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# F=MA watch

1. A trolley is being pulled at a constant force with an acceleration of 0.60m/s^2
A mass of 0.5kg is then mixed to the trolley and nows accelerates at 0.48m/s^2
What is the mass of the trolley?
2. If the force stays the same then you have
Original ma=new ma
Call x the mass of the trolley (or the original mass)

Can you create an equation from this (remember the new mass is the trolley plus 0.5kg)
3. (Original post by Pastelx)
If the force stays the same then you have
Original ma=new ma
Call x the mass of the trolley (or the original mass)

Can you create an equation from this (remember the new mass is the trolley plus 0.5kg)
I get up to F = M x 0.60ms^-2 = F = (M+0.5)0.48ms^-2
4. Okay if you expand that bracket and make the equations equal each other, you should be able to take the M out of the equation and get a suitable answer!
5. (Original post by Pastelx)
Okay if you expand that bracket and make the equations equal each other, you should be able to take the M out of the equation and get a suitable answer!
M x 0.60 = (0.48M + 0.5) x 0.48
= 0.60M = 0.48M + 0.24
= 0.60M - 0.48M = 0.24
= 0.12M = 0.24
= 0.24/0.12
= 2kg
6. Seams to work.
Test 1 = 0.60ms^-2
F= 2 x 0.60 = 1.20N
Test 2 = 0.48ms^-2
F= (2+0.5)0.48 = 1.20N
7. (Original post by Lokum)
M x 0.60 = (0.48M + 0.5) x 0.48
= 0.60M = 0.48M + 0.24
= 0.60M - 0.48M = 0.24
= 0.12M = 0.24
= 0.24/0.12
= 2kg
8. Isn't there an easier method to get the answer?
9. (Original post by Lokum)
Isn't there an easier method to get the answer?
Nope just do simultaneous equations for m*a I believe

Personally I’d do less lines of working but that’s the easiest method to work it out as far as I can tell
10. Just made up a formula!

a2 x added mass / change in a

e.g 048 x 0.5 / 0.12 = 2kg

Trolley 1
A = 0.70m/s^2
Trolley 2
A = 0.50m/s^2
M2 = M1 + 1.5

M1 = 0.50 x 1.5 / (0.70-0.50) = 3.75 Kg

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