Video Transcript
Two bodies of masses 644 grams and
156 grams were connected to the ends of a light inextensible string passing over a
smooth pulley. The system was released from rest,
and two seconds later the larger mass hit the ground. Find the maximum height the smaller
mass reached above its initial position. Take the acceleration due to
gravity π equal to 9.8 meters per second squared.
We will begin by sketching a
diagram to model the situation. We have two bodies A and B of
masses 644 grams and 156 grams. They are connected by a light
inextensible string that passes over a smooth pulley. As the pulley is smooth, we know
that the tension in the string will be constant throughout. The bodies will exert a downward
force equal to their mass multiplied by the acceleration due to gravity. We are told that this is equal to
9.8 meters per second squared, which is the same as 980 centimeters per second
squared. Body A exerts a downward force of
944 grams multiplied by 980 centimeters per second squared. This is equal to 631,120 dynes. And body B exerts a downward force
of 156 grams multiplied by 980 centimeters per second squared, which is equal to
152,880 dynes.
When the system is released from
rest, body A will accelerate downwards, and body B will accelerate upwards. As the string is inextensible, the
acceleration of the system will be constant. We are told that the larger body
hits the ground after two seconds. During this period, both bodies
will have traveled the same distance. And we are asked to calculate the
maximum height that the smaller mass reaches above its initial position. Once body A hits the ground, the
string will become slack and body B will continue to move under the action of
gravity.
We begin to solve this problem
using Newtonβs second law, which states that πΉ equals ππ. The vector sum of the forces is
equal to the mass multiplied by the acceleration. There are two forces acting on body
A. And if we let the positive
direction be vertically downwards, they have a sum of 631,120 minus π. This is equal to the mass of 644
grams multiplied by π. We will call this equation one.
Repeating this process for body B,
where we take the positive direction to be vertically upwards, we have π minus
152,880 is equal to 156π. We will call this equation two. We now have a pair of simultaneous
equations. And we can eliminate the tension
force π by adding equation one to equation two. This gives us 478,240 is equal to
800π. We can then divide through by 800,
giving us π is equal to 597.8. This is the acceleration of the
system in centimeters per second squared.
We can now use the equations of
motion or SUVAT equations to find the displacement of body B. As already mentioned, there are two
parts to this, firstly, the displacement of the whole system in the first two
seconds. We know that the initial speed π’
is zero centimeters per second. We have just calculated the
acceleration π is 597.8 centimeters per second squared. As π‘ is equal to two, we can use
the equation π is equal to π’π‘ plus a half ππ‘ squared. Since π’ equals zero, this
simplifies to π is equal to a half multiplied by 597.8 multiplied by two
squared. π is therefore equal to
1,195.6. In the first two seconds, both
bodies travel a distance of 1,195.6 centimeters.
We can calculate the velocity of
the bodies at this point using the equation π£ is equal to π’ plus ππ‘. Since π’ is zero, π£ is equal to
597.8 multiplied by two. This is also equal to 1,195.6. At the point at which body A hits
the ground, the system is moving with velocity 1,195.6 centimeters per second. This is the initial velocity at the
point at which the string becomes slack. We know that when body B reaches
its maximum height, its velocity will be equal to zero. The acceleration of the body doing
this part of the motion is negative 980 centimeters per second squared, as gravity
is acting against the direction of motion. We can use the equation π£ squared
is equal to π’ squared plus two ππ to calculate the displacement in this part of
the journey.
Substituting in our values, we have
zero is equal to 1,195.6 squared plus two multiplied by negative 980 multiplied by
π . This can be simplified to 1,960π
is equal to 1,429,459.36. Dividing through by 1,960, we have
π is equal to 729.316. Body B travels a further distance
of 729.316 centimeters after body A hits the ground. We can now calculate the maximum
height that the smaller mass reaches above its initial position by adding 1,195.6 to
729.316. This is equal to 1,924.916. The smaller mass reaches a height
of 1,924.916 centimeters above its initial position. Alternatively, we could give our
answer in meters by dividing by 100, giving us 19.24916 meters.
