Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    A mass of 2kg moving at 35ms-1 catches up and collides with a mass of 10kg moving in the same direction at 20ms-1. Five seconds after the impact the 10kg mass encounters a fixed barrier which reduces it to rest. Assuming the coefficient of restitution between the masses is 3/5, find the further time that will elapse before the 2kg mass strikes the 10kg mass again. You may assume that the masses are moving on a smooth surface and have constant velocity between collisions.

    Book has answer as 3s, I got 6.38s.

    I found the speeds after collision as 65/3ms-1 for the 2kg mass and 83/3ms-1 for the 10kg mass.

    Then using 5s and the speed of the 10kg mass, I found the distance to the barrier.

    Finally knowing the 2kg and 10kg mass travel the same distance to the barrier where they collide, I used the distance and the speed of the 2kg mass to calculate the time taken to reach the barrier and so collide again.

    Can anybody see where I have gone wrong? Thanks...
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by fayled)
    Can anybody see where I have gone wrong? Thanks...
    Your velocities after collision are incorrect, and should be 15 and 24 m/s.
    • Political Ambassador
    Offline

    3
    ReputationRep:
    Political Ambassador
    (Original post by fayled)
    A mass of 2kg moving at 35ms-1 catches up and collides with a mass of 10kg moving in the same direction at 20ms-1. Five seconds after the impact the 10kg mass encounters a fixed barrier which reduces it to rest. Assuming the coefficient of restitution between the masses is 3/5, find the further time that will elapse before the 2kg mass strikes the 10kg mass again. You may assume that the masses are moving on a smooth surface and have constant velocity between collisions.

    Book has answer as 3s, I got 6.38s.

    I found the speeds after collision as 65/3ms-1 for the 2kg mass and 83/3ms-1 for the 10kg mass.

    Then using 5s and the speed of the 10kg mass, I found the distance to the barrier.

    Finally knowing the 2kg and 10kg mass travel the same distance to the barrier where they collide, I used the distance and the speed of the 2kg mass to calculate the time taken to reach the barrier and so collide again.

    Can anybody see where I have gone wrong? Thanks...
    Well, taking the speed of the 2kg mass after the collision to be x, and the same for the 10kg mass to be y, momentum gives

    5y +x = 135.

    Your values give 160 when added in this way.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Your velocities after collision are incorrect, and should be 15 and 24 m/s.
    (Original post by Indeterminate)
    Well, taking the speed of the 2kg mass after the collision to be x, and the same for the 10kg mass to be y, momentum gives

    5y +x = 135.

    Your values give 160 when added in this way.

    Yeah that was a poor mistake, checked my answer so many times and still never realised.

    However even with these values I get the answer as 8s...
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by fayled)
    Yeah that was a poor mistake, checked my answer so many times and still never realised.

    However even with these values I get the answer as 8s...
    And subtract the 5 seconds it took the 10kg mass to get to the wall....
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    And subtract the 5 seconds it took the 10kg mass to get to the wall....
    Ohhhhhhhhhh. Thanks.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    And subtract the 5 seconds it took the 10kg mass to get to the wall....
    Actually why would you do that? The question asks for the time from when they collide and then meet again, which would be 8s from the first collision when they both get to the wall...
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by fayled)
    Actually why would you do that? The question asks for the time from when they collide and then meet again, which would be 8s from the first collision when they both get to the wall...
    Nope, the question asks for the further time from when the 10kg hits to wall, to when the 2kg meets it.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Nope, the question asks for the further time from when the 10kg hits to wall, to when the 2kg meets it.
    Oh yes, I don't think it's written very clearly - glad you saw it that way.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.