During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away
from another trolley, Y, of mass (2.561 ± 0.001) kg with a speed of (3.2 ± 0.01) ms
-1
. The
second trolley is moving away with a distance of (2.5 ± 0.01) ms-1
.
What is the absolute uncertainty of the ratio of momentum of the two trolleys
X
Y
?
Please I really need the answer of this question. Can somebody here do it for me?

The ratio X/Y is (1.172 ± 97.667)

Explanation: Absolute uncertainty is the value that when combined with a reported number, gives the range of the number.

Momentum is a quantity of motion an object has. It is calculated as a relation between mass of an object and its velocity: p = m.v

For the trolley X, momentum is:

= (2.34 ± 0.01)*(3.2 ± 0.01)

and for trolley Y, momentum is:

= (2.561 ± 0.001)*(2.5 ± 0.01)

To solve the multiplications:

For x:

= 2.34*3.2 = 7.5

relative uncertainty = = 0.0074

absolute uncertainty = 7.5*0.0074 = 0.055

= 7.5 ± 0.055

For y:

= 2.561*2.5 = 6.4

relative uncertainty = = 0.0044

absolute uncertainty = 6.4*0.0044 = 0.028

= 6.4 ± 0.028

The ratio of momentum:

= (7.5 ± 0.055) ÷ (6.4 ± 0.028)

Dividing absolute uncertainty, the rules are the same for multiplication.

= = 1.172

relative uncertainty = = 0.012

absolute uncertainty = 1.172*0.012 = 97.667

= 1.172 ± 97.667

The ratio of momentum of the 2 trolleys is 1.172 ± 97.667.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y