# Two blocks with masses m1 = 3.00 kg and m2 = 5.00 kg are connected by a light string that slides over two frictionless pulleys as shown.Initially m2 is held 5.00 m off the floor while m1 is on the floor. The system is then released. ?

## a) At what speed does m2 hit the floor? b) Find the speed of m1 when it is at height of 1.80 m? c) What is the maximum height attained by m1?

Dec 5, 2015

(a)

$4.95 \text{m/s}$

(b)

$2.97 \text{m/s}$

(c)

$5 \text{m}$

#### Explanation:

$\left(a\right)$

Mass ${m}_{2}$ experiences $5 g \text{N}$ downwards and $3 g \text{N}$ upwards giving a net force of $2 g \text{N}$ downwards.

The masses are connected so we can regard them as acting as a single 8kg mass.

Since $F = m a$ we can write:

$2 g = \left(5 + 3\right) a$

$\therefore a = \frac{2 g}{8} = 2.45 {\text{m/s}}^{2}$

If you like to learn formulae the expression for 2 connected masses in a pulley system like this is:

$a = \frac{\left({m}_{2} - {m}_{1}\right) g}{\left({m}_{1} + {m}_{2}\right)}$

Now we can use the equations of motion since we know the acceleration of the system $a$.

So we can get the speed that ${m}_{2}$ hits the ground $\Rightarrow$

${v}^{2} = {u}^{2} + 2 a s$

${v}^{2} = 0 + 2 \times 2.45 \times 5$

${v}^{2} = 24.5$

$\therefore v = 4.95 \text{m/s}$

$\left(b\right)$

${v}^{2} = {u}^{2} + 2 a s$

$\therefore {v}^{2} = 0 + 2 \times 2.45 \times 1.8$

${v}^{2} = 8.82$

$\therefore v = 2.97 \text{m/s}$

(c)

Since ${m}_{2}$ cannot drop any further than 5m I reason that ${m}_{1}$ cannot go any higher than 5m.