As shown in Figure P8.46, a light string that does not...

As shown in Figure P8.46, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects $m_{1},$ a 3.50 -kg block originally at reston the horizontal table at a height $h=1.20 \mathrm{m}$ above the floor, to $m_{2},$ a hanging $1.90-\mathrm{kg}$ block originally a distance $d=0.900 \mathrm{m}$ above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block $m_{1}$ is projected horizontally after reaching the edge of the table. The hanging block $m_{2}$ stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which $m_{1}$ leaves the edge of the table. (b) Find the impact speed of m1 on the floor. (c) What is the shortest length of the string so that it does not go taut while $m_{1}$ is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before $m_{1}$ strikes the ground? (e) Why or why not?

As shown in Figure P8.46, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects $m_{1},$ a 3.50 -kg block originally
at reston the horizontal table at a height $h=1.20 \mathrm{m}$ above the floor, to $m_{2},$ a hanging
$1.90-\mathrm{kg}$ block originally a distance $d=0.900 \mathrm{m}$ above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block $m_{1}$ is projected horizontally after reaching the edge of the table. The hanging block $m_{2}$ stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which $m_{1}$ leaves the edge of the table. (b) Find the impact speed of m1 on the floor. (c) What is the shortest length of the string so that it does not go taut
while $m_{1}$ is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before $m_{1}$ strikes the ground? (e) Why or why not?

Key Concepts

Projectile Motion

Projectile motion describes the two-dimensional motion of an object under the influence of gravity alone once it is in free flight. This concept is crucial for analyzing the trajectory, flight time, and impact speed of objects that are projected, such as a block launched horizontally from a table edge.

Inelastic Collisions and Energy Dissipation

In inelastic collisions, kinetic energy is not conserved because some of the energy is transformed into other forms, such as heat or sound, or is absorbed during deformation. This concept is key when assessing scenarios where an object impacts the floor and comes to a stop without bouncing, indicating energy loss from the system.

Conservation of Mechanical Energy

This concept involves understanding that in the absence of non?conservative forces (like friction), the total mechanical energy (the sum of potential and kinetic energy) of a system remains constant. It is used to relate changes in the heights of objects to their speeds, such as when a block slides off a table converting gravitational potential energy into kinetic energy.

String Constraint and Tension

The behavior of a massless, non-stretchable string imposes constraints on the motion of connected objects, ensuring that their accelerations are related. Understanding how the string changes direction over a table edge and how its length affects motion is important for determining conditions like when the string goes taut or remains slack.

Transcript

00:01 So we have a system of blocks set up like this.

00:05 We have one on a table and one that's hanging off the edge.

00:10 And they're connected by a string and there's some sort of pulley here.

00:14 Don't need to be that exact.

00:16 But we're given some measurements.

00:19 This is m1, which is 3 .5 kilograms.

00:26 And this is m2, which is 1 .9 kilograms.

00:32 Luckily, everything we're given is in si units already, so we don't have to do any conversions.

00:40 The height of this block above the ground is d, which is .9 meters, and then the height of the table from the ground, h is 1 .2 meters.

01:03 So first we want to figure out the speed at which m1 leaves the edge of the table.

01:09 I guess there's no pulley here.

01:11 It doesn't really matter.

01:15 And so basically we can just figure out the final speed of m2, and then since there's no friction, m1 is just going to have that speed and continue at that speed until it leaves the edge of the table.

01:28 So i'll set up a big energy equation.

01:32 So i have the changing kinetic energy for both of them, so for m1 and m2.

01:40 And then i'll do the same thing for the gravitational potential energy.

01:45 And since i'm including that, we don't need to worry about any external work, since there's no friction or anything like that.

01:55 So changing kinetic energy is going to be one -half times m1 times v -final minus v initial squared.

02:05 Now their velocities are always going to be the same since they're connected, at least until m -2 hits the ground.

02:12 But that's all we're worrying about for this part, so i can just say that the velocities are the same, and i don't have to put any subscripts on them.

02:23 Oops, this should be squared.

02:27 Okay, and now the change in potential for the first one, it's not changing in height at all, so that's just going to be zero.

02:34 And for the second one, it's going to be minus m g times d, the distance it's falling.

02:43 That all equals zero.

02:46 Now, the initial velocity, well, they're starting from rest, so we can get rid of that.

02:54 Now i want to solve for the final velocity, right? so i have right now, i have one -half m -1 v -final squared plus one -half m2, v -final -squared minus m -g -d equals 0.

03:13 So i can factor out one half v.

03:20 Final squared.

03:21 Let me get the sum of the masses, which equals m.

03:25 Oh, this should be m2, right? and then solve for v.

03:34 Final.

03:34 So i'll divide by the sum of the masses and multiply by 2.

03:44 And then take the square root of the whole thing.

03:48 So this is going to be square root of 2 times.

03:55 M2 is 1 .9 is that right yes 1 .9 times 9 .8 times 0 .9 and then divided by 3 .5 plus 1 .9 and we get 2 .49 meters per second so that's for that's the final velocity of m2 but it also happens to be the final velocity of m1, and it's going to be the same when it leaves the edge of the table...

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