vB > u

=> 6u/5 - 3eu/10 > u

=> 6u/5 - u > 3eu/10

=> u/5 > 3eu/10

=> 2u > 3eu

=> 3eu < 2u

=> 3e < 2

=> e < 2/3

A particle A of mass 2m is moving with speed 2u on a smooth horizontal table. The particle collides directly with a particle B of mass 4m moving with speed u in the same direction as A. The coefficient of restitution between A and B is 0.5.

(a) Show that the speed of B after the collision is 2/3u

(a) Show that the speed of B after the collision is 2/3u

m

m

u

u

v

v

Conservation of momentum

2m x 2u + 4m x u = 2m x vA + 4m x vB

4mu + 4mu = 2mvA + 4mvB

8mu = 2mvA + 4mvB

8u = 2vA + 4vB (1)

Coefficient of restitution

e = (vB - vA)/(2u - u) = (vB - vA)/u

vB - vA = eu

vA = vB - eu (2)

(2) into (1)

8u = 2(vB - eu) + 4vB

8u = 2vB - 2eu + 4vB

6vB = 8u + 2eu

6vB = 8u + 2(½)u

6vB = 9u

vB = 9u/6 = 3u/2

I think this is right.

(b) Find the speed of A after the collision.

vA = vB - eu

vA = vB - ½u

vA = 3u/2 - ½u

vA = u

Subsequently B collides directly with a particle C of mass m which is at rest on the table. The coefficient of restitution between B and C is e. Given that there are no further collisions,

(c) find the range of possible values for e.

(c) find the range of possible values for e.

mB = 4m

uB = 3u/2

vB = vB

mC = m

uC = 0

vC = vC

e = e

If there are no further collisions after this collision vB > u because otherwise A would catch up with B and collide again.

Conservation of momentum

4m x 3u/2 + m x 0 = 4m x vB + m x vC

6mu = 4mvB + mvC

6u = 4vB + vC

vC = 6u - 4vB (3)

Coefficient of restitution

e = (vC - vB)/3u/2

e = 2(vC - vB)/3u

3eu/2 = vC - vB

3eu/2 = 6u - 4vB - vB

5vB = 6u - 3eu/2

vB = 6u/5 - 3eu/10

vB > u

=> 6u/5 - 3eu/10 > u

=> 6u/5 - u > 3eu/10

=> u/5 > 3eu/10

=> 2u > 3eu

=> 3eu < 2u

=> 3e < 2

=> e < 2/3

- the issac physics astronomy question
- Device for 6th form and GCSEs
- What units do I need for Edexcel IAL or IAS Further Maths
- Isaac Physics Essential Pre-Uni Physics C2.5 part E
- Amount of Substance question
- My family say I’m stupid for not wanting to spend more than £1.1k on a laptop
- Weighing at Sea
- GCSE
- Type of ipad and keyboard for uni
- Edexcel IAL Maths & Further Mathematics
- nuclear physics inverse square law
- Best Laptops for uni?
- ipad air or macbook air m2 for sixth form?
- 1st year of uni- iPad dilemma
- a level physics- Engineering option question (1)
- Recommend macbook
- Stationary Points 5
- iPad gen 10
- Deciding Edexcel IAL FM Cash-in Unit Combinations
- Biological Natural Sciences @ Cambridge (2025 entry)

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