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# M2 rev ex 2 q2b watch

1. Could ne1 do this question for me from the heinamann book rev ex2, q2b
2. (Original post by lgs98jonee)
Could ne1 do this question for me from the heinamann book rev ex2, q2b
come on sum1 out there must be able to help me

Two particles A and B have masses kM and M, where k is a +ve constant. Particles r moving in same direction, along same straight line, with speeds u and 2u. There is a collision, which halves the speed of A. After collision, both A and B continue to move in same direction as before.

a. Find in terms of k and u, the speed of B after collision.
b=(k+1)u which is correct

b. In the collision, one twelfth of the total kinetic energy is lost.
Find the possible values of k
3. (Original post by lgs98jonee)
come on sum1 out there must be able to help me

Two particles A and B have masses kM and M, where k is a +ve constant. Particles r moving in same direction, along same straight line, with speeds u and 2u. There is a collision, which halves the speed of A. After collision, both A and B continue to move in same direction as before.

a. Find in terms of k and u, the speed of B after collision.
b=(k+1)u which is correct

b. In the collision, one twelfth of the total kinetic energy is lost.
Find the possible values of k
can no1 do this question??
4. (Original post by lgs98jonee)
can no1 do this question??
1/2 or 1/6
5. (Original post by It'sPhil...)
1/2 or 1/6
could i have a little of ur working pls??
6. (Original post by lgs98jonee)
could i have a little of ur working pls??
Initial KE = km/2(2u)² + mu²/2 = mu²/2(4k + 1)

Final KE = kmu²/2 + u²(1 + k)²m/2 = mu²/2(k + (1 + k)²)

Final KE = 11/12 (initial KE)

so, 12(3k + 1 + k²) = 11(4k + 1)

so 12k² - 8k + 1 = 0

(6k - 1)(2k - 1) = 0
7. (Original post by It'sPhil...)
Initial KE = km/2(2u)² + mu²/2 = mu²/2(4k + 1)

Final KE = kmu²/2 + u²(1 + k)²m/2 = mu²/2(k + (1 + k)²)

Final KE = 11/12 (initial KE)

so, 12(3k + 1 + k²) = 11(4k + 1)

so 12k² - 8k + 1 = 0

(6k - 1)(2k - 1) = 0
jeeeeez...thnx for that...did 11(4k+1)=44k+1

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