Mathematics · Oct / Nov 2022 · Paper 51 · Question 1
YMA01
As printed in the paper
The question as text
1. A railway truck S of mass 20 tonnes is moving along a straight horizontal track when it collides with another railway truck T of mass 30 tonnes which is at rest. Immediately before the collision the speed of S is 4 m s–1 As a result of the collision, the two railway trucks join together. Find
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1(a)[2]
v
v
=
1.6(m s )
−
1
v
1(b)[3]
±
20000(1.6
−
4)
OR 30000 1.6
±
×
48000 N s or 48 kN s
.
Notes for question 1
M1 for a CLM equation, condone sign errors and extra g’s and any equivalent
1(a)
equation (e.g. 2 x 4 = 5v, 20 x 4 = 50v, 200 x 4 = 500v,… etc)
OR : for equating impulses
A1 oe Units not needed but must be positive.
M1 impulsemomentum equation, dimensionally correct, correct no. of
1(b)
terms, condone sign errors but must be attempting a difference of momenta
(allow 20 or 30 for the mass, M0 if g included or mass omitted)
A1ft a correct equation, follow through on their v
(allow 20 or 30 for the mass)
N.B. If using S to find the impulse, 4 and their v must have opposite signs
when awarding the A1ft.
m s−) and must be positive.
1
A1 cao units needed (allow kg
3
2
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