Question

The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls loose from the system. What is the speed of the right-hand mass when it returns to its original position?​

Answer 1

Let $a$ be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

$1.3mg - T = 1.3ma$

where $T$ is the magnitude of tension in the pulley cord, and

• for the mass on the right,

$T - mg = ma$

Eliminate $T$ to get

$(1.3mg - T) + (T - mg) = 1.3ma + ma$

$0.3mg = 2.3ma$

$\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}$

Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity $v$ such that

$v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}$

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

$mg - T' = ma'$

where $T'$ and $a'$ are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

$T' - 0.8mg = 0.8ma'$

Eliminate $T'$.

$(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'$

$0.2mg = 1.8 ma'$

$\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}$

Now, the right-hand mass has an initial upward velocity of $v$, but we're now treating down as the positive direction. As it returns to its starting position, its speed $v'$ at that point is such that

${v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}$