
Mechanics problem
A 2kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.15m, to a hanging book with mass 3kg. The system is released from rest, and the books are observed to move 1.2m in 0.8s.
My solution so far:
using constant acceleration formula
$\displaystyle s = ut+\frac{1}{2}at^2 $
$\displaystyle 1.2 = 0 +\frac{1}{2}(a)(0.8)^2$
$\displaystyle a = 3.75$
For the hanging part:
$\displaystyle 3g  T_1 = ma$
$\displaystyle T_1 = 3gma$
$\displaystyle T_1 = 3g3(3.75) = 18.2$(rounded) This agrees with my textbook
however for the horizontal tension in the cord
I make it
$\displaystyle T_1  T_2 = ma$
$\displaystyle T_2 = T_1  ma$
$\displaystyle T_2 = 18.2  2(3.75)$
$\displaystyle T_2 = 10.7N $
however the answer in my textbook is 7.5 which would suggest
$\displaystyle T_2=2(3.75)$
where has the $\displaystyle T_1$ force gone?
Have I missed out a force
Also how can you have different tensions in the cord surely it would snap?

If you draw the free body diagram for the 2kg book. $\displaystyle T_1 = (2kg)a$
The equation $\displaystyle T_1  T_2 = ma$ is incorrect.
If your pulley has no mass or is frictionless, then $\displaystyle T_1 = T_2$.
However, this is not the case, so there is some inertia in the pulley.
$\displaystyle (T_2  T_1)(0.15m/2)$ is the torque on the pulley.

why wouldnt the cord snap if it has different tensions in it?

Tension is just a force, it can be balanced by other forces such as friction.
Think about dragging a really long heavy chain on the ground.
The front link of the chain has the most tension and friction.
The back link of the chain may have only a little tension and a little friction.
However, you are right, the net acceleration at each link must be the same (if it has fixed length).