Two masses 5kg and 3kg are attached with massless, flexible and inextensible string and passed over a pulley (frictionless). After 4s of release of system, the string breaks. Find how much higher the 3 kg mass would go after the string breaks?

Solution

Given,
Mass of first body, $m subscript italic 1 italic space italic equals italic space italic 5 italic space k g$
Mass of second body, $m subscript 2 space equals space 3 space k g$
The acceleration of system of masses is,
$a equals fraction numerator m subscript 1 minus m subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction g space equals space fraction numerator 5 minus 3 over denominator 5 plus 3 end fraction cross times 10 space equals space 2.5 space m divided by s squared$
i.e. The mass 3 kg accelerates upward with acceleration 2.5 m/s².

Let the motion of 3 kg mass start from A and accelerate for 4s and reach the point B.
At point B the string breaks and 3kg mass rises to point C.
Motion from A to B:
Initial velocity, "<pre
Rate of acceleration, $a subscript 1 equals 2.5 space m divided by s squared$
Time of travel, $space straight t equals 4 straight s$
∴ Velocity attained during the motion is given by first equation of motion,
$space space space space space space space space space space space v subscript B equals u subscript A plus a subscript 1 t equals 0 plus 2.5 cross times 4 equals 10 m divided by s$
Motion from B to C:
Velocity, $V e l o c i t y comma space v subscript B equals 10 space m divided by s comma$
$v subscript c equals 0 comma$
Acceleration, $a subscript 2 equals negative 10 m divided by s squared$
∴ Distance travelled by the mass,
$BC equals fraction numerator straight v subscript straight C superscript 2 minus straight v subscript straight B superscript 2 over denominator 2 straight a subscript 2 end fraction$
$space space space equals fraction numerator 0 minus 100 over denominator negative 2 cross times 10 end fraction equals 5 straight m$
The mass would go 5m high.